Answer:
(a) emf = 1.18 mV
(b) counter-clockwise sense
Explanation:
(a) The induced emf is given by the following formula:
[tex]emf=-\frac{d\Phi_B}{dt}[/tex] (1)
where:
ФB: magnetic flux = AB = (area of the loop)*(magnitude of the magnetic field)
A = πr^2
B = 0.800 T
You replace the expression for the magnetic flux in the equation (1):
[tex]emf=-B\frac{\Delta A}{\Delta t}=-B\frac{A_2-A_1}{t_2-t_1}[/tex]
A1: initial area
A2: final area
t2-t1: time interval = 9.0s
Then you have to calculate the change in the area of the loop, by using the information about the circumference of the loop. First you calculate the radius of the loop for a circumference of 165 cm = 1.65m
[tex]s=1.65m=2\pi r\\\\r=\frac{1.65m}{2\pi}=0.262m[/tex]
You calculate the initial area A1:
[tex]A_1=\pi (0.262m)^2=0.215m^2[/tex]
After 9.0 second the circumference will be:
[tex]s'=1.65m-0.14\frac{m}{s}(9.0s)=0.39m[/tex]
the new radius and the final area is:
[tex]r=\frac{0.39m}{2\pi}=0.062m[/tex]
[tex]A_2=\pi(0.062m)^2=0.012m^2[/tex]
Finally, you replace in the equation (1):
[tex]emf=-(0.800T)\frac{0.012m^2-0.215m^2}{9.0s}=1.8*10^{-3}V=1.8mV[/tex]
The induced emf in the circular loop is 1.18mV
(b) The induced emf generates an electric current, which produces a magnetic field that is opposite to the direction of the constant magnetic field of 0.800T. Due to this magnetic field point into the loop. The current has to have a direction in a counter-clockwise sense.
1. Calculate the centripetal force exerted on a 900kg900kg car that rounds a 600m600m radius curve on horizontal ground at 25.0m/s25.0m/s. 2. Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
Explanation:
It is given that,
Mass of a car is 900 kg
Radius of curve is 600 m
Speed of the car in the curve is 25 m/s
We need to find the centripetal force exerted on a car. The formula used to find the centripetal force is given by :
[tex]F=\dfrac{mv^2}{r}\\\\F=\dfrac{900\times (25)^2}{600}\\\\F=937.5\ N[/tex]
So, the centripetal force exerted on a car is 937.5 N.
Static friction prevents the car from slipping. It means that the magnitude of centripetal force is balanced by the frictional force. So, the frictional force of 937.5 N is acting on the car.
A glass sphere carrying a uniformly distributed charge of +Q is surrounded by an initially neutralspherical plastic shell. (Assume the charge +Q is uniformly distributed across thesurface of the glass sphere.)
Required:
a. Qualitatively, indicate the polarization of the plastic.
1. The plastic will polarize so as to have positive charge +Qon its inner surface and negativecharge −Q on its outer surface.
2. Dipoles in the plastic will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center.
4. Dipoles in the plastic will polarize and orient themselves radially, with their positiveendspointing toward the center.
b. Qualitatively, indicate the polarization of the inner glass sphere. Explain briefly.A net charge −Q from the dipoles will be uniformly distributed through the volume of the sphere.
1. There will be no polarization inside the glass sphere since the net electric field there iszero.
2. Dipoles in the glass will polarize and orient themselves perpendicular to the radial electricfield due to the charge +Q.
3. Dipoles in the glass will polarize and orient themselves radially, with their positive endspointing toward the center.
c. Is the electric field at location Poutside the plastic shell larger, smaller, or the same as itwould be if the plastic weren't there? Explain briefly.
1. Larger, because a net positive charge is created from the polarization of the shell.
2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
3. Smaller, because the negative charges displaced during polarization are closer to Pthanthe positive charges.
4. Smaller, because the plastic shell shields location Pfrom the charge +Q, such that the netfield at Pis zero.
5. The same, because no net charge is created from the polarization of the field.
Answer:
(A) 3. Dipoles in the plastic will polarize and orient themselves radially, with their negativeends pointing toward the center
(B) 2. There will be no polarization inside the glass sphere since the net electric field there is zero.
Explanation: charges are only distributed on the surface of the charged hollow conductor. The core must have zero charge.
