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
HELPP ?Air at a temperature of 27 C and 1 atm pressure in a 4 liter cylinder of a diesel engine There. By pushing the piston, the volume of air shrinks 16 times and the pressure increases 40 times. a) How many moles of air are in the cylinder. b) What is the final temperature of the air?
Answer:
a. 0.16240664737515434 moles
b. 67.5 degrees Celcius
Explanation:
a. Use Ideal Gas Equation
PV=nRT
Where P = pressure in pascals, V=Volume in cubic meters, n=number of moles, R is a constant=8.314 J/mol.K and T is temperature in Kelvin.
27C = 273+27=300Kelvin
volume 4L = 0.004m^3
Pressure = 1atm = 101325 Pascal
PV=nRT
101325Pa*0.004m^3=n*8.314J/mol.K*300K
Solving for n from the above you get n=0.16240664737515434 moles
b.Use combined gas law equation
P1*V1/T1=P2*V2/T2
P1= 1atm
V1=4L
T1=27C
P2= 4/16 L =0.25L
P=1*40 atm = 40atm
We do not know T2
USING THE FORMULA
(1atm*4L)/27C = (40atm*0.25L)/T2
(1*4)/27=(40*0.25)/T2
IF you simplify for T2, you get 67.5
Hence final temperature = 67.5 degrees Celcius
In a 2 dimensional Cartesian system, the x-component of a vector is known, and the angle between vector and x-axis is known. Which operation is used to calculate the magnitude of the vector? (taken with respect to the x-component)
a. dividing by cosine
b. dividing by sine
c. multiplying by cosine
d. multiplying by sine
Answer:
The correct answer is a
Explanation:
The cosine function is
cos θ = ca / H
done ca is the adjacent leg (x-axis) and H is the hypotenuse (vector module)
we clear
H = ca / cos θ
therefore, to find the magnitude of the vector, the cathete is divided into the cosine.
The correct answer is a
To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant angular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.7m. If the thrower takes 1.2s to complete one revolution, starting from rest, what will be the speed of the discus at release?
Answer:
4.437 m/s
Explanation:
Diameter of rotation d is 1.7 m
Radius of rotation = d/2 = 1.7/2 = 0.85 m
If he takes 1.2 sec to complete one revolution, then his angular speed is 1/1.2 = 0.83 rev/s
We convert to rad/s
Angular speed = 2 x pi x 0.83
= 2 x 3.142 x 0.83 = 5.22 rad/s
Speed is equal to the angular speed times the radius of rotation
Speed = 5.22 x 0.85 = 4.437 m/s
In the given case, the speed of the discus at release, If the thrower takes 1.2s to complete one revolution, starting from rest would be - 8.90 m/s.
Given:
diameter of the circle = 1.7 mradius f the circle would be = 1.7/2 = 0.85 m
time taken for one revolution t = 1.2 sThis rotation exercise can be treated using the rotation kinematics.
Angular acceleration:
θ = w₀ t + ½ α t²
t = 1.2 s to give a revolution (T = 2π rad) and with part of the rest the initial angular velocity is zero (wo = 0)
=> θ = 0 + ½ α t²
=> α = 2θ / t²
=> α= 2 × 2π / 1.2²
=> α = 4π = 8.7266 rad / s²
Let's calculate the angular velocity:
=> w = wo + α t
=> w = 0 + α t
=> w = 8.7266 × 1.2
=> w = 10.47192 rad / s
The relationship between linear and angular velocity is
=> r = d / 2
=> r = 1.7 / 2 = 0.85 m
=> v = w r
=> v = 10.47192 × 0.85
=> v = 8.90 m / s
Thus, the correct speed would be - 8.90 m/s
Learn more:
https://brainly.com/question/14663644
cellus
An object ends up at a position of
327 m after a displacement of -144 m.
What was its initial position?
(Unit = m)
Answer:
Its initial position was 471 m.
