Answer:
Approximately [tex]0.608[/tex] (assuming that [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex].)
Explanation:
The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.
Let [tex]m[/tex] represent the mass of this car.Let [tex]r[/tex] represent the radius of the circular track.This answer will approach this question in two steps:
Step one: determine the centripetal force when the car is about to skid.Step two: calculate the coefficient of static friction.For simplicity, let [tex]a_{T}[/tex] represent the tangential acceleration ([tex]1.90\; \rm m \cdot s^{-2}[/tex]) of this car.
Centripetal Force when the car is about to skidThe question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to [tex]90^\circ[/tex] or [tex]\displaystyle \frac{\pi}{2}[/tex] radians.
The angular acceleration of this car can be found as [tex]\displaystyle \alpha = \frac{a_{T}}{r}[/tex]. ([tex]a_T[/tex] is the tangential acceleration of the car, and [tex]r[/tex] is the radius of this circular track.)
Consider the SUVAT equation that relates initial and final (tangential) velocity ([tex]u[/tex] and [tex]v[/tex]) to (tangential) acceleration [tex]a_{T}[/tex] and displacement [tex]x[/tex]:
[tex]v^2 - u^2 = 2\, a_{T}\cdot x[/tex].
The idea is to solve for the final angular velocity using the angular analogy of that equation:
[tex]\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta[/tex].
In this equation, [tex]\theta[/tex] represents angular displacement. For this motion in particular:
[tex]\omega(\text{initial}) = 0[/tex] since the car was initially not moving.[tex]\theta = \displaystyle \frac{\pi}{2}[/tex] since the car travelled one-quarter of the circle.Solve this equation for [tex]\omega(\text{final})[/tex] in terms of [tex]a_T[/tex] and [tex]r[/tex]:
[tex]\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}[/tex].
Let [tex]m[/tex] represent the mass of this car. The centripetal force at this moment would be:
[tex]\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}[/tex].
Coefficient of static friction between the car and the trackSince the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, [tex]m\, g[/tex].
Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.
Let [tex]\mu_s[/tex] denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:
[tex]F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g[/tex].
The size of this force should be equal to that of the centripetal force when the car is about to skid:
[tex]\mu_s\, m\, g = \pi\, m\, a_{T}[/tex].
Solve this equation for [tex]\mu_s[/tex]:
[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g}[/tex].
Indeed, the expression for [tex]\mu_s[/tex] does not include any unknown letter. Let [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex]. Evaluate this expression for [tex]a_T = 1.90\;\rm m \cdot s^{-2}[/tex]:
[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608[/tex].
(Three significant figures.)
What is the relationship between electric force and distance between charged objects and the amount of charge?
Explanation:
The relationship between electric force and distance between charged objects is given by the formula as follows :
[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]
k is electrostatic constant and d is distance between charges
The electric force between charges is inversely proportional to the square of distance between them.
A note on a piano vibrates 262 times per second . What is the period of the wave ?
A wire of length L is made up of two sections of two different materials connected in series. The first section of length L1 = 17.7 m is made of steel and the second section of length L2 = 28.5 m is made of iron. Both wires have the same radius of 5.30 ✕ 10−4 m. If the compound wire is subjected to a tension of 148 N, determine the time taken for a transverse pulse to move from one end of the wire to the other. The density of steel is 7.75 ✕ 103 kg/m3 and the density of iron is 7.86 ✕ 103 kg/m3.