(C) 2. Larger, because the positive charges displaced during polarization are closer to P than thenegative charges.
Identify the five categories of stressors.
Answer:
The five kinds of stressors are:
Acute time-limited
Brief naturalistic
Stressful events sequences
Chronic
Distant
Explanation:
yeah
When you take your 1900-kg car out for a spin, you go around a corner of radius 56 m with a speed of 14 m/s. The coefficient of static friction between the car and the road is 0.88. Part A Assuming your car doesn't skid, what is the force exerted on it by static friction
Answer:
6,650 newtons
Explanation:
The computation of the force exerted on it by static friction is shown below:
Data provided in the question
Mass of car = m = 1,900 kg
speed = v = 14 m/s
radius = r = 56 m
Let us assume friction force be f
And, the Coefficient of friction = [tex]\mu[/tex]= 0.88
As we know that
[tex]f = \frac{mv^2}{r}[/tex]
[tex]= \frac{1,900 \times 14^2}{56}[/tex]
= 6,650 newtons
We simply applied the above formula so that the force exerted could come
please help
Complete the first and second sentences, choosing the correct answer from the given ones.
1. A temperature of 100 K corresponds on a Celsius scale to 100 ° C / 0 ° C / 173 ° C / –173 ° C.
2. At 50 ° C, it corresponds to a Kelvin scale of 150 K / 323 K / 273 K / 223 K.
1) 100 ° C
2) 323 K
hope it helps youuuuuu
You are at a stop light in your car, stuck behind a red light. Just before the light is supposed to change, a fire engine comes zooming up towards you traveling at a horrendous 85.0 km/h. If the siren has a rated frequency 665 Hz, what frequency of the sound do you hear
Answer:
The frequency of the sound you will hear is 713.85 Hz
Explanation:
Given;
speed of your car, [tex]v_s[/tex] = 85.0 km/h
frequency of the siren, f = 665 Hz
Speed of sound in air, v = 345 m/s
The frequency of the sound you hear, can be calculated as;
[tex]f' = f(\frac{v}{v-v_s})[/tex]
Convert the speed of the car to m/s
[tex]85 \ km/h =\frac{85 \ km}{h} (\frac{1000\ m}{1 \ km})(\frac{1 \ h}{3600 \ s} ) = 23.61 \ m/s[/tex]
[tex]f' = f(\frac{v}{v-v_s} )\\\\f' = 665(\frac{345}{345-23.61} )\\\\f' = 665 (1.07346)\\\\f' = 713.85 \ Hz[/tex]
Therefore, the frequency of the sound you will hear is 713.85 Hz
A 1,269 kg rocket is traveling at 413 m/s with 2,660 kg of fuel on board. If the rocket fuel travels at 1,614 m/s relative to the rocket, what is the rockets final velocity after it uses half of its fuel?
Answer:
About 2104m/s
Explanation:
[tex]F=ma \\\\F=\dfrac{2660kg}{2}\cdot 1614m/s=2,146,620N \\\\2,146,620N=1,269kg\cdot a \\\\a\approx 1691m/s \\\\v_f=v_o+at=413m/s+1691m/s=2104m/s[/tex]
Hope this helps!
What percent of our solar system's mass is in the sun?
Answer:
99.8
Explanation:
most massive the sun is at the center of the universe
The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and velocity vectors of the plane at a later time are given by r=(1.21x103i+3.45x104;)m and v= (2 i-3.5j) m/s The magnitude, in meters, of the plane's displacement from the origin is:_________
a. 2.50 x104
b. 1.45 x 104
c. 3.45x104
d. 2.5x103
e. none of the above
Answer:
d = 3.5*10^4 m
Explanation:
In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:
[tex]\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m[/tex]
The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:
[tex]d=\sqrt{(x-x_o)^2+(y-y_o)^2}[/tex] (1)
where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):
[tex]d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m[/tex]
hence, the displacement of the airplane is 3.45*10^4 m
A hornet circles around a pop can at increasing speed while flying in a path with a 12-cm diameter. We can conclude that the hornet's wings must push on the air with force components that are Group of answer choices down and backwards. down, backwards, and outwards. down and inwards. down and outwards. straight down.
Answer:
down, backwards, and outwards.