Explanation:
We have,
Final position of the object is 327 m
Displacement of the object is -144 m
It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,
[tex]d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m[/tex]
So, its initial position was 471 m.
When the distance between a point source of light and a light meter is reduced from 6.0m to 2.0 m, the intensity of illumination at the meter will be the original value multiplied by _____.
Answer:
Explanation:
Let the point source have power P .
At distance r , the intensity I
I = P / 4πr² . If intensity at 6 m and 2 m be I₁ and I₂
I₁ = P / 4π x 6²
I₂ = P / 4π x 2²
I₁ / I₂ = 2² / 6²
= 1 / 9
I₂ = 9 I₁
Intensity will be 9 times that at 6 m .
A camera takes a picture that is the correct brightness and the correct zoom level, but the depth-of-focus is too small. One way to increase the depth-of-focus is to increase the f-number. Assuming that we will make changes that have the overall effect to:
1. increase the f-number, and
2. keep the brightness and the zoom level the same, which changes should we make to the aperture diameter and to the shutter time? (keep in mind we're talking about the time the shutter is open; we aren't talking about the shutter speed)
a. Increase the aperture diameter, decrease the shutter time
b. Decrease the aperture diameter, increase the shutter time
c. Increase both the aperture diameter as well as the shutter time
d. Decrease both the aperture diameter as well as the shutter time
Two identical charges,2.0m apart,exert forces of magnitude 4.0 N on each other.What is the value of either charge?
Answer:
[tex]\large \boxed{42\, \mu \text{C}}$[/tex]
Explanation:
The formula for the force exerted between two charges is
[tex]F=k \dfrac{ q_1q_2}{r^2}[/tex]
where k is the Coulomb constant.
The charges are identical, so we can write the formula as
[tex]F=k\dfrac{q^{2}}{r^2}[/tex]
[tex]\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}[/tex]
A worker with spikes on his shoes pulls on rope that is attached to a box that is resting on a flat, frictionless frozen lake. The box has mass m, and the worker pulls with a constant tension T at an angle θ = 40 ∘ above the horizontal. There is a strong headwind on the lake, which produces a horizontal force Fw that is pointed in the opposite direction than the box is being pulled. Draw a free-body diagram for this system. Assume that the worker pulls the box to the right. If the wind force has a magnitude of 30 N, with what tension must the worker pull in order to move the box at a constant velocity?
Answer:
a
The free body diagram is shown on the first uploaded image
b
The tension on the rope is [tex]T=39.16 \ N[/tex]
Explanation:
From the question we are told that
The mass of the box is m
The tension on the box is T
The angle at which it is pulled is [tex]\theta = 40^o[/tex]
The force produced by the strong head wind is [tex]Fw = 30 \ N[/tex]
At equilibrium the net force acting on the block along the horizontal axis is zero i.e
[tex]Tcos \theta -F_w = 0[/tex]
substituting values
[tex]Tcos (40) -30 = 0[/tex]
[tex]Tcos (40) = 30[/tex]
[tex]T(0.76604)) = 30[/tex]
[tex]T=39.16 \ N[/tex]
b. A locomotive of a train exerts a constant force of 280KN on a train while pulling
it at 50 km/h along a level track. What is:
[4 marks)
i. Workdone in quarter an hour and
[4 marks]
Answer:
Work-done in quarter an hour = 3.5 × 10⁶ J
Explanation:
Given:
Force (F) = 280 KN = 280,000 N
Velocity (V) = 50 km / h
Time (t) = 1 / 4 = 0.25 hour
Find:
Work-done in quarter an hour
Computation:
⇒ Displacement = Velocity (V) × Time
⇒ Displacement = 50 × 0.25
⇒ Displacement = 12.5 km
⇒ Work-done = Force (F) × Displacement
⇒ Work-done in quarter an hour = 280,000 × 12.5
⇒ Work-done in quarter an hour = 3,500,000
Work-done in quarter an hour = 3.5 × 10⁶ J
b) A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected
in series in order that the same current shall be supplied from 240 V, 50 Hz mains.
