Answer:
a) 5.143 m/s^2
b) T = 15.43 N
c) Fr = 10.29 N
d) 5.143 m/s^2 , T = 15.43 N
Explanation:
Given:-
- The mass of left most block, m1 = 1.0 kg
- The mass of center block, m2 = 2.0 kg
- The mass of right most block, m3 = 4.0 kg
- A force that acts on the right most block, F = 36 N
Solution:-
a)
- For the first part we will consider the three blocks with masses ( m1 , m2 , and m3 ) as one system on which a force of F = 36 N is acted upon. The masses m1 and m3 are connected with a string with tension ( T ) and the m1 and m2 are in contact.
- We apply the Newton's second law of motion to the system with acceleration ( a ) and the combined mass ( M ) of the three blocks as follows:
[tex]F = M*a\\\\36 = ( 1 + 2 + 4 )*a\\\\a = \frac{36}{7}\\\\a = 5.143 \frac{m}{s^2}[/tex]
Answer: The system moves in the direction of external force ( F ) i.e to the right with an acceleration of 5.143 m/s^2
b)
- The blocks with mass ( m1 and m3 ) are connected with a string with tension ( T ) with a combined acceleration of ( a ).
- We will isolate the massive block ( m3 ) and notice that two opposing forces ( F and T ) act on the block.
- We will again apply the Newton's 2nd law of motion for the block m3 as follows:
[tex]F_n_e_t = m_3 * a\\\\F - T = m_3 * a\\\\36 - T = 4*5.143\\\\T = 36 - 20.5714\\\\T = 15.43 N[/tex]
Answer:- A tension of T = 15.43 Newtons acts on both blocks ( m1 and m3 )
c)
- We will now isolate the left most block ( m1 ) and draw a free body diagram. This block experiences two forces that is due to tension ( T ) and a reaction force ( Fr ) exerted by block ( m2 ) onto ( m3 ).
- Again we will apply the the Newton's 2nd law of motion for the block m3 as follows:
[tex]F_n_e_t = m_1*a\\\\T - F_r = m_1*a\\\\15.43 - F_r = 1*5.143\\\\F_r = 15.43 - 5.143\\\\F_r = 10.29 N[/tex]
- The reaction force ( Fr ) is contact between masses ( m1 and m2 ) exists as a pair of equal magnitude and opposite direction acting on both the masses. ( Newton's Third Law of motion )
Answer: The block m2 experiences a contact force of ( Fr = 10.29 N ) to the right.
d)
- If we were to stack the block ( m2 ) on-top of block ( m1 ) such that block ( m2 ) does not slip we the initial system would remain the same and move with the same acceleration calculated in part a) i.e 5.143 m/s^2
- We will check to see if the tension ( T ) differs or not as the two block ( m1 and m2 ) both experience the same Tension force ( T ) as a sub-system. with a combined mass of ( m1 + m2 ).
- We apply the Newton's 2nd law of motion for the block m3 as follows:
[tex]T = ( m_1 + m_2 ) *a\\\\T = ( 1 + 2 ) * 5.143\\\\T = 15.43 N[/tex]
Answer: The acceleration of the whole system remains the same at a = 5.143 m/s^2 and the tension T = 15.43 N also remains the same.
A surveyor measures the distance across a river that flows straight north by the following method. Starting directly across from a tree on the opposite bank, the surveyor walks distance, D = 130 m along the river to establish a baseline. She then sights across to the tree and reads that the angle from the baseline to the tree is an angle θ = 25°. How wide is the river?
Answer:
The width of the river is [tex]z = 60.62 \ m[/tex]
Explanation:
From the question we are told that
The distance of the base line is D = 130 m
The angle is [tex]\theta = 25^o[/tex]
A diagram illustration the question is shown on the first uploaded image
Applying Trigonometric Rules for Right-angled Triangle,
[tex]tan 25 = \frac{z}{130}[/tex]
Now making z the subject
[tex]z = 130 * tan (25)[/tex]
[tex]z = 60.62 \ m[/tex]
In an RC-circuit, a resistance of R=1.0 "Giga Ohms" is connected to an air-filled circular-parallel-plate capacitor of diameter 12.0 mm with a separation distance of 1.0 mm. What is the time constant of the system?
Answer:
[tex]\tau = 1\ ms[/tex]
Explanation:
First we need to find the capacitance of the capacitor.