Explanation:
For a hornet that is accelerating in flight, this means that there is a net forward motion at a relatively constant vertical height above the ground.
For this flight, the wings beat downwards to counter the weight of the hornet due to gravity, keeping it at that height above the floor.
For the hornet to accelerate forward, there has to be a net backwards force by the wing on the air. This backwards force accelerates tr forward due to the absence of an equal opposing force in the opposite direction save for a little drag.
The wings also beat with forces directed outwards to provide centripetal force to keep the hornet stable. The absence of this would cause it to spiral out of control.
Question 10
Air with a density of 1.20 kg/m3 flows through a 75.0 cm diameter pipe with a velocity of 2.00 m/s. What is the mass flow rate?
Answer:
75.0 cm
Explanation:
becouse i don,t no the right answer
An arrow is shot from a height of 1.55 m toward a cliff of height H. It is shot with a velocity of 26 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.99 s later.
(a) Draw a sketch of the given example. Include the x-y coordinate system.
(b) What is the height of the cliff?
(c) What is the maximum height reached by the arrow along its trajectory?
(d) What is the arrow's impact speed just before hitting the cliff?
Answer:
Explanation:
vertical component of the velocity of arrow
= 26 sin 60 = 22.516 m
height reached by it after 3.99 s
h = ut - 1/2 g t²
= 22.516 x 3.99 - .5 x 9.8 x 3.99²
= 89.83 - 78
11.83 m
Total height of cliff = 1.55 + 11.83
= 13.38 m
c ) maximum height covered s
v² = u² - 2gs
0 = u² - 2gs
s = u² / 2g
= 22.516² / 2 x 9.8
= 25.86
maximum height reached
= 25.86 + 1.55
= 27.41 m
d )
vertical speed after 3.99 s
v = u - gt
= 22.516 - 9.8 x 3.99
= -16.586
Horizontal component will remain unchanged
Horizontal component = 26 cos 60
= 13 m /s
Resultant of two velocities
= √ 13²+ 16.568²
= 21 m /s
"A trooper is moving due south along the freeway at a speed of 28 m/s. At time t = 0, a red car passes the trooper. The red car moves with constant velocity of 40 m/s southward. At the instant the trooper's car is passed, the trooper begins to speed up at a constant rate of 2.9 m/s2. What is the maximum distance ahead of the trooper that is reached by the red car?"
Answer:
24.83 m
Explanation:
Applying the equation of motion;
d = vt + 0.5at^2 ......1
Where;
d = distance
v = velocity
t = time
a = acceleration
For the trooper;
v = 28 m/s
a = 2.9 m/s^2
Substituting into equation 1;
d1 = 28t + 0.5(2.9t^2)
d1 = 28t + 1.45t^2
For the red car;
v = 40 m/s
a = 0
Substituting into equation 1
d2 = 40t
The difference in distance is;
d = d2 - d1
d = 40t - (28t + 1.45t^2)
d = 12t - 1.45t^2
The maximum distance is at d(d)/dt = 0
differentiating d;
d' = 12 - 2.9t = 0
2.9t = 12
t = 12/2.9 = 4.137931034482
t = 4.138 s
Substituting t into function d;
d(max) = 12(4.138) - 1.45(4.138^2)
d(max) = 24.8275862 = 24.83 m
the maximum distance ahead of the trooper that is reached by the red car is 24.83 m
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed
Complete Question
An infinite sheet carries a uniform, positive charge per unit area. The electric field produced by the sheet is represented by parallel lines drawn with a density N lines per m2 that are perpendicular to and away from the sheet. The charge per unit area on the sheet is doubled. How should the density of the electric field lines be changed?
A It should stay the same
B It should be quadrupled.
C It should be quintupled
D It should be doubled.
E It should be tripled
Answer:
Option D is the correct option
Explanation:
Generally electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon_o}[/tex]
Where [tex]\sigma[/tex] is the charge per unit area (Charge density )
From the question we are told that [tex]\sigma[/tex] is doubled hence the
[tex]E = \frac{2 \sigma }{\epsilon_o}[/tex]
Looking the equation above we see that the value of the electric field will also double given that it is directly proportional to the charge density
The red giant Betelgeuse has a surface temperature of 3000 K and is 600 times the diameter of our sun. (If our sun were that large, we would be inside it!) Assume that it radiates like an ideal blackbody.a) If Betelgeuse were to radiate all of its energy at the peak-intensity wavelength, how many photons per second would it radiate?b) Find the ratio of the power radiated by Betelgeuse to the power radiated by our sun (at 5800 K).