Ignore the resistance of the inductor and calculate:
i. the inductance of the inductor;
ii. the impedance of the circuit;
iii. the phase difference between the current and the applied voltage.
Assume the waveform to be sinusoidal.
Answer:
i. 43.5 mH ii. 16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°
Explanation:
i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω
The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.
So, x = 2π(50)L = 100πL Ω = 314.16L Ω
Since the current is the same when the 240 V supply is applied, then
the impedance Z = √(R² + X²) = 240 V/15 A
√(R² + X²) = 16 Ω
8.33² + X² = 16²
69.3889 + X² = 256
X² = 256 - 69.3889
X² = 186.6111
X = √186.6111
X = 13.66 Ω
Since X = 314.16L = 13.66 Ω
L = 13.66/314.16
= 0.0435 H
= 43.5 mH
ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.
So in phasor form Z = (8.33 + j13.66) Ω
iii. The phase difference θ between the current and voltage is
θ = tan⁻¹X/R
= tan⁻¹(314.16L/R)
= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)
= tan⁻¹(13.66/8.33)
= tan⁻¹(1.6406)
= 58.64°
A student in the front of a school bus tosses a ball to another student in the back of the bus while the bus is moving forward at constant velocity. The speed of the ball as seen by a stationary observer in the street:_________
a. is less than that observed inside the bus.
b. is the same as that observed inside the bus
c. may be either greater or smaller than that observed inside the bus.
d. may be either greater, smaller, or equal to that observed inside the bus.
e. is greator than that observed inside the bus
Answer:
d
Explanation:
good question. now the bus is moving in constant velocity . a student in front tosses a ball to the student in back. but we dont know the speed at which the student tosses a ball. we have to assume the speed
assume the speed of ball is slightly less than the speed of bus. in this case the stationary observer sees the ball in slower speed than the one inside the bus.
so a is correct
now assume the speed of ball is 1/2 the speed of bus. here stationary observer sees the ball the same speed as the one in bus observe
b is correct
assume the speed of ball is very small than the speed of bus . in this case the stationary observer see in grater speed than the student in bus
e also correct
so correct answer is d. it depends on the speed of ball tossed by the student in front.
n astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a time interval 11.2 s . What are the average acceleration in each interval? Assume that the positive direction is to the right.
Answer:-
-1 m/s^2
Explanation:
The average acceleration is given by dividing the change in velocity by change in time;
[tex]a_f=\frac{v_f-v_i}{t_f-t_i}[/tex]
[tex]=\frac{(0-11.2)}{(11.2-0)}=-1 m/s^2[/tex]
the point to be noted here is if the velocity is to the left we substitute it with a negative sign and if it is to the right we substitute it with a positive sign.
A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic field that produces on it an upward force exactly equal in magnitude to the wire's weight, causing the wire to "levitate."
Required:
a. What is the field's magnitude?
b. What is the field's direction?
Answer:
The classification of that same issue in question is characterized below.
Explanation:
The given values are:
Current, I = 50.0 A
Diameter, d = 0.10 cm
(a)...
As we know,
⇒ Magnetic force = Copper wire's weight
So,
⇒ [tex]B\times I\times L=M\times g[/tex]
On putting the estimated values, we get
⇒ [tex]B\times 50\times 1=7.037\times 10^{-3}\times 9.81[/tex]
⇒ [tex]50B=69.03297\times 10^{-3}[/tex]
⇒ [tex]B=1.38\times 10^{-3} \ T[/tex]
(b)...