Answer:
Explanation:
velocity of wave in a tense wire is given by the expression
[tex]v= \sqrt{\frac{T}{m} }[/tex]
v is velocity . T is tension and m is mass per unit length .
for steel wire
m = π r² ρ where r is radius and ρ is density
= 3.14 x (5.3 x 10⁻⁴)²x7.75 x 10³
= 683.57 x 10⁻⁵ kg/m
v = [tex]\sqrt{\frac{148}{683.57\times 10^{-5}} }[/tex]
= 1.47 x 10² m /s
= 147 m /s
for iron wire
m = π r² ρ where r is radius and ρ is density
= 3.14 x (5.3 x 10⁻⁴)²x7.86 x 10³
= 693.27 x 10⁻⁵ kg/m
[tex]v = \sqrt{\frac{148}{693.27\times 10^{-5}} }[/tex]
= 146 m /s
Time taken to move from one end to another
= 17.7 / 147 + 28.5 / 146
= .12 + .195
= .315 s .
A sample of silver (with work function Φ=4.52 eV ) is exposed to an ultraviolet light source (????=200 nm), which results in the ejection of photoelectrons. What changes will be observed if:
1. The silver is replaced with copper (Φ= 5.10 eV)?
a. more energetic photoelectrons (on average)
b. no photoelectrons are emitted more photoelectrons ejected
c. less energetic photoelectrons (on average)
d. fewer photoelectrons ejected
2. A second (identical) light source also shines on the metal?
a. fewer photoelectrons ejected
b. no photoelectrons are emitted more
c. energetic photoelectrons (on average)
d. less energetic photoelectrons (on average)
e. more photoelectrons ejected
3. The ultraviolet source is replaced with an X-ray source that emits the same number of photons per unit time as the original ultraviolet source?
a. no photoelectrons are emitted
b. less energetic photoelectrons (on average)
c. fewer photoelectrons ejected
d. more energetic photoelectrons (on average)
e. more photoelectrons ejected
Answer:
1. c
2. e
3. d
Explanation:
1.
From Einstein's Photoelectric Equation, we know that:
Energy given up by photon = Work Function + K.E of Electron
hc/λ = φ + K.E
where,
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light source = 200 nm = 2 x 10⁻⁷ m
φ = (5.1 eV)(1.6 x 10⁻¹⁹ J/eV) = 8.16 x 10⁻¹⁹ J
Therefore,
(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2 x 10⁻⁷ m) - 8.16 x 10⁻¹⁹ = K.E
K.E = (9.939 - 8.16) x 10⁻¹⁹ J
K.E = 1.778 x 10⁻¹⁹ J
The positive answer shows that electrons will be emitted. Since it is clear from the equation the the K.E of electron decreases with the increase in work function. Therefore:
c. less energetic photo-electrons (on average)
2.
The increase in light sources means an increase in the intensity of light. The no. of photons are increased, due to increase of intensity. Thus, more photons hit the metal and they eject greater no. of electrons. Therefore,
e. more photo-electrons ejected
3.
X-rays have smaller wavelength and greater energy than ultraviolet rays. Thus, the photons with greater energy will strike the metal and as a result, electrons with higher energy will be ejected.
d. more energetic photo-electrons (on average)
A rod of mass M = 2.95 kg and length L can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 45 g, moving with speed v = 4.23 m/s, strikes the rod at angle θ = 46° from the normal at a distance D = 2/3 L, where L = 0.95 m, from the point of rotation and sticks to the rod after the collision.
Required:
What is the angular speed ωf of the system immediately after the collision, in terms of system parameters and I?
Answer:
Explanation:
angular momentum of the putty about the point of rotation
= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .
= .045 x 4.23 x 2/3 x .95 cos46
= .0837 units
moment of inertia of rod = ml² / 3 , m is mass of rod and l is length
= 2.95 x .95² / 3
I₁ = .8874 units
moment of inertia of rod + putty
I₁ + mr²
m is mass of putty and r is distance where it sticks
I₂ = .8874 + .045 x (2 x .95 / 3)²
I₂ = .905
Applying conservation of angular momentum
angular momentum of putty = final angular momentum of rod+ putty
.0837 = .905 ω
ω is final angular velocity of rod + putty
ω = .092 rad /s .