Answer:
Explanation:
a )
Radius of the sun = .69645 x 10⁹ m .
600 times = 600 x .69645 x 10⁹ m
= 4.1787 x 10¹¹ m .
surface area A = 4π (4.1787 x 10¹¹)²
= 219.317 x 10²²
energy radiated E = σ A Τ⁴
= 5.67 x 10⁻⁸ x 219.317 x 10²² x (3000)⁴
= 100695 x 10²⁶ J
To know the wavelength of photon emitted
[tex]\lambda_mT= b[/tex]
[tex]\lambda_m= \frac{b}{T}[/tex]
= 2.89777 x 10⁻³ / 3000
= 966 nm
= 1275 /966 eV
1.32 x 1.6 x 10⁻¹⁹ J
= 2.112 x 10⁻¹⁹ J
No of photons radiated = 100695 x 10²⁶ / 2.112 x 10⁻¹⁹
= 47677.5 x 10⁴⁵
= .476 x 10⁵⁰ .
b )
energy radiated by our sun per second
E₂ = σ A 5800⁴
energy radiated by Betelgeuse per second
E₁ = σ x 600²A x 3000⁴
E₁ / E₂ = σ x 600²A x 3000⁴ / σ A 5800⁴
= 36 X 10⁴ x 3⁴ x 10¹² / 58⁴ x 10⁸
= 25.76 x 10⁸ x 10⁻⁵
= 25760 times .
Two carts undergo an inelastic collision where they stick together. Cart A has an initial velocity v0, and the second cart B is initially at rest. After the collision, it is observed that the ratio of the final kinetic energy system to its initial kinetic energy is KfK0= 1/6. Determine the ratio of the carts' masses, mBmA. (Assume the track is frictionless.)
Answer:
Explanation:
Initial kinetic energy of the system = 1/2 mA v0²
If Vf be the final velocity of both the carts
applying conservation of momentum
final velocity
Vf = mAvo / ( mA +mB)
kinetic energy ( final ) = 1/2 (mA +mB)mA²vo² / ( mA +mB)²
= mA²vo² / 2( mA +mB)
Given 1/2 mA v0² / mA²vo² / 2( mA +mB) = 6
mA v0² x ( mA +mB) / mA²vo² = 6
( mA +mB) / mA = 6
mA + mB = 6 mA
5 mA = mB
mB / mA = 5 .
I really need help with this question someone plz help !
Answer:weight
Explanation:weight
What is the highest point at which weather will generally occur?
Answer:
At thestratosphere: it 20- 25km
Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 11.3 m/s. Olaf's mass is 75.0 kg. (a) If Olaf catches the ball, with what speed v_f do Olaf and the ball move afterward
Answer:
v = 0.059 m/s
Explanation:
To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:
[tex]mv_{1i}+Mv_{2i}=(m+M)v[/tex] (1)
m: mass of the ball = 0.400kg
M: mass of Olaf = 75.0 kg
v1i: initial velocity of the ball = 11.3m/s
v2i: initial velocity of Olaf = 0m/s
v: final velocity of Olaf and the ball
You solve the equation (1) for v and replace the values of all variables:
[tex]v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}[/tex]
Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s
You are helping your friend prepare for the next skateboard exhibition by determining if the planned program will work. Your friend will take a running start and then jump onto a heavy-duty 13-lb stationary skateboard. The skateboard will glide in a straight line along a short, level section of track, then up a curved concrete wall. The goal is to reach a height of at least 10 feet above the starting point before coming back down the slope. Your friend's maximum running speed to safely jump on the skateboard is 24 feet/second. Your friend weighs 155-lbs. What is the height hf that your friend will reach according to his plan?
Answer:
8.3 feet
Explanation:
The kinetic energy of the system on the ground is ...
KE = Σ(1/2)(mv^2) = (1/2)(155)(24^2) +(1/2)(13)(0^2) = 44640 lb·ft²/s²
The potential energy at the highest point is the same:
PE = mgh
44640 = (155 +13)(32)h
h = 44640/5376 = 8.30 . . . . feet
_____
We haven't worried too much about the conversion between pounds mass and pounds force. Whatever factor may be involved will divide out when computing the maximum height. We have used g=32 ft/s².