The capacitance is given by:
[tex]C = \epsilon_0 * area / distance[/tex]
Where [tex]\epsilon_0[/tex] is the air permittivity, which is approximately 8.85 * 10^(-12)
The radius is 12/2 = 6 mm = 0.006 m, so the area of the capacitor is:
[tex]Area = \pi * radius^{2}\\Area = \pi * 0.006^2\\Area = 113.1 * 10^{-6}\ m^2[/tex]
So the capacitance is:
[tex]C = \frac{8.85 * 10^{-12} * 113.1 * 10^{-6}}{0.001}[/tex]
[tex]C = 10^{-12}\ F = 1\ pF[/tex]
The time constant of a rc-circuit is given by:
[tex]\tau = RC[/tex]
So we have that:
[tex]\tau = 10^{9} * 10^{-12} = 10^{-3}\ s = 1\ ms[/tex]
A uniform disk with a 25 cm radius swings without friction about a nail through the rim. If it is released from rest from a position with the center level with the nail, then what is its angular velocity as it swings through the point where the center is below the na
Answer:
Explanation:
During the swing , the center of mass will go down due to which disc will lose potential energy which will be converted into rotational kinetic energy
mgh = 1/2 I ω² where m is mass of the disc , h is height by which c.m goes down which will be equal to radius of disc , I is moment of inertia of disc about the nail at rim , ω is angular velocity .
mgr = 1/2 x ( 1/2 m r²+ mr²) x ω²
gr = 1/2 x 1/2 r² x ω² + 1/2r² x ω²
g = 1 / 4 x ω² r + 1 / 2 x ω² r
g = 3 x ω² r/ 4
ω² = 4g /3 r
= 4 x 9.8 / 3 x .25
= 52.26
ω = 7.23 rad / s .
. A ball weighs 120g on the earth surface,
i) What is its mass on the surface of the moon? 1mk
Answer:
WEIGHT ON MOON IS 0.2004N
Explanation:
mass of the body=120g=[tex]\frac{120}{1000}[/tex]kg=0.12kg (we will convert g into kg)
gravity on moon=1.67m/s²( to find the mass of anybody on another we should know its gravity)
as we know that (from the formula of weight)
weight=mass×gravity
w=mg
w=0.12kg²×1.67m/s²
w=0.2004N
The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A If the suitcase has a mass of 80.0 kg , how far can it be pushed across the level floor with 660 J of work
Answer:
Explanation:
The work required to push will be equal to work done by friction . Let d be the displacement required .
force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction
work done by force of friction
mg x μ x d = 660
80 x 9.8 x .272 x d = 660
d = 3 .1 m .
Astrophysicist Neil deGrasse Tyson steps into an elevator on the 29th floor of a skyscraper. For some odd reason, there is a scale on the floor of the elevator. Neil, whose mass is about 115 kg, decides to step on the scale and presses the button for a lower floor. The elevator starts traveling downwards with a constant acceleration of 1.5 m/s2 for 6.0 seconds, and then travels at a constant velocity for 6.0 seconds. Finally, the elevator has an upward acceleration of 1.5 m/s2 for 6.0 seconds as it comes to a stop.
A. If each floor is approximately 4 m tall, which floor does the elevator stop at?
B. If the mass of the elevator is 1,200 kg, what is the maximum tension of the elevator cable?
Answer:
A. Final Floor = 15.5 = 15 (Considering downward portion of elevator)
B. T = 14859.5 N = 14.89 KN
Explanation:
A.
First we calculate distance covered by the elevator during downward motion. Downward motion consists of two parts. First one is uniformly accelerated. For that part we use 2nd equation of motion:
s₁ = Vi t + (0.5)at²
where,
s₁ = distance covered during accelerated downward motion = ?
Vi = initial speed = 0 m/s (since elevator is initially at rest)
t = time taken = 6 s
a = acceleration = 1.5 m/s²
Therefore,
s₁ = (0 m/s)(6 s) + (0.5)(1.5 m/s²)(6 s)²
s₁ = 4.5 m
also we find the final velocity using 1st equation of motion:
Vf = Vi + at
Vf = 0 m/s + (1.5 m/s²)(6 s)
Vf = 9 m/s
Now, the second part of downward motion is with constant velocity. So:
s₂ = vt
where,
s₂ = distance covered during constant speed downward motion = ?
v = Vf = 9 m/s
t = 6 s
Therefore,
s₂ = (9 m/s)(6 s)
s₂ = 54 m
Now for distance covered during upward motion is given by the 2nd equation of motion. Since the values of acceleration and time are same. Therefore, it will be equal in magnitude to s₁:
s₃ = s₁ = 4.5 m
Therefore, the total distance covered by elevator is given by following equation:
s = s₁ + s₂ - s₃ (Downward motion taken positive)
s = 4.5 m + 54 m - 4.5 m
s = 54 m
Therefore, net motion of the elevator was 54 m downwards.