A merry-go-round is shaped like a uniform disk and has moment of inertia of 50,000 kg m 2 . It is rotating so that it has an angular momentum of 10,000 (kg m 2 radians/s) and its outer edge has a speed of 2 m/s. What is its radius, in m
Answer:
r = 20 m
Explanation:
The formula for the angular momentum of a rotating body is given as:
L = mvr
where,
L = Angular Momentum = 10000 kgm²/s
m = mass
v = speed = 2 m/s
r = radius of merry-go-round
Therefore,
10000 kg.m²/s = mr(2 m/s)
m r = (10000 kg.m²/s)/(2 m/s)
m r = 5000 kg.m ------------- equation 1
Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:
I = (1/2) m r²
where,
I = moment of inertia = 50000 kg.m²
Therefore,
50000 kg.m² = (1/2)(m r)(r)
using equation 1, we get:
50000 kg.m² = (1/2)(5000 kg.m)(r)
(50000 kg.m²)/(2500 kg.m) = r
r = 20 m
4) (7 pts.) A water molecule is centered at the origin of a coordinate system with its dipole moment vector aligned with the x axis. The magnitude of a water molecule dipole is 6.16 × 10−30 C·m. What is the magnitude of the electric field at x = 3.00 × 10−9 m?
Answer:
[tex]E=3.69*10^{-11}\frac{V}{m}[/tex]
Explanation:
To solve this problem you use the following formula, for the calculation of the electric field along the axis of the dipole.
[tex]E=\frac{p}{2\pi \epsilon_ox^3}[/tex] (1)
p: dipole moment = 6.16*10^-30 Cm
x: distance to the center of mass of the dipole = 3.00*10^-9m
eo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
You replace the values of the variables in the equation (1):
[tex]E=\frac{6.16*10^{-30}Cm}{2\pi(8.85*10^{-12}C^2/Nm^2)(3.00*10^{-9}m)^3}\\\\E=3.69*10^{-11}\frac{V}{m}[/tex]
describe Piaget's four stages of cognitive development. Include the major hallmarks of each stage.
Answer:
Explanation:
Sensorimotor Infants "think" by acting on the world with their eyes, ears, hands, and mouth.
Preoperational. Development of language and make-believe play takes place.
Concrete Operational children think in a logical, organized fashion only when dealing with concrete information they can perceive directly.
Formal Operational. Adolescences can also evaluate the logic of verbal statements without referring to real-world circumstances.
Sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.
What is cognitive development?The way youngsters think, investigate, and figure things out is referred to as cognitive development.
Piaget defined four stages of cognitive development:
1. Sensorimotor. From birth through the age of 18-24 months.
2. Preoperational.From infancy (18-24 months) until toddlerhood (age 7)
3. Operational concrete. 7 to 11 years old
4. Formal operational. From adolescence to adulthood
Hence, sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.
To learn more about the cognitive development refer to:
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While her kid brother is on a wooden horse at the edge of a merry-go-round, Sheila rides her bicycle parallel to its edge. The wooden horses have a tangential speed of 6 m/s. Sheila rides at 4 m/s. The radius of the merry-go-round is 8 m. At what time intervals does Sheila encounter her brother, if she rides opposite to the direction of rotation of the merry-go-round?
a. 5.03 s
b. 8.37 s
c. 12.6 s
d. 25.1 s
e. 50.2 s
Answer:
t = 5.03 s
Explanation:
To find the time interval when Sheila encounter her brother, you first calculate the angular speed of both Sheila and her brother.