As we know,
⇒ [tex]m=\delta\times L\times \frac{\pi \ d^2}{4}[/tex]
⇒ [tex]=8960\times 1\times \frac{\pi \ (0.001)^2}{4}[/tex]
⇒ [tex]=2240\times \pi \ 0.000001[/tex]
⇒ [tex]=7.037\times 10^{-3} \ kg[/tex]
A CD is spinning on a CD player. In 220 radians, the cd has reached an angular speed of 92 r a d s by accelerating with a constant acceleration of 14 r a d s 2 . What was the initial angular speed of the CD
Answer:
The initial angular speed is [tex]w_i = 48 \ rad/s[/tex]
Explanation:
From the question we are told that
The angular displacement is [tex]\theta = 220 \ rad[/tex]
The angular speed is [tex]w_f = 92 \ rad/s[/tex]
The acceleration is [tex]\alpha = 14 \ rad/s^2[/tex]
Generally the initial angular speed can be evaluated as
[tex]w_f ^2 = w_i ^2 + 2 * \alpha * \theta[/tex]
=> [tex]w_i ^2 = w_f ^2 - 2 * \alpha * \theta[/tex]
substituting values
=> [tex]w_i ^2 = 92 ^2 - 2 * 14 * 220[/tex]
=> [tex]w_i ^2 = 2304[/tex]
=> [tex]w_i = 48 \ rad/s[/tex]
The Gulf Stream off the east coast of the United States can flow at a rapid 3.8 m/s to the north. A ship in this current has a cruising speed of 8.0 m/s . The captain would like to reach land at a point due west from the current position.
At this heading, what is the ship's speed with respect to land?
Answer:
61.6° west of South
Explanation:
The ship goes to the south at an equal rate just like water flows to the north. Thus, the velocities would balance making the ship move towards the west.
Since we're dealing with water, the ship goes 3.8 m / s to the South, but a lot still remains to the west. Finding this would require us drawing a triangle. 3.8 m/s point down side and the hypotenuse is 8
cos(θ) = [adjacent/hypotenuse]
Cos θ = 3.8/8
Cos θ = 0.475
θ = cos^-1 (0.475)
θ = 61.6°
Therefore the angle is 61.6° west of South.
A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of the circle. (A) What is the speed of the stopper at the bottom of the circle? (HINT: Use energy conservation principles!) (10.2 m/s) (B) What is the tension in the string when the stopper is at the top of the circle? (0.27 N) (C) What is the tension in the string when the stopper is at the bottom of the circle?
Answer:
Explanation:
A )
At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy
1/2 m V² = mg x 2r + 1/2 mv²
m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top
1/2 V² = g x 2r + 1/2 v²
V² = g x 4r + v²
V² = 9.8 x 4 + 8²
V² = 103.2
V = 10.16 m/s
B )
If T be the tension at the top
Net downward force
= mg + T . This force provides centripetal force for the circular motion
mg +T = mv² / r
T = mv²/r -mg
= m ( v²/r - g )
= .005 ( 8²/1 -g )
= .005 x 54.2
= .27 N .
C ) At the bottom
Net force = T - mg , T is tension at the bottom , V is velocity at bottom
T-mg = mV²/r
T = m ( V²/r +g )
= .005 ( 10.16²/1 +9.8)
= .005 x 113
= .56 N .
A kicked ball rolls across the grass and eventually comes to a stop in 4.0 sec. When the ball was kicked, its initial velocity was 20 mi/ hr. What is the acceleration of the ball as it rolls across the grass?
Answer:
-2.24 m/s²
Explanation:
Given:
v₀ = 20 mi/hr = 8.94 m/s
v = 0 m/s
t = 4.0 s
Find: a
v = v₀ + at
0 m/s = 8.94 m/s + a (4.0 s)
a = -2.24 m/s²
Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy of the surrounding electrons). The following masses are given:
5727Co: 56.936296u
5726Fe: 56.935399u
Express your answer in millions of electron volts (1u=931.5MeV/c2) to three significant figures.
A negligible amount of this energy goes to the resulting 5726Fe atom as kinetic energy. About 90 percent of the time, after the electron-capture process, the 5726Fe nucleus emits two successive gamma-ray photons of energies 0.140MeV and 1.70 102MeV in decaying to its ground state. The electron-capture process itself emits a massless neutrino, which also carries off kinetic energy. What is the energy of the neutrino emitted in this case?