So the final floor will be:
Final Floor = Initial Floor - Distance Covered/Length of a floor
Final Floor = 29 - 54 m/4m
Final Floor = 15.5 = 15 (Considering the downward portion or floor of elevator)
B.
The maximum tension will occur during the upward accelerated motion. It is given by the formula:
T = m(g + a)
where,
T = Max. Tension in Cable = ?
m = total mass of person and elevator = 115 kg + 1200 kg = 1315 kg
g = 9.8 m/s²
a = acceleration = 1.5 m/s²
Therefore,
T = (1315 kg)(9.8 m/s² + 1.5 m/s²)
T = 14859.5 N = 14.89 KN
John heats 1 kg of soup from 25 °C to 70 °C for 15 minutes by a heater. How long does the same heater take to heat 1.5 kg of the same kind of soup from 20 °C to 80 °C? The energy output per unit time by the heater is constant.
Answer:
30 minutes
Explanation:
Energy per time is constant, so:
E₁ / t₁ = E₂ / t₂
m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂
(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t
(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t
3/min = 90 / t
t = 30 min
A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
E = -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E = -4556.18 N/m
The value of Ey such that the particle will cross the x axis at x=1.5 cm is -4556.18 N/m.
What is electric field?The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.
Given is a particle leaves the origin with a speed of 3.6 x 10⁶ m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis.
The distance x = 1.5 cm = 1.5×10⁻² m (assumed, not given in question)
The horizontal distance traveled by particle is
x = ucosθt
t = x/ucosθ
The force in an electric field is F = qE...................(1)
where, q is charge , E is the strength of electric field
From, newton 2nd law of motion, Force F = ma.................(2)
Equating both the equations, we get
ma = qE
a = qE/m..................(3)
The vertical distance, y =usinθt - 1/2at²
From equation 3, we have
y = usinθt - 1/2 (qE/m) t²
if y = 0, t = 2musinθ/(qE) = x / (ucosθ)
The electric field is represented as
Also, E = 2mu²×sinθ×cosθ/(xq)
Plug the values, we get
E = 2×(9.1×10⁻³¹)×(3.6 x 10⁶)²×sin34°×cos34°/( 1.5×10⁻² ×(-1.6)×10⁻¹⁹)
E = -4556.18 N/m
Thus, the electric field of the particle is -4556.18 N/m.
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The velocity of an object is given by the expression v (t) = 3.00 m / s + (2.00 m / s ^ 3) t ^ 2. Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.00 s.
Answer: [tex]x=\frac{2}{3}t^3+3t+1[/tex]
Explanation:
Given
velocity of object is given by
[tex]v(t)=3+2t^2[/tex]
and we know change of position w.r.t time is velocity
[tex]\Rightarrow \dfrac{dx}{dt}=v[/tex]
[tex]\Rightarrow \dfrac{dx}{dt}=3+2t^2[/tex]
[tex]\Rightarrow dx=(3+2t^2)dt[/tex]
Integrating both sides we get
[tex]\Rightarrow \int_{1}^{x}dx=\int_{0}^{t}(3+2t^2)dt[/tex]
[tex]\Rightarrow x\mid _{1}^{x}=(3t+\frac{2}{3}t^3)\mid _{0}^{t}[/tex]
[tex]\Rightarrow x-1=3(t-0)+\frac{2}{3}(t^3-0)[/tex]
[tex]\Rightarrow x=\frac{2}{3}t^3+3t+1[/tex]
Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (mb). Note how the air pressure changes as you move StationB towards the center of the high-pressure system.
A. What do you notice?
B. Why do you think this is called a high-pressure system?
Answer:
a) When moving towards a high pressure center the pressure values increase in the equipment
b) This area is called high prison since the weight of the atmosphere on top is maximum
Explanation:
A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.
When moving towards a high pressure center the pressure values increase in the equipment
B) This area is called high prison since the weight of the atmosphere on top is maximum
in general they are areas of good weather
What is the resistance of a circuit with a voltage of 10 V in a current of 5 A use almond law to create the resistance
Answer:
2Ω
Explanation:
Ohm's law:
V = IR
10 V = (5 A) R
R = 2 Ω
1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to
estimate the length of time a fully charged battery could deliver any particular current
before discharge. Approximately how much energy is stored by a 50 ampere-minute 12
volt battery?