You use the following formula:
[tex]\omega = \frac{v}{r}[/tex]
w: angular speed
v: tangential speed
r: radius of the trajectory = 8 m
For you have:
[tex]\omega=\frac{4m/s}{8m}=0.5\frac{rad}{s}[/tex]
For her brother:
[tex]\omega'=\frac{6m/s}{8m}=0.75\frac{rad}{s}[/tex]
Next, they will encounter to each other when the angular distance of the Brother of sheila equals the angular distance of Sheila in the opposite direction. This can be written as follow:
[tex]\theta=\omega t\\\\\theta'=\omega ' t[/tex]
They encounter for θ = 2π-θ':
[tex]\omega t=2\pi-\omega' t[/tex]
You replace the values of the parameters in the previous equation and solve for t:
[tex]0.5t=2\pi-0.75t\\\\1.25t=2\pi\\\\t=5.026\approx5.03[/tex]
Hence, Sheila encounter her brother in 5.03 s
Assume the three blocks (m. = 1.0 kg, m = 20 kg and m = 40 ko) portrayed in the figure below move on a frictionless surface and a force F: 36w acts as shown on the 4.0 kg block.
a) Determine the acceleration given this system (in m/s2 to the right). m/s2 (to the right)
b) Determine the tension in the cord connecting the 4.0 kg and the 1.0 kg blocks in N). Determine the force exerted by the 1.0 kg block on the 2.0 kg block (in N). N (a) What If How would your answers to parts (a) and (b) of this problem change if the 2.0 kg block was now stacked on top of the 1.0 kg block? Assume that the 2.0 kg block sticks to and does not slide on the 1.0 kg block when the system is accelerated.
(Enter the acceleration in m/s2 to the right and the tension in N.) acceleration m/s (to the right) tension
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 550 kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s
Answer:
[tex]52.25\times10^4W\\699.1 hp[/tex]
Explanation:
According to the energy conversation:
ΔK=[tex]-f_kd+W[/tex]
ΔK=[tex]K_f-K_i ; K=1/2 mv^2[/tex]
where,
[tex]k_i, k_f[/tex] are initial and final kinetic energy of the system.
[tex]v_i[/tex]= initial velocity of the system
[tex]v_f[/tex]=final velocity of the system
W= total work done on the system
[tex]f_k[/tex]= friction force
d= distance traveled
Given: [tex]v_f[/tex]=110m/s
d=400m
[tex]f_k[/tex]=1200N
[tex]v_i[/tex]=0m/s
t=7.3s
ΔK=[tex]-f_kd+W[/tex]
W= ΔK + [tex]f_kd[/tex]
=[tex]K_f-K_i+f_kd\\[/tex]
[tex]=1/2 mv_f^2-1/2 mv_i^2+f_kd\\=\frac{1}{2} \times 550\times110^2 - \frac{1}{2} \times 550\times0^2+ (1200\times400)\\=3807500[/tex]
[tex]P=\frac{W}{t} =\frac{3807500}{7.3} \\P=52.15 \times10^4w\\P=\frac{52.15 \times10^4}{746} \\P=699.1 hp[/tex]
A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. (a) Determine the time necessary for the projectile to reach its maximum height. (b) Determine the maximum height reached by the projectile. (c) Determine the horizontal and vertical components of the velocity vector at the maximum height. (d) Determine the horizontal and vertical components of the acceleration vector at the maximum heigh
Answer:
(a) t = 3.74 s
(b) H = 136.86 m
(c) Vₓ = 41.83 m/s, Vy = 0 m/s
(d) ax = 0 m/s², ay = 9.8 m/s²
Explanation:
(a)
Time to reach maximum height by the projectile is given as:
t = V₀ Sinθ/g
where,
V₀ = Launching Speed = 55.6 m/s
Angle with Horizontal = θ = 41.2°
g = 9.8 m/s²
Therefore,
t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)
t = 3.74 s
(b)
Maximum height reached by projectile is:
H = V₀² Sin²θ/g
H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)
H = 136.86 m
(c)
Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:
Vₓ = V₀ₓ = V₀ Cos θ
Vₓ = (55.6 m/s)(Cos 41.2°)
Vₓ = 41.83 m/s
Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.
Vy = 0 m/s
(d)
Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.
ax = 0 m/s²
The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:
ay = 9.8 m/s²
The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm1cm and the IMA of the machine is 66 , what is the radius of the handle
Question: The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm and the IMA of the machine is 6, what is the radius of the handle?