Express your answer in millions of electron volts.
Answer:
Explanation:
⁵⁷Co₂₇ + e⁻¹ = ²⁷Fe₂₆
mass defect = 56.936296 + .00055 - 56.935399
= .001447 u
equivalent energy
= 931.5 x .001447 MeV
= 1.3479 MeV .
= 1.35 MeV
energy of gamma ray photons = .14 + .017
= .157 MeV .
Rest of the energy goes to neutrino .
energy going to neutrino .
= 1.35 - .157
= 1.193 MeV.
Espresso is a coffee beverage made by forcing steam through finely ground coffee beans. Modern espresso makers generate steam at very high pressures and temperatures, but in this problem we'll consider a low-tech espresso machine that only generates steam at 100?C and atomospheric pressure--not much good for making your favorite coffee beverage.The amount of heat Q needed to turn a mass m of room temperature ( T1) water into steam at 100?C ( T2) can be found using the specific heat c of water and the heat of vaporization Hv of water at 1 atmosphere of pressure.Suppose that a commercial espresso machine in a coffee shop turns 1.50 kg of water at 22.0?C into steam at 100?C. If c=4187J/(kg??C) and Hv=2,258kJ/kg, how much heat Q is absorbed by the water from the heating resistor inside the machine?Assume that this is a closed and isolated system.Express your answer in joules to three significant figures.Q = _________________ J
Answer:
Q = 3877 KJ
Explanation:
Since, the system is closed and isolated. Therefore, the law of conservation of energy can be written as:
Heat Absorbed By Water (Q) = Heat required to raise the temperature of water (Q₁) + Heat required to convert water to steam (Q₂)
Q = Q₁ + Q₂ ----- equation (1)
Now, for Q₁:
Q₁ = m C ΔT
where,
m = Mass of Water = 1.5 kg
C = Specific Heat of Water = 4187 J/kg.°C
ΔT = Change in Temperature of Water = T₂ - T₁ = 100°C - 22°C = 78°C
Therefore,
Q₁ = (1.5 kg)(4187 J/kg.°C)(78°C)
Q₁ = 490 x 10³ J =490 KJ
Now, for Q₂:
Q₂ = m H
where,
m = Mass of Water = 1.5 kg
H = Heat of Vaporization of Water = 2258 KJ/kg
Therefore,
Q₂ = (1.5 kg)(2258 KJ/kg)
Q₂ = 3387 KJ
Substituting the values in equation (1), we get:
Q = Q₁ + Q₂
Q = 490 KJ + 3387 KJ
Q = 3877 KJ
An object, with mass 70 kg and speed 21 m/s relative to an observer, explodes into two pieces, one 4 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame
Answer:
K = 3.9 kJ
Explanation:
The kinetic energy ([tex]K_{T}[/tex]) added is given by the difference between the final kinetic energy and the initial kinetic energy:
[tex] K_{T} = K_{f} - K_{i} [/tex]
The initial kinetic energy is:
[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} [/tex]
Where m₁ is the mass of the object before the explosion and v₁ is its velocity
[tex] K_{i} = \frac{1}{2}m_{1}v_{1}^{2} = \frac{1}{2}70 kg*(21 m/s)^{2} = 1.54 \cdot 10^{4} J [/tex]
Now, the final kinetic energy is:
[tex] K_{f} = \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2} [/tex]
Where m₂ and m₃ are the masses of the 2 pieces produced by the explosion and v₁ and v₂ are the speeds of these pieces
Since m₂ is 4 times as massive as m₃ and v₃ = 0, we have:
[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}v_{2}^{2} + \frac{1}{2}*\frac{1}{5}m_{1}*0 [/tex] (1)
By conservation of momentum we have:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]
[tex] m_{1}v_{1} = \frac{4}{5}m_{1}v_{2} + \frac{1}{5}m_{1}*0 [/tex]
[tex] v_{2} = \frac{5}{4}v_{1} [/tex] (2)
By entering (2) into (1) we have:
[tex] K_{f} = \frac{1}{2}*\frac{4}{5}m_{1}(\frac{5}{4}v_{1})^{2} = \frac{1}{2}*\frac{4}{5}70 kg(\frac{5}{4}*21 m/s)^{2} = 1.93 \cdot 10^{4} J [/tex]
Hence, the kinetic energy added is:
[tex] K_{T} = K_{f} - K_{i} = 1.93 \cdot 10^{4} J - 1.54 \cdot 10^{4} J = 3.9 \cdot 10^{3} J [/tex]
Therefore, the kinetic energy added to the system during the explosion is 3.9 kJ.