__
To achieve a 10 ft height, the running speed would need to be 26.34 ft/s, about 10% higher.
At an accident scene on a level road, investigators measure a car’s skid mark (mass of car is M) to be of length d. It was a rainy day and the coefficient of friction was estimated to be μk.
A) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.B) Why does the car's mass not matter?1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation.2) Since the work done by friction does not depend on mass.3) Since the change in kinetic energy and the work done by friction do not depend on mass.
Answer:
1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation
Explanation:
The kinetic friction works against the kinetic energy of the car and the car stops when these two equalises .
friction force = μk x R , μk is coefficient of kinetic friction and R is reaction from the ground.
= μk x mg
work done by friction
= force x displacement
= μk x mg x d
kinetic energy of car at the time of accident = 1/2 m v²
kinetic energy = work done by friction
1/2 m v² = μk x mg x d
d = v² / (2 μk x g)
v² = 2dμk g
v = √(2dμk g)
Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation
A 20 g "bouncy ball" is dropped from a height of 1.8 m. It rebounds from the ground with 80% of the speed it had just before it hit the ground. Assume that during the bounce the ground causes a constant force on the ball for 75 ms. What is the force applied to the ball by the ground in N?
The following are not correct: 0.513 N, 0.317 N, 0.121 N. Please show your work so I can understand!
Answer:
F = 0.314 N
Explanation:
In order to calculate the applied force to the ball by the ground, you first calculate the speed of the ball just before it hits the ground. You use the following formula:
[tex]v^2=v_o^2+2gy[/tex] (1)
y: height from the ball starts its motion = 1.8 m
vo: initial velocity = 0 m/s
g: gravitational acceleration = 9.8 m/s^2
v: final velocity of the ball = ?
You replace the values of the parameters in the equation (1):
[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(1.8m)}=5.93\frac{m}{s}[/tex]
Next, you take into account that the force exerted by the ground on the ball is given by the change, on time, of the linear momentum of the ball, that is:
[tex]F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}=m\frac{v_2-v_1}{\Delta t}[/tex] (2)
m: mass of the ball = 20g = 20*10^-3 kg
v1: velocity of the ball just before it hits the ground = 5.93m/s
v2: velocity of the ball after it impacts the ground (80% of v1):
0.8(5.93m/s) = 4.75 m/s
Δt: time interval o which the ground applies the force on the ball = 75*10^-3 s
You replace the values of the parameters in the equation (2):
[tex]F=(20*10^{-3}kg)\frac{4.75m/s-5.93m/s}{75*10^{-3}s}=-0.314N[/tex]
The minus sign means that the force is applied against the initial direction of the motion of the ball.
The applied force by the ground on the bouncy ball is 0.314 N
g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positive constant. Part A What is the force that one atom exerts on the other? Express your answer in terms of C6 and x. Fx = nothing Request Answer Part B Is this force attractive or repulsive? Is this force attractive or repulsive? attractive repulsive
Answer:
[tex]F_x = -\frac{6 C_6}{2^7}[/tex]
Attractive
Explanation:
Data provided in the question
The potential energy of a pair of hydrogen atoms given by [tex]\frac{C_6}{X_6}[/tex]
Based on the given information, the force that one atom exerts on the other is
Potential energy μ = [tex]\frac{C_6}{X_6}[/tex]
Force exerted by one atom upon another
[tex]F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X} (-\frac{C_6}{X^6})[/tex]
or
[tex]F_x = \frac{\partial}{\partial X} (\frac{C_6}{X^6})[/tex]
or
[tex]F_x = -\frac{6 C_6}{2^7}[/tex]
As we can see that the [tex]C_6[/tex] comes in positive and constant which represents that the force is negative that means the force is attractive in nature
first law of equilibrium
Answer:
For an object to be an equilibrium it must be experiencing no acceleration.
Explanation:
Hope it helps.