Answer:
Radius of the handle = 3 cm = 0.03 m
Explanation:
Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod).......(1)
Diameter of the rod = 1 cm
Radius of the rod = Diameter/2
Radius of the rod = 1/2
Radius of the rod = 0.5 cm
Mechanical Accuracy of the machine, MA = 6
Substitute the values into equation (1)
6 = (Radius of the handle)/0.5
Radius of the handle = 6 * 0.5
Radius of the handle = 3 cm
The Radius of the handle is = 3 cm = 0.03 m
Calculation of the radius of the handle:Since
Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod)
Here,
Diameter of the rod = 1 cm
We know that
The radius of the rod = Diameter/2
So,
Radius of the rod = 1/2
So,
Radius of the rod = 0.5 cm
Now
Mechanical Accuracy of the machine, MA = 6
Now
6 = (Radius of the handle)/0.5
Radius of the handle = 6 * 0.5
Radius of the handle = 3 cm
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If two twins (54 kg each) were 0.02 m apart, what is the force of gravity between them?
Answer:
Force, [tex]F=4.86\times 10^{-4}\ N[/tex]
Explanation:
We have,
Masses of two twins are 54 kg each
They are placed at a distance of 0.02 m
It is required to find the force of gravity between them. The formula used to find the gravitational force between masses is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
plugging all the known values:
[tex]F=6.67\times 10^{-11}\times \dfrac{54^2}{(0.02)^2}\\\\F=4.86\times 10^{-4}\ N[/tex]
So, the force of gravity between them is [tex]4.86\times 10^{-4}\ N[/tex].
The average, year-after-year conditions of temperature, precipitation, winds, and cloud in an area are known as its
A.climate.
b.weather.
C. global warming
d. seasons
Answer:
a. global warming
Explanation:
that's the definitain of global warming
Answer:
A climate
Explanation:
A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 3.30 m/s and rebounds with a speed of 1.60 m/s, determine the following. (a) magnitude of the change in the ball's momentum in kg · m/s (Let up be in the positive direction.)
Answer:
[tex]\Delta p=1.3475\ kg-m/s[/tex]
Explanation:
The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-
We represent
The ball mass = m = 275 g = 0.275 kg
Thus it goes to the floor and resurfaces upward.
The ball hits the ground at 3.30 m/s speed that is
u = -3.30 m/s which represents the Negative since the ball hits the ground)
It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)
[tex]\Delta p=p_f-p_i[/tex]
[tex]\Delta p=m(v-u)[/tex]
[tex]\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))[/tex]
[tex]\Delta p=1.3475\ kg-m/s[/tex]
ii.
The drift velocity
(b) A 1800w toaster, a 1.3KW electric frying pan, and a 100w lamp are plugged to the same
20A, 120V circuit.
i.
What current is drawn by each device and what is the resistance of each device?
State whether this combination will blow the fuse or not.
Answer:
toaster -- 15 A, 8 Ωfry pan -- 10.83 A, 11.08 Ωlamp -- 0.83 A, 144 Ωfuse will blowExplanation:
P = VI
I = P/V = P/120
R = V/I = V/(P/V) = V^2/P = 14400/P
Toaster: I = 1800/120 = 15 . . . amps
R = 14400/1800 = 8 . . . ohms
Fry pan: I = 1300/120 = 10.833 . . . amps
R = 14400/1300 = 11.08 . . . ohms
Lamp: I = 100/120 = 0.833 . . . amps
R = 14400/100 = 144 . . . ohms
The total current exceeds 20 A, so will blow the fuse.
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 Stored = 36000 J = 36 KJ
Explanation:
The power of a battery is given by the formula:
P = IV
where,
P = Power delivered by the battery
I = Current Supplied to the battery
V = Potential Difference between terminals of battery = 12 volt
Now, we multiply both sides by the time period (t):
Pt = VIt
where,
Pt = (Power)(Time) = Energy Stored = E = ?