I hope it helps you!
the heat capacity of 0.125Kg of water is measured to be 523j/k at a room temperature.Hence, calculate the heat capacity of water
(a) per unit mass
(b) per unit volume
Answer:
A. 4148 J/K/Kg
B. 4148 J/K/L
Explanation:
A. Heat capacity per unit mass is known as the specific heat capacity, c.
C = Heat capacity/mass(kg)
C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg
B. Volume of water = mass/density
Density of water = 1 Kg/L
Volume of water = 0.125 Kg/ 1Kg/L
Volume of water = 0.125 L
Heat capacity per unit volume = (523 J/K) / 0.125 L
Heat capacity per unit volume = 4148 J/K/L
At an instant when a soccer ball is in contact with the foot of the player kicking it, the horizontal or x component of the ball's acceleration is 950 m/s2 and the vertical or y component of its acceleration is 750 m/s2. The ball's mass is 0.35 kg. What is the magnitude of the net force acting on the soccer ball at this instant?
Answer:
F = 423.63 N
Explanation:
Since, the x-component and y-components of the acceleration of ball are given. Therefore, we need to find the resultant or net acceleration of the soccer ball first. For that purpose we use to the formula for the resultant of rectangular components of a vector:
a = √(ax² + ay²)
where,
a = net acceleration = ?
ax = x - component of acceleration = 950 m/s²
ay = y - component of acceleration = 750 m/s²
Therefore,
a = √[(950 m/s²)² + (750 m/s²)²]
a = 1210.4 m/s²
Now, from Newton's Second Law, we know that:
F = ma
where,
m = mass of ball = 0.35 kg
F = Net force acting on ball = ?
F = (0.35 kg)(1210.4 m/s²)
F = 423.63 N
1. Describe what must happen to an atom to make it
A. A cation
B. An anion
2. Describe why some acids are strong while other acids are weak
3. Compare protons, neutrons and electron, listing their similarities and differences
4. Explain why you breathe faster and deeper when exercising
Answer:
Explanation:
Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 × 10−23 g, and an electron ... The amu was originally defined based on hydrogen, the lightest element, ... but three-letter symbols have been used to describe some elements that have ...
Protons: Protons are positively charged particles that are also found in the nucleus. Like neutrons, protons give mass to the atom but do not participate in ... 3) Electrons: Electrons are negatively charged particles that are found in ... pair of electrons with 4 different hydrogen atoms, forming a molecule of CH4 (methane).Elements differ from each other in the number of protons they have, e.g. ... Atoms of an element that have differing numbers of neutrons (but a constant atomic ... Electrons, because they move so fast (approximately at the speed of light), ...toms are made up of particles called protons, neutrons, and electrons, which ... Therefore, they do not contribute much to an element's overall atomic mass. ... For instance, iron, Fe, can exist in its neutral state, or in the +2 and +3 ionic states. ... Isotopes of the same element will have the same atomic number but different ...