Answer:
Energy is stored by a 50 ampere-minute 12
volt battery is approximately = 36,000 J = 36 kJ
Explanation:
Power in electrical circuits is given as
Power = IV
But power generally is defined as energy expended per unit time
Power = (Energy/time)
Energy = Power × Time
Energy = IV × Time
Energy = (I.t × V)
I.t = 50 Ampere-minute = 50 × 60 = 3000 Ampere-seconds
V = 12 V
Energy = 3,000 × 12 = 36,000 J = 36 kJ
Hope this Helps!!!
The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as measured from the planet's surface). What type of relationship exists between the altitude and the atmospheric density, and what would the atmospheric density be at an altitude of 1,291 kilometers?
A.
inverse plot, 0.45 kilograms/meter3
B.
inverse plot, 0.51 kilograms/meter3
C.
quadratic plot, 1.05 kilograms/meter3
D.
inverse plot, 1.23 kilograms/meter3
E.
inverse plot, 0.95 kilograms/meter3
' A ' looks like the best choice.
Answer:
B. inverse plot, 0.51 kilograms/meter3
Explanation:
A population _____ follows a period of
Answer:
a population increase
Explanation:
During the 20th century, the world population increased from 1.65 billion to 6 billion. In 1970, the world's population was half that of today. In less than 15 years, 47% of the population will live in areas already under heavy water stress. In Africa, between 75 and 250 million people will face growing shortages in 2020 due to climate change. The scarcity of some arid and semi-arid regions will have a decisive impact on migration.
what is the speed of light in quartz
Answer:
1.95 x 10^8 m/s.
Explanation:
Answer:
the answer is 1.95 x 10^8 m/s
Explanation:
Which factor caused higher oil prices to directly lead to inflation?
It increased demand for cars, leading to higher automobile prices.
Companies passed on production and transportation costs to consumers.
The government began to print more money.
Gas prices declined too quickly, leading to oversupply
Answer: B, Companies passed on production and transportation costs to consumers
Explanation:
A higher oil price occurred when companies passed on production and transportation costs to consumers.
Cause of high price of oilThe oil producing companies spend so much money in producing crude oil from the reservoirs to the surface. They also spend money in processing and transporting the crude oil to the end users or consumers.
The final price of the oil depends on the total amount spent by these companies in producing the hydrocarbons.
Thus, a higher oil price occurred when companies passed on production and transportation costs to consumers.
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A certain metal with work function of Φ = 1.7 eV is illuminated in vacuum by 1.4 x 10-6 W of light with a wavelength of λ = 600 nm. 1)How many photons per second, N, are incident on the metal? N = photons per second Submit 2)What is KEmax, the maximum kinetic energy of the electron that is emitted from the metal? KEmax = eV
Answer:
1) n = 4.47*10^12 photons
2) K = 0.25 eV
Explanation:
This is a problem about the photoelectric effect.
1) In order to calculate the number of photons that impact the metal, you take into account the power of the light, which is given by:
[tex]P=\frac{E}{t}=1.4*10^{-6}\frac{J}{s}[/tex] (1)
Furthermore you calculate the energy of a photon with a wavelength of 600nm, by using the following formula:
[tex]E_p=h\frac{c}{\lambda}[/tex] (2)
c: speed of light = 3*10^8 m/s
h: Planck's constant = 6.626*10^-34 Js
λ: wavelength = 600*10^-9 m
You replace the values of the parameters in the equation (2):
[tex]E_p=(6.626*10^{-34}Js)\frac{3*10^8m/s}{600*10^{-9}m}=3.131*10^{-19}J[/tex]
Next, you calculate the quotient between the power of the light (equation (1)) and the energy of the photon:
[tex]n=\frac{P}{E_p}=\frac{1.4*10^{-6}J/s}{3.131*10^{-19}J}=4.47*10^{12}photons[/tex]
The number of photons is 4.47*10^12
2) The kinetic energy of the electrons emitted by the metal is given by the following formula:
[tex]K=E_p-\Phi[/tex] (3)
Ep: energy of the photons
Φ: work function of the metal = 1.7 eV
You first convert the energy of the photons to eV:
[tex]E_p=3.131*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.954eV[/tex]
You replace in the equation (3):
[tex]K=1.95eV-1.7eV=0.25eV[/tex]
The kinetic energy of the electrons emitted by the metal is 0.25 eV
(1). The Number of photons per second is,[tex]4.23*10^{12}[/tex]
(2). The maximum kinetic energy of the electron is 0.37eV.