It = Battery Current Rating = 50 A.min
Converting this to A.sec;
It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
E = 36000 J = 36 KJ
A hydraulic lift is made by sealing an ideal fluid inside a container with an input piston of cross-sectional area 0.004 m2 , and an output piston of cross-sectional area 1.2 m2 . The pistons can slide up or down without friction while keeping the fluid sealed inside. What is the maximum weight that can be lifted when a force of 60 N is applied to the input piston
Answer:
Maximum weight that can be lifted = 18,000 N
Explanation:
Given:
Cross-sectional area of input (A1) = 0.004 m²
Cross-sectional area of the output (A2) = 1.2 m ²
Force (F) = 60 N
Computation:
Pressure on input piston (P1) = F / A1
Assume,
Maximum weight lifted by piston = W
Pressure on output piston (P2) = W / A2
We, know that
P1 = P2
[F / A1] = [W / A2]
[60 / 0.004] = [W / 1.2]
150,00 = W / 1.2
Weight = 18,000 N
Maximum weight that can be lifted = 18,000 N
Work out the velocity v at the end of a rollercoaster ride (0). (rearrange the equation for KE to make velocity v the subject)
KE=1/2mv^2
Explanation:
If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :
[tex]K=\dfrac{1}{2}mv^2[/tex]
m is mass of the object
We can rearrange the above equation such that,
[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]
Hence, this is the velocity at the end of a rollercoaster ride.
A book of 500 leaves has a mass of 1kg if its thickness is 5cm what are the mass and thickness of each leaf
Answer:
0.002kg and 0.01cm
Explanation:
500 leaves has a thickness is 5cm
Means I leaf has a thickness of 5/500= 0.01cm
Similarly the mass of one leaf would be 1/500 =0.002kg
A layer of ethyl alcohol (n = 1.361) is on top of water (n = 1.333). To the nearest degree, at what angle relative to the normal to the interface of the two liquids is light totally reflected?
a. 78 degree
b. 88 degree
c. 68 degree
d. 49 degree
e. the critical angle isundefined
Answer:
a. 78 degree
Explanation:
According to Snell's Law, we have:
(ni)(Sin θi) = (nr)(Sin θr)
where,
ni = Refractive index of medium on which light is incident
ni = Refractive index of ethyl alcohol = 1.361
nr = Refractive index of medium from which light is refracted
nr = Refractive index of ethyl alcohol = 1.333
θi = Angle of Incidence
θr = Angle of refraction
So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:
θi = θc
when, θr = 90°
Therefore, Snell's Law becomes:
(1.361)(Sin θc) = (1.333)(Sin 90°)
Sin θc = 1.333/1.361
θc = Sin⁻¹ (0.9794)
θc = 78.35° = 78° (Approximately)
Therefore, correct answer will be:
a. 78 degree
The angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
From the information given;
the refractive index of the ethyl alcohol [tex]\mathbf{n_1= 1.361}[/tex]the refractive index of the water [tex]\mathbf{n_2 = 1.333}[/tex] the angle of incidence is the critical angle [tex]\theta_i = \theta_c[/tex] the angle of refraction [tex]\theta _r = 90^0[/tex]According to Snell's Law of refraction;
[tex]\mathbf{n_1 sin \theta _c = n_2 sin \theta_r}[/tex]
[tex]\mathbf{1.361 \times sin \theta _c = 1.333 \times sin 90}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times sin 90}{1.361}}[/tex]
[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times 1}{1.361}}[/tex]
[tex]\mathbf{ \theta _c = sin^{-1} (0.9794)}[/tex]
[tex]\mathbf{ \theta _c =78.35^0}[/tex]
[tex]\mathbf{ \theta _c \simeq78^0}[/tex]
Therefore, we can conclude that the angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.