Find the equivalent resistance from the indicated terminal pair of the networks in the attached doc
Answer:
a) R = 2.5 Ω, b) R = 1 Ω, c) R = 2R / 3 Ω
Explanation:
The resistance configuration can be in series or in parallel, for each one the equivalent resistance can be calculated
series, the equivalent resistance is the sum of the resistances
parallel, the inverse of the equivalent resistance is the inverse of the sum of the resistances
let's apply these principles to each case
case a)
equivalent series resistance
R₁ = 1 +4 = 5 ohm
R₂ = 2 +3 = 5 ohn
these two are in parallel
1 / R = 1/5 +1/5
1 / R = 2/5
R = 2.5 Ω
case B
we solve the parallel
1 / R₁ = ½ + ½ = 1
R₁ = 1 Ω
we solve the resistors in series
R₂ = 1 + 1
R₂ = 2 Ω
finally we solve the last parallel
1 / R = ½ +1/2 = 1
R = 1 Ω
case C
we solve house resistance pair in series
R₁ = R + 2R = 3R
we go to the next mesh
R₂ = R + 2R = 3R
R₃ = R + 2R = 3R
last mesh
R₄ = R + R = 2R
now we solve the parallel of this equivalent resistance
1 / R = 1 / R₁ + 1 / R₂ + 1 / R₃ + 1 / R₄
1 / R = 1 / 3R + 1 / 3R + 1 / 3R + 1 / 2R
1 / R = 3 / 3R + 1 / 2R = 1 / R + 1 / 2R
1 / R = 3 / 2R
R = 2R / 3 Ω
A small block with a mass of 0.120 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. 6.34). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is observed to be 2.80 m/s.
(a) What is the tension in the cord in the original situation when the block has speed v = 0.70 m/s? (b) What is the tension in the cord in the final situation when the block has speed v = 2.80 m/s? (c) How much work was done by the person who pulled on the cord?
Answer:
a) 0.147 N
b) 9.408 N
c) 9.261 N
Explanation:
The tension on the cord is the only force keeping the block in circular motion, thus representing the entirety of its centripetal force [tex]\frac{mv^{2} }{r}[/tex]. Plugging in values for initial and final states and we get answers for a and b. The work done by the person causes the centripetal force to increase, and thus is the difference between the final tension and the initial tension.
28 points!! please help
A certain type of laser emits light that has a frequency of 4.6 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 3.1 x 10-11s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse
Answer:
14,260
Explanation:
Relevant data provided for computing the wavelengths are in one pulse is here below:-
The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]
Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]
The number of wavelengths are in one pulse is shown below:-
[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]
[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]
= 14,260
Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.
A rigid tank contains 2 kg of an ideal gas at 4 atm and 40 C. Now a valve is opened, and half of the mass of the gas is allowed to escape. if the final pressure in the tank is 2.2 atm. The final temperature in the tank is: Hint: make sure you convert the units of temperature and pressure to the proper units
Answer:
Final Temperature = 71 °C
Explanation:
In this case, the ideal gas equation is written as;
PV = mRT
Where;
P is pressure
V is volume
m is mass
R is gas constant
T is temperature
We will take the volume to be constant.
So, in the initial state, we have;
P1•V = m1•R•T1 - - - eq(1)
In the final state, we have;
P2•V = m2•R•T2 - - - - eq(2)
Combining eq (1) and eq(2),we have;
P1•m2•R•T2 = P2•m1•R•T1
Dividing both sides by R gives;
P1•m2•T2 = P2•m1•T1
Making T2 the subject gives;
T2 = (P2•m1•T1)/(P1•m2)
Now, we are given;
m1 = 2kg
m2 = ½*2 = 1kg
P1 = 4 atm
P2 = 2.2 atm
T1 = 40°C = 273 + 40 K = 313K
Plugging in this values into the T2 equation, we have;
T2 = (2.2 × 2 × 313)/(4 × 1)
T2 = 344 K
Converting to °C, we have;
T2 = 344 - 273 = 71 °C
a 15-nC point charge is at the center of a thin spherical shell of radius 10cm, carrying -22nC of charge distributed uniformly over its surface. find the magnitude and direction of the electric field (a) 2.2cm,(b)5.6cm,and (c)14 cm from the point charge.