(1). The power of light is given as,
[tex]P=1.4*10^{-6}W[/tex]
Energy is given as,
[tex]E=\frac{hc}{\lambda} =\frac{6.626*10^{-34}*3*10^{8} }{600*10^{-9} } \\\\E=3.313*10^{-19} Joule\\\\E=\frac{3.313*10^{-19}}{1.6*10^{-19} }=2.07eV[/tex]
Number of photons per second is,
[tex]N=\frac{P}{E}=\frac{1.4*10^{-6} }{3.313*10^{-19} } =4.23*10^{12}[/tex]
(2). the maximum kinetic energy of the electron is,
[tex]K.E=E-\phi[/tex]
Where [tex]\phi[/tex] is work function.
[tex]K.E=2.07-1.7=0.37eV[/tex]
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A person is swimming in a river with a current that has speed vR with respect to the shore. The swimmer first swims downstream (i.e. in the direction of the current) at a constant speed, vS, with respect to the water. The swimmer travels a distance D in a time tOut. The swimmer then changes direction to swim upstream (i.e. against the direction of the current) at a constant speed, vS, with respect to the water and returns to her original starting point (located a distance D from her turn-around point) in a time tIn. What is tOut in terms of vR, vS, and D, as needed?
Answer:
The time taken is [tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]
Explanation:
From the question we are told that
The speed of the current is [tex]v__{R}}[/tex]
The speed of the swimmer in direction of current is [tex]v__{S}}[/tex]
The distance traveled by the swimmer is [tex]D[/tex]
The time taken to travel this distance is [tex]t_{out}[/tex]
The speed of the swimmer against direction of current is [tex]v__{s}}[/tex]
The resultant speed for downstream current is
[tex]V_{r} = v__{S}} +v__{R}}[/tex]
The time taken can be mathematically represented as
[tex]t_{out} = \frac{D}{V_{r}}[/tex]
[tex]t_{out} = \frac{D}{v__{R}} + v__{S}}}[/tex]
In order to determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an incline. In one experiment, the eraser begins to slip down the incline when the angle of inclination is 35.6° and then moves down the incline with constant speed when the angle is reduced to 30.8°. From these data, determine the coefficients of static and kinetic friction for this experiment.
Answer:
The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.
Explanation:
The Free Body Diagram associated with the experiment is presented as attachment included below.
Friction is a contact force that occurs as a reaction against any change in state of motion, which is fostered by gravity.
Normal force is another contact force that appears as a reaction to the component of weight perpendicular to the direction of motion. Let consider a framework of reference consisting in two orthogonal axes, one being parallel to the direction of motion (x-axis) and the other one normal to it (y-axis). Equations of motion are described herein:
[tex]\Sigma F_{x} = W \cdot \sin \theta - f = 0[/tex]
[tex]\Sigma F_{y} = N - W \cdot \cos \theta = 0[/tex]
Where:
[tex]W[/tex] - Weight of the eraser, measured in newtons.
[tex]f[/tex] - Friction force, measured in newtons.
[tex]N[/tex] - Normal force, measured in newtons.
[tex]\theta[/tex] - Angle of the incline, measured in degrees.
The maximum allowable static friction force is:
[tex]f = \mu_{s} \cdot N[/tex]
Where:
[tex]\mu_{s}[/tex] - Coefficient of static friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
Likewise, the kinetic friction force is described by the following model:
[tex]f = \mu_{k} \cdot N[/tex]
Where:
[tex]\mu_{k}[/tex] - Coefficient of static friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
And weight is equal to the product of the mass of eraser and gravitational constant ([tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
In this exercise, coefficients of static and kinetic friction must be determined. First equation of equilibrium has to be expanded and coefficient of friction cleared:
[tex]m\cdot g \cdot \sin \theta - \mu\cdot N = 0[/tex]
[tex]\mu = \frac{m\cdot g \cdot \sin \theta}{N}[/tex]
But [tex]N = m\cdot g \cos \theta[/tex], so that:
[tex]\mu = \tan \theta[/tex]
Now, coefficients of static and kinetic friction are, respectively:
[tex]\mu_{s} = \tan 35.6^{\circ}[/tex]
[tex]\mu_{s} \approx 0.716[/tex]
[tex]\mu_{k} \approx \tan 30.8^{\circ}[/tex]
[tex]\mu_{k} \approx 0.596[/tex]
The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.
A sound level of 96 dB is how many times as intense as one of 90 dB?