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Leah is moving in a spaceship at a constant velocity away from a group of stars. Which one of the following statements indicates a method by which she can determine her absolute velocity through space?
A) She can measure her increases in mass.
B) She can measure the contraction of her ship.
C) She can measure the vibration frequency of a quartz crystal.
D) She can measure the changes in total energy of her ship.
E) She can perform no measurement to determine this quantity.
Answer:
E) She can perform no measurement to determine this quantity.
Explanation:
A spacecraft is a machine used to fly in outer space.
According to Isaac Newton's third law of motion, every action produces an equal and opposite reaction. When fuel is shoot out of one end of the rocket, the rocket moves forward for which no air is required.
As Leah is moving in a spaceship at a constant velocity away from a group of stars, she cannot measure to determine this quantity.
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal range of the ball from the base of the platform is 20.0m. What is the velocity of the ball just before it touches the ground
Answer:
v = 46.99 m/s
Explanation:
The velocity of the ball just before it touches the ground, is given by the following formula:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex] (1)
vx: horizontal component of the velocity
vy: vertical component of the velocity
The vertical component vy is calculated by using the following formula:
[tex]v_y^2=v_{oy}^2+2gh[/tex] (2)
vy: final velocity
voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)
g: gravitational acceleration = 9.8 m/s^2
h: height = 1.60m
You replace the values of the parameters in the equation (2):
[tex]v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}[/tex]
vx is calculated by using the information about the horizontal range of the ball:
[tex]R=v_o\sqrt{\frac{2h}{g}}[/tex] (3)
R: horizontal range of the ball = 20.0 m
You solve the previous equation for vo, the initial horizontal velocity:
[tex]v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}[/tex]
The horizontal component of the velocity is constant in the complete trajectory, hence, you have that
vx = vo = 35 m/s
Finally, you replace the values of vx and vy in the equation (1):
[tex]v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}[/tex]
The velocity of the ball just before it touches the ground is 46.99 m/s
1. Which of the following is NOT a vector quantity? (a) Displacement. (b) Energy. (c) Force. (d) Momentum. (e) Velocity.
Answer:
B. energy
Explanation:
A vector has direction.
Energy does not have a direction.
19. After a snowstorm, you put on your frictionless skis and tie a rope to the back of your friend’s truck. Your total mass is 70 kg and the truck exerts a constant force of 20 N. How fast will you be going after 15 seconds, in m/s and MPH?
Explanation:
It is given that,
Total mass is 70 kg
The truck exerts a constant force of 20 N.
Then the net force is given by :
F = ma
a is acceleration of rider
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2[/tex]
Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :
[tex]v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s[/tex]
Since, 1 m/s = 2.23 mph
4.28 m/s = 9.57 mph
So, the speed of the rider is 4.28 m/s or 9.57 mph.
Superman is jogging alongside the railroad tracks on the outskirts of Metropolis at 100 km/h. He overtakes the caboose of a 500-m-long freight train traveling at 50 km/h. At that moment he begins to accelerate at 10 m/s2. How far will the train have traveled before Superman passes the locomotive?