A man pushes a 25kg box up an incline 2.0m by applying a steady force of 95N parallel to the incline. The box moves up the incline at a steady speed. The incline makes an angle 15 degrees to the horizontal
a) What is the force of friction between the incline and the box
b)If the box is released at the top of the incline, what will its speed be at the bottom
Answer:
a) Ff = 19.29 N
b) v = 3.00 m/s
Explanation:
a) To calculate the friction force you use the second Newton Law in the incline plane, with an acceleration equal to zero, because the motion of the box has a constant velocity:
[tex]F-F_f-Wsin(\theta)=0\\\\[/tex] (1)
F: force applied by the man = 95N
Ff: friction force
W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N
θ: degree of the inclined plane = 15°
You solve the equation (1) for Ff and you replace the values of all variables in the equation (1):
[tex]F_f=-Wsin(\theta)+F\\\\F_f=-(245N)sin18\°+95N=19.29N[/tex]
b) To fins the velocity of the box at the bottom you use the following formula:
[tex]W_N=\Delta K[/tex] (2)
That is, the net work over the box is equal to the change in the kinetic energy of the box.
The net work is:
[tex]W_N=Mgsin(18\°)d-Ffd[/tex]
d: distance traveled by the box = 2.0m
[tex]W_N=245sin18\°(2.0m)N-19.29(2.0m)N=112.83J[/tex]
You use this value of the net work to find the final velocity of the box, by using the equation (2):
[tex]112.8J=\frac{1}{2}m[v^2-v_o^2]\\\\v_o=0m/s\\\\v=\sqrt{\frac{2(112.8J)}{m}}=\sqrt{\frac{225.67J}{25kg}}=3.00\frac{m}{s}[/tex]
The speed of the box, at the bottom of the incline plane is 3.00 m/s
A uniform rod of mass 2.30 kg and length 2.00 m is capable of rotating about an axis passing through its center and perpendicular to its length. A mass m1 = 5.30 kg is attached to one end and a second mass m2 = 3.50 kg is attached to the other end of the rod. Treat the two masses as point particles.
A) What is the moment of inertia of the system?B) If the rod rotates with an angular speed of 2.00 rad/s, how much kinetic energy does the system have?C) Now consider the rod to be of negligible mass. What is the moment of inertia of the rod and masses combined?D) If the rod is of negligible mass, what is the kinetic energy when the angular speed is 2.00 rad/s?
Answer:
Explanation:
Moment of inertia of the rod = 1/12 m L²
m is mass of the rod and L is its length
= 1/2 x 2.3 x 2 x 2
= 4.6 kg m²
Moment of inertia of masses attached with the rod
= m₁ d² + m₂ d²
m₁ and m₂ are masses attached , and d is their distance from the axis of rotation
= 5.3 x 1² + 3.5 x 1²
= 8.8 kg m²
Total moment of inertia = 13.4 kg m²
B )
Rotational kinetic energy = 1/2 I ω²
I is total moment of inertia and ω is angular velocity
= .5 x 13.4 x 2²
= 26.8 J .
C )
when mass of rod is negligible , moment of inertia will be due to masses only
Total moment of inertia of masses
= 8.8 kg m²
D )
kinetic energy of the system
= .5 x 8.8 x 2²
= 17.6 J .