Answer:
A sound level of 96 dB is 4 times as intense as one of 90 dB
Explanation:
The formula of the intensity level of sound in decibels is given as follows:
Intensity Level = 10 log₁₀(I/I₀)
where,
I = Intensity of Sound
I₀ = Reference Intensity Level = 10⁻¹² W/m²
Therefore, for 96 dB sound level:
96 = 10 log₁₀(I₁/10⁻¹²)
log₁₀(I₁/10⁻¹²) = 96/10
I₁/10⁻¹² = 10^9.6
I₁ = (10⁻¹²)(4 x 10⁹)
I₁ = 0.004 W/m²
For 90 dB sound level:
90 = 10 log₁₀(I₂/10⁻¹²)
log₁₀(I₂/10⁻¹²) = 90/10
I₂/10⁻¹² = 10^9
I₂ = (10⁻¹²)(10⁹)
I₂ = 0.001 W/m²
Therefore,
I₁/I₂ = 0.004/0.001
I₁ = 4 I₂
Hence, the sound level of 96 dB is 4 times as intense as one of 90 dB.
A student drives 105.0 mi with an average speed of 61.0 mi/h for exactly 1 hour and 30
minutes for the first part of the trip. What is the distance in miles traveled during this
time?
Answer:
91.5 miles
Explanation:
61 miles per hour so 61(x amount of hours)
so 61 x 1.5 hours is 91.5 miles
A flat coil of wire is used with an LC-tuned circuit as a receiving antenna. The coil has a radius of 0.30 m and consists of 420 turns. The transmitted radio wave has a frequency of 1.3 MHz. The magnetic field of the wave is parallel to the normal of the coil and has a maximum value of 1.7 x 10-13 T. Using Faraday's Law of electromagnetic induction and the fact that the magnetic field changes from zero to its maximum value in one-quarter of a wave period, find the magnitude of the average emf induced in the antenna in this time.
Answer:
The average emf induce is [tex]V = 2.625 * 10^{-5} \ V[/tex]
Explanation:
From the question we are told that
The radius of the coil is [tex]r = 0.30 \ m[/tex]
The number of turns is [tex]N = 420 \ turns[/tex]
The frequency of the transition radio wave is [tex]f = 1.3\ MHz = 1.3 *10^{6} Hz[/tex]
The magnetic field is [tex]B_,{max} = 1.7 * 10^{-13} \ T[/tex]
The time taken for the magnetic field to go from zero to maximum is [tex]\Delta T = \frac{T}{4}[/tex]
The period of the transmitted radio wave is [tex]T = \frac{1}{f}[/tex]
So
[tex]\Delta T = \frac{T}{4} = \frac{1}{4 f}[/tex]
The potential difference can be mathematically represented as
[tex]V = NA (\frac{\Delta B}{\Delta T} )[/tex]
[tex]V = NA ([B_{max} - B_{min} ] * 4f)[/tex]
Where [tex]B_{min} = 0T[/tex]
substituting values
[tex]V = 420 * (\pi *(0.30)^2) * (1.7 *10^{-13} * 4 * 1.3 *10^{6})[/tex]
[tex]V = 2.625 * 10^{-5} \ V[/tex]
student conducted an experiment and find the density of an ICEBERGE. A students than recorded the following readings. Mass 425 25 g Volume 405 15 mL What experimental value should be quoted for the density of the ICEBERG? Compare your answer with the density of water, which is 3 1.00 10 kg . Show any calculations necessary to justify your answer
Complete Question
The complete question is shown on the first uploaded image
Answer:
The experimental value of density is [tex]\rho = 1.05*10^{3} \ kg/m^3 \pm 101 \ kg/m^3[/tex]
Comparing it with the value of density of water ([tex]1.0*10^{3} \ kg/m^3[/tex]) we can see that the density of ice is greater
Explanation:
From the question we are told
The mass is [tex]M = (425 \pm 25) \ g =(0.425 \pm 0.025) \ kg[/tex]
The volume is [tex]V = (405 \pm 15 ) \ mL = (0.000405 \pm 1.5*10^{-5}) \ m^3[/tex]
The experimental value of density is mathematically evaluated as
[tex]\rho = \frac{M}{V}[/tex]
[tex]\rho = \frac{0.425}{0.000405}[/tex]
[tex]\rho = 1.05 *10^{3} \ kg/m^3[/tex]
The possible error in this experimental value of density is mathematically evaluated as
[tex]\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} +\frac{\Delta V}{V}[/tex]
substituting value
[tex]\frac{\Delta \rho}{1.05*10^{3}} = \frac{0.025}{0.425} +\frac{1.5*10^{-5}}{0.000405}[/tex]
[tex]\Delta \rho = 101 \ kgm^{-3}[/tex]
Thus the experimental value of density is
[tex]\rho = 1.05*10^{3} \ kg/m^3 \pm 101 \ kg/m^3[/tex]
A conducting sphere contains positive charge distributed uniformly over its surface. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity
a. The potential at the center of the sphere is zero.
b.The potential is lowest, but not zero, at the center of the sphere.
c. The potential at the center of the sphere is the same as the potential at the surface.
d. The potential at the center is the same as the potential at infinity.
e. The potential at the surface is higher than the potential at the center.