Answer:
A) E = 278925.62 N/C with direction; radially out.
B) E = 43048.47 N/C with direction radially out.
C) E = -3214.29 N/C with direction radially in.
Explanation:
From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;
E = kQ/r²
where;
Q is the net charge within the distance r.
We are given the charge Q = 15-nC and
spherical shell of radius 10cm
A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C
While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²
E = 278925.62 N/C
This will be radially out ,since the net charge is positive.
B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C
While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²
E = 43048.47 N/C
This will be radially out ,since the net charge is positive.
C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;
Q = 15 nC - 22 nC
Q = -7 nC = -7 x 10^(-9) C
and;
E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²
E = -3214.29 N/C
This will be radially in, since the net charge is negative. You can indicate this with a negative answer.
A) When The distance r is = 2.2 cm = 0.022 m is between the surface and also the point charge, also that so only the point charge lies within this distance and also Q = 15 NC = 15 x 10^(-9) C
Then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²Then E = 278925.62 N/CThen This will be radially out since the net charge is positive.
B) When The distance r = 5.6 cm = 0.056 m is between the surface and also the point charge, so only the point charge lies within this distance and also Q = 15 nC = 15 x 10^(-9) C
then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²Then E = 43048.47 N/CAfter that This will be radially out since the net charge is positive.
C) Then when The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;
Then Q = 15 nC - 22 nCAfter that Q = -7 nC = -7 x 10^(-9) CWhen E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²Then E = -3214.29 N/C Thus, This will be radially in, since the net charge is negative.Find out more information about magnitude here:
https://brainly.com/question/13502329
A silver rod having a length of 83.0 cm and a cross-sectional diameter of 2.40 cm is used to conduct heat from a reservoir at a temperature of 540 oC into an otherwise completely thermally insulated chamber that contains 1.43 kg of ice at 0 oC. How much time is required for the ice to melt completely
Answer:
3985 s or 66.42 mins
Explanation:
Given:-
- The length of the rod, L = 83.0 cm
- The cross sectional diameter of rod , d = 2.4 cm
- The temperature of reservoir, Tr = 540°C
- The amount of ice in chamber, m = 1.43 kg
- The temperature of ice, Ti = 0°C
- Thermal conductivity of silver, k = 406 W / m.K
- The latent heat of fusion of water, Lf = 3.33 * 10^5 J / kg
Find:-
How much time is required for the ice to melt completely
Solution:-
- We will first determine the amount of heat ( Q ) required to melt 1.43 kg of ice.
- The heat required would be used as latent heat for which we require the latent heat of fusion of ice ( Lf ). We will employ the first law of thermodynamics assuming no heat is lost from the chamber ( perfectly insulated ):
[tex]Q = m*L_f\\\\Q = ( 1.43 ) * ( 3.33 * 10 ^5 )\\\\Q = 476190 J[/tex]
- The heat is supplied from the hot reservoir at the temperature of 540°C by conduction through the silver rod.
- We will assume that the heat transfer through the silver rod is one dimensional i.e along the length ( L ) of the rod.
- We will employ the ( heat equation ) to determine the rate of heat transfer through the rod as follows:
[tex]\frac{dQ}{dt} = \frac{k.A.dT}{dx}[/tex]
Where,
A: the cross sectional area of the rod
dT: The temperature difference at the two ends of the rod
dx: The differential element along the length of rod ( 1 - D )
t: Time ( s )
- The integrated form of the heat equation is expressed as:
[tex]Q = \frac{k*A*( T_r - T_i)}{L}*t[/tex]
- Plug in the respective parameters in the equation above and solve for time ( t ):
[tex]476190 = \frac{406*\pi*0.024^2 * ( 540 - 0 ) }{0.83*4}*t \\\\t = \frac{476190}{119.49619} \\\\t = 3985 s = 66.42 mins[/tex]
Answer: It would take 66.42 minutes to completely melt the ice
