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lesantik [10]
2 years ago
14

2. A solenoid. Suppose the south end of a bar magnet was introduced to the right end of this solenoid at a constant velocity. Wh

at direction would you expect the induced magnetic field to be in?4. To decrease the magnitude of current induced in the inductor using a bar magnet, you could (select all that apply):a. Decrease the strength of the magnet.b. Decrease the velocity of the magnet going into the solenoid.c. Decrease the number of coils in the solenoid.d. Increase the cross sectional area of the solenoid.6. List the TWO values you need to record to measure a voltage on the oscilloscope
Physics
1 answer:
Studentka2010 [4]2 years ago
5 0

Answer:

2) deflection must be towards the negative side of the voltage.

4) the correct statements are: b and c

Explanation:

2) This question is based on Faraday's law of induction, when we introduce a magnet in a solenoid an induced current is produced that generates a voltage that is given by

           E = - N d \phi_{B} / dt

where \phi_{B} = B. A

The bold are vectors

Therefore, when applying this formula to our case, the induction lines of the magnetic field increase as we approach the solenoid, as the South pole approaches the lines are in the direction of the magnet, therefore the normal to the solenoid that has an outgoing direction and the magnetic field has 180º between them and the cos 180 = -1; consequently the deflection must be towards the negative side of the voltage.

4) From the Faraday equation we can see that the inductive electromotive force depends

* The magnitude of B that changes over time

* The area of ​​the loop that changes over time

* The angle between B and the area that changes over time

* A combination of the above

With this analysis we will review the different alternatives given

a) False. It takes a temporary change and an absolute value of B

b) True. As the speed decreases, the change in B decreases, that is, dB / dt decreases

c) True. The current is induced in each turn, if there is a smaller number the total current will be smaller

d) False. A temporary change of area is needed, in addition to increasing the area the current increases

We can see that the correct statements are: b and c

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A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the
Elis [28]

Answer:

a)R = 171μΩ

b)E = 1.7 *10^{-4} V/m

c)R_{2} = 1.16 *10^{-4}Ω

here * stand for multiplication

Explanation:

length of cylinder = 1.5 m

radius of cylinder  =  1.1 cm

resistivity depends on the distance x from the left

p(x)=a+bx^2 ............(i)

using equation

R = \frac{pl}{a}

let dR is the resistance of thickness dx

dR =\frac{p(x)dx}{a}

where p(x) is resistivity  l is length

a is area

\int\limits^R_0 {dR}  =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx  \\.........................(2)

after integration

R = \frac{[aL+\frac{bL^3}{3}] }{\pi  r^2}  ...............(3)

it is given p(0) = a = 2.25 * 10 ^{-8}Ωm

p(L) = a + b(L)^2  = 8.5 * 10 ^{-8} Ωm

8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\

(here * stand for multiplication )

on solving we get

b = 2.78* 10^{-8} Ωm

put each value of a  and b and r value in equation 3rd we get

R = \frac{[aL+\frac{bL^3}{3}] }{\pi  r^2}

R = 1.71 * 10^{-4}Ω

R = 171μΩ

FOR (b)

for mid point  x = L/2

E = p(x)L

for x = L/2

p(L/2) = a+b(L/2)^2

for given current  I = 1.75 A

so electric field

 

E = \frac{[a+b(L/2)^2]I }{\pi  r^2}

by substitute the values

we get;

E = 1.7 *10^{-4} V/m

(here * stand for multiplication )

c ).

75 cm means length will be half

 that is   x =  L/2

integrate  the second equation with upper limit  L/2  

Let resistance is R_{1}

so after integration we get

R_{1}  =  \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}

substitute the value of a , b and L we get

R_{1} = 5.47 * 10 ^{-5}Ω

for second half resistance

R_{2} =  R- R_{1}

R_{2}  = 1.7 *10^{-4} -5.47 *10^{-5}

R_{2} = 1.16 *10^{-4}Ω

(here * stand for multiplication )

5 0
2 years ago
An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t
marissa [1.9K]
The energy transferred to the spring is given by:
U= \frac{1}{2}kx^2
where 
k is the spring constant
x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
k=52.9 N/cm = 5290 N/m
x=40.0 cm=0.4 m

so now we can use these data inside the equation ,to find the energy transferred to the spring:
U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J
4 0
2 years ago
Denise is conducting a physics experiment to measure the acceleration of a falling object when it slows down and comes to a stop
iren [92.7K]
We need a and we have m and F . Now a = f÷m so therefore a = 4,9 ÷ 0,5 which is 0,98 metres per cubic second
4 0
2 years ago
Read 2 more answers
Jeremy accidently dropped his toy stuffed animal from the balcony of his apartment on the fourth floor. The toy hit the ground a
LekaFEV [45]
According to the second law of motion, force is the product of mass times acceleration. If we were to solve for the force, we would need the acceleration and the mass. 

Acceleration was not given in the problem, but we can solve for it by using the velocity and the time because by definition, acceleration is the change in velocity over time. 

a =  \frac{vf-vi}{t}

The initial velocity is 16.0m/s because it is the first recorded time. The final is 0 m/s or the last recorded velocity. The time we will use is 2.0 s because it is the time the change in velocity occurred. 

a = \frac{vf-vi}{t}
a = \frac{0m/s-16.0m/s}{2.0s}
a = \frac{-16.0m/s}{2.0s}
a=-8.0m/ s^{2}

Now that we have acceleration, we can now solve for the Force. Again Force is the product of mass and acceleration. 

F=ma
F=(0.25kg)(-8.0m/ s^{2} )
F=-2kg.m/ s^{2} ) or -2N

The force is -2N. Now if you are looking for the magnitude of the force then you can just put 2N.

The negative sign indicates the direction of the force, if it is negative this means that the force applied was in the opposite direction.  

4 0
2 years ago
A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin
KengaRu [80]

a) 219.8 rad/s

b) 20.0 rad/s^2

c) 2.9 m/s^2

d) 7005 m/s^2

e) Towards the axis of rotation

f) 0 m/s^2

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

\omega=\frac{\theta}{t}

where

\theta is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

\omega_{max}=35 rev/s

The angle covered during 1 revolution is

\theta=2\pi rad

Therefore, the