Answer:
a. FALSE
b. FALSE
c. TRUTH
d. FALSE
e. FALSE
Explanation:
To determine which statements are truth or false you focus in the following formula, for the electric potential generated by a conducting sphere:
[tex]V=\frac{Q}{4\pi \epsilon_o R}[/tex] inside the sphere
[tex]V'=\frac{Q}{4\pi \epsilon_o r}[/tex] for r > R (outside the sphere)
R: radius of the sphere
ε0: dielectric permittivity of vacuum
Q: charge of the sphere
As you can notice, inside the sphere the potential is constant. Inside the sphere, the potential is the same. Outside the surface the potential decreases as 1/r, being r the distance to the center of the sphere.
Hence, you can conclude:
a. The potential at the center of the sphere is zero. FALSE
b.The potential is lowest, but not zero, at the center of the sphere. FALSE
c. The potential at the center of the sphere is the same as the potential at the surface. TRUTH
d. The potential at the center is the same as the potential at infinity. FALSE
e. The potential at the surface is higher than the potential at the center. FALSE
A scuba diver and her gear displace a volume of 68.5 L and have a total mass of 71.8 kg . Part A What is the buoyant force on the diver in sea water? FB = nothing N Request Answer Part B Will the diver sink or float?
Answer:
A) Fb = 671.3 N
B) The diver will sink.
Explanation:
A)
The buoyant force applied on an object by a fluid is given by the following formula:
Fb = Vρg
where,
Fb = Buoyant Force = ?
V = Volume of the water displaced by the object = 68.5 L = 0.0685 m³
ρ = Density of Water = 1000 kg/m³
g = 9.8 m/s²
Therefore,
Fb = (0.0685 m³)(1000 kg/m³)(9.8 m/s²)
Fb = 671.3 N
B)
Now, in order to find out whether the diver sinks or float, we need to find weight of the diver with gear.
W = mg = (71.8 kg)(9.8 m/s²)
W = 703.64 N
Since, W > Fb. Therefore, the downward force of weight will make the diver sink.
The diver will sink.
Block 1, of mass m1 = 2.50 kg , moves along a frictionless air track with speed v1 = 27.0 m/s. It collides with block 2, of mass m2 = 33.0 kg , which was initially at rest. The blocks stick together after the collision.A. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.B. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.C. what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.
Answer:
a
The total initial momentum of the two-block system is [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
b
The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]
c
the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is
[tex]\Delta KE =- 847.08 \ J[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]m_1 = 2.50 \ kg[/tex]
The initial velocity of first block is [tex]u_1 = 27.0 \ m/s[/tex]
The mass of second block is [tex]m_2 = 33.0\ kg[/tex]
initial velocity of second block is [tex]u_2 = 0 \ m/s[/tex]
The magnitude of the of the total initial momentum of the two-block system is mathematically repented as
[tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]
substituting values
[tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]
[tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
According to the law of linear momentum conservation
[tex]p_i = p_f[/tex]
Where [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as
[tex]p_f = (m_+m_2) * v_f[/tex]
Where [tex]v_f[/tex] is the final velocity of the system
[tex]p_i = (m_1 +m_2 ) v_f[/tex]
substituting values
[tex]67.5 = (2.50+33 ) v_f[/tex]
[tex]v_f = 1.9014 \ m/s[/tex]
The change in kinetic energy is mathematically represented as
[tex]\Delta KE = KE_f -KE_i[/tex]
Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system which is mathematically represented as
[tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]
substituting values
[tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]
[tex]KE_f =64.17 J[/tex]
While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system
[tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]
substituting values
[tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]
[tex]KE_i = 911.25 \ J[/tex]
So
[tex]\Delta KE = 64.17 -911.25[/tex]
[tex]\Delta KE =- 847.08 \ J[/tex]
6. When a positive charge is released and moves along an electric field line, it moves to a position of A) lower potential and lower potential energy. B) lower potential and higher potential energy. C) higher potential and lower potential energy. D) higher potential and higher potential energy.