Answer:
d = 41.91 m
Explanation:
In order to calculate the distance traveled by the train while superman passes it, you write the equations of motion for both superman and train:
For train, you have a motion with constant speed. You write the equation of motion of the position of the front of the train:
[tex]x=x_o+v_1t[/tex] (1)
xo: initial position of the front of the train = 500m
v1: speed of the train = 50km/h
For superman, you take into account that the motion is an accelerated motion (you assume superman is at the origin of coordinates):
[tex]x'=v_2t+\frac{1}{2}at^2[/tex] (1)
v2: initial speed of superman = 100km/h
a: acceleration = 10m/s^2
When superman passes the train, both positions x and x' will be equal. Hence, you equal the equations (1) and (2) and you calculate the time t. But before you convert the units of the velocities v1 and v2 to m/s:
[tex]v_1=50\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=13.88\frac{m}{s}\\\\v_2=100\frac{km}{h}=\frac{1000m}{1km}*\frac{1h}{3600s}=27.77\frac{m}{s}[/tex]
Thus, you equal x=x'
[tex]x=x'\\\\x_o+v_1t=v_2t+\frac{1}{2}at^2\\\\500m+(13.88m/s)t=(27.77m/s)t+\frac{1}{2}(10m/s^2)t^2\\\\(50\frac{m}{s^2})t^2+(13.89\frac{m}{s})t-500m=0[/tex]
You solve the last equation for t by using the quadratic formula:
[tex]t_{1,2}=\frac{-13.89\pm \sqrt{(13.89)^2-4(50)(-500)}}{2(50)}\\\\t_{1,2}=\frac{-13.89\pm 316.53}{100}\\\\t_1=3.02s\\\\t_2=-3.30s[/tex]
You only use t1 = 3.02s because negative times do not have physical meaning.
Next, you replace this value of t in the equation (1) to calculate the position of the train (for when superman just passed it):
[tex]x=500m+(13.88m/s)(3.02s)=541.91m[/tex]
x is the position of the front of the train, then, the dstance traveled by the train is:
d = 541.91m - 500m = 41.91 m
A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 592 N. As the elevator later stops, the scale reading is 400 N. Assume the magnitude of the acceleration is the same during starting and stopping. (a) Determine the weight of the person. N (b) Determine the person's mass. kg (c) Determine the magnitude of acceleration of the elevator. m/s2
Answer:
a) 496Nb) 50.56kgc) 1.90m/s²Explanation:
According to newton's secomd law, ∑F = ma
∑F is the summation of the force acting on the body
m is the mass of the body
a is the acceleration
Given the normal force when the elevator starts N1 = 592N
Normal force after the elevator stopped N2 = 400N
When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)
When moving up;
N1 - Fg = ma
N1 = ma + Fg ...(1)
Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;
N2 - Fg = -ma
N2 = -ma+Fg ...(2)
Adding equation 1 and 2 we will have;
N1+N2 = 2Fg
592N + 400N = 2Fg
992N 2Fg
Fg = 992/2
Fg = 496N
The weight of the person is 496N
\b) To get the person mass, we will use the relationship Fg = mg
g = 9.81m/s
496 = 9.81m
mass m = 496/9.81
mass = 50.56kg
c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;
N1-N2 = 2ma
592-400 = 2(50.56)a
192 = 101.12a
a = 192/101.12
a = 1.90m/s²
Convert from scientific notation to standard form
9.512 x 10-8
Answer:
0.00000009512
Explanation:
Scientific notation is a very useful and abbreviated way of writing quantities that are very large or small. It consists of placing the number with an integer and multiplying by an exponent to arrive at the same number.
let's pass the number 9,512 10⁻⁸ to decimal notation
9,512 / 10⁸ = 9,512 / 100000000
0.00000009512
As we see writing this number, it is very easy to make mistakes
If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _____________. 4% Between 4% and 48% 48% More than 48%
Answer:
More than 48%
Explanation:
If the interest is computed monthly on the outstanding balance, it has an effective annual rate of ...
(1 +4%)^12 -1 = 60.1% . . . . more than 48%
The effective annual or yearly interest rate would be=30.56% which is Between 4% and 48%
Calculation of Annual Interest rateThe formula used to calculate annual Interest rate =
[tex](1+ \frac{i}{n} ) {}^{n} - 1[/tex]
where i= nominal interest rate = 4%
n= number of periods= 12 months
Annual Interest rate=
[tex](1 + \frac{4\%}{12} ) {}^{12} - 1[/tex]
= (1+0.333)^12 -1
= (1.333)^12-1
= 31.56 - 1
= 30.56%
Therefore, the effective annual or yearly interest rate would be= 30.56%
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