(A) Total moment of inertia is 13.4 kgm²
(B) Total kinetic energy is 26.8J
(C) Moment of inertia is 8.8 kgm²
(D) Kinetic energy is 17.6J
Rotational motion:
(A) The moment of inertia of the rod is given by:
I = 1/12 mL²
where m is the mass of the rod
and L is the length
So,
I = (1/12) × 2.3 × 2²
I = 4.6 kgm²
Now, the moment of inertia of masses attached to the rod is given by:
I' = m₁ d² + m₂d²
where m₁ and m₂ are masses
and d is their distance from the axis of rotation
I' = 5.3 × 1² + 3.5 × 1²
I' = 8.8 kgm²
The total moment of inertia of the system is given by:
I(tot) = I + I'
I(tot) = 13.4 kgm²
(B) The rotational kinetic energy of an object with a moment of inertia I and angular velocity ω is given by:
KE = 1/2 I(tot)ω²
KE = 0.5 × 13.4 × 2²
KE = 26.8J
(C) If the mass of the rod is negligible, then the moment of inertia of the rod will be zero. So the total moment of inertia will be
I(tot) = I' = 8.8 kgm²
(D) the kinetic energy of the system when the mass of the rod is negligible and the angular speed is 2 rad/s is given by:
KE = 1/2 I'ω²
KE = 0.5 × 8.8 × 2²
KE = 17.6J
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Blocks of mass 10, 30, and 90 kg are lined up from left to right in that order on a frictionless surface so each block is touching the next one. A rightward-pointing force of magnitude 32 N is applied to the left-most block. 1) What is the magnitude of the force that the left block exerts on the middle one
Answer:
32N
Explanation:
The Left force exerts an opposite horizontal force of 32N on the middle object
Your new toaster has two separate toasting units, each of which consumes 600 watts of power when it is in use. When you operate one unit, a current of 5 amperes flowsthrough the wiring in your home and the wires waste about 1 watt of power handling that current. If you operate both toasting units at once, your toaster consumes 1200 watts and the current flowing through the wiring in your home doubles to 10 amperes. How much power will the wires in your home waste now
Answer:
1.92 Watt lost
Explanation:
Power rating of each toaster = 600 Watts
Current that flows = 5 Amperes
Wasted power = 1 Watt
Voltage of toaster can be gotten from P = [tex]I^{2}[/tex]R
where I = current
and R = Resistance
600 = [tex]5^{2}[/tex] x R
R = 600/25 = 24 Ohms.
According to joules loss due to heating of wire
Power loss P ∝ [tex]I^{2}[/tex]R
imputing values,
1 ∝ [tex]5^{2}[/tex] x 24
1 ∝ 600
to remove the proportionality sign, we introduce a constant k
1 = 600k
k = 1/600 = 0.00167
For the case where the current is doubled to 10 ampere, as the power doubles to 1200 W.
The resistance across the wire becomes
1200 = [tex]10^{2}[/tex]R
R = 1200/100 = 12 Ohms
power loss P = k x [tex]I^{2}[/tex]R
P = 0.0016 x [tex]10^{2}[/tex] x 12
P = 1.92 Watt lost
This question involves the concepts of power, current, and resistance.
The power wasted by the wires in the home for two units will be "4 watt".
POWER WASTAGEThe power wasted by the wires can be given in terms of current and resistance by the following formula:
[tex]P=I^2R\\\\\frac{P}{I^2}=R=Constant\\\\\frac{P_1}{I_1^2}=\frac{P_2}{I_2^2}[/tex]
where,
P₁ = Power wasted for one unit = 1 wattI₁ = current through wires for one unit = 5 AR = Resistance of wires = constantP₂ = Power wasted for two units = ?I₂ = Current through wires for two units = 10 ATherefore,
[tex]\frac{1\ watt}{(5\ A)^2}=\frac{P_2}{(10\ A)^2}\\\\P_2=\frac{(1\ watt)(100\ A^2)}{25\ A^2}[/tex]
P₂ = 4 watt
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A 330-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,110 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (Use 3.156 107 for the number of seconds in a year.)
Answer:
t = 402 years
Explanation:
To find the number of year that electrons take in crossing the complete transmission line, you first calculate the drift speed of the electrons. Then, you use the following formula for the current in a wire:
[tex]I=nqv_dA[/tex] (1)
n: number of mobile charge carrier per volume = 8.50*10^28 e/m^3
q: charge of the electron = 1.6*10^-19 C
vd: drift velocity of electron in the metal = ?
A: cross sectional area of the wire = π r^2 = π (0.02m/2)^2 = 3.1415*10^-4 m^2
I: current in the wire = 1110 A
You solve the equation (1) for vd:
[tex]v_d=\frac{I}{nqA}=\frac{110A}{(8.50*10^{28}m^{-3})(1.6*10^{-19}C)(3.1415*10^{-4}m^2)}\\\\v_d=2.59*10^{-4}m/s[/tex]
Next, you calculate the time by using the information about the length of the line transmission:
[tex]x=v_dt\\\\x=330km=330000m\\\\t=\frac{x}{v_d}=\frac{330000m}{2.59*10^{-4}m/s}=1,270,184,865s\\\\1,270,184,865s*\frac{1\ year}{3,156,107}=402.45\ years[/tex]
hence, the electrons will take aproximately 402 years in crossing the line of transmission
Why do bears activity increase as certain points during the day
Because they are well rested and have to work to get food in their system.