Answer:
Since you would have to do work on the charge to bring it back to its original position, the charge moves to a position of lower potential and lower potential energy.
The positive charge is released from a point such that it will move along an uniform electric field to the position of lower potential and lower potential energy. Therefore, option (A) is correct,
When a positive charge (say +Q) is released from a point (say A) and moves in an uniform electric field to reach the point (say B), then some work is done on the charge. This work done is given as,
[tex]W=+Q(V_{A}-V_{B})[/tex]
Here, [tex]V_{A}[/tex] and [tex]V_{B}[/tex] are the potential differences between the points A and B respectively..
This means the charge is moving from higher potential to lower potential. And since it is moving along the uniform electric field, therefore the electric potential energy of charged system is decreased.
Thus, we conclude that on releasing the positive charge from a point, it starts moving along the electric field towards the direction of lower electric potential and lower electric potential energy. Hence, option (A) is correct.
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An organism has 20 chromosomes after fertilization.after meiosis, how many chromosomes would each sex cell have?
Answer:
EACH SEX CELL WILL HAVE 10 CHROMOSOMES BECAUSE n+n=2n
means haploid parent cells join or fuse to form diploid zygote
Answer:
10
Explanation:
A student is given a small object that is hanging from a ring stand on a nylon thread. The student attempts to charge the object electrically in several ways. Based upon his results, he concludes the object is made of an insulating material. Which set of results must he have collected?
A. The object could be charged only by contact.
B. The object could be charged by either contact or induction.
C. The object could be charged by either contact or polarization.
D. The object could be charged only by polarization.
Answer:(a)
Explanation:
Student must have known that insulators can only be charged when they are rubbed against each other. In this process, one becomes electrically negative while other becomes electrically positive such that both have the same magnitude. The one which gains electrons becomes electrically negative due to the transfer of electrons while others lose the electron becomes positive due to the transfer of an electron to another body.
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B suspended by a cord of length l =2.2m. The block then swings through a maximum angle of theta = 60. Determine (a) the initial speed of the bullet vo, (b) the impulse imparted by the bullet on the block, (c) the force on the cord immediately after the impact
Answer:
(a) v-bullet = 399.04 m/s
(b) I = 2.38 kg m/s
(c) T = 2.59 N
Explanation:
(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.
The potential energy is given by:
[tex]U=(M+m)gh[/tex] (1)
U: potential energy
M: mass of the wood block = 1 kg
m: mass of the bullet = 6g = 6.0*10^-3 kg
g: gravitational constant = 9.8m/s^2
h: distance to the ground
The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:
[tex]h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m[/tex]
The potential energy is:
[tex]U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J[/tex]
Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:
[tex]U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}[/tex]
Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:
[tex]Mv_1+mv_2=(M+m)v[/tex] (2)
v1: initial velocity of the wood block = 0m/s
v2: initial speed of the bullet
v: speed of bullet and block = 2.38m/s
You solve the equation (2) for v2:
[tex]M(0)+mv_2=(M+m)v[/tex]
[tex]v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}[/tex]
The speed of the bullet before the impact with the wood block is 399.04 m/s
(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:
[tex]I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}[/tex]
The impulse is 2.38 kgm/s
(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:
[tex]T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N[/tex]
The force on the cord after the impact is 2.59N
Answer:
The initial speed of the bullet [tex]V_o = 777.97m/s[/tex]The force on the cord immediately after the impact = [tex]19.71N[/tex]Explanation:
Apply the law of conversion of energy
[tex]V_f = \sqrt{2gh}[/tex]
where,
h = height of which the bullet and block rise after impact
[tex]h = L - Lcos\theta\\\\h = 2.2 - (2.2*cos60)\\\\h = 1.1m[/tex]
Therefore,
[tex]V_f = \sqrt{2gh}\\\\V_f = \sqrt{2*9.8*1.1}\\\\V_f = 4.64m/s[/tex]
From conservation of momentum principle, [tex]m_Bv_B = 0[/tex]
[tex]m_ov_o + m_Bv_B = (m_b+m_B)V_f\\\\0.006V_o = (0.006+1)*4.64\\\\V_o = 777.97m/s[/tex]
C) The force in the cable is due to the centrfugal force of the system, which is due to the motion of the system is a curved path and weight of the system
[tex]F = \frac{m_b+m_B}{L}V_f^2 + (m_b+m_B)g\\\\F = \frac{0.006+1}{2.2}*4.64^2 + (0.006+1)9.81\\\\F = 19.71N[/tex]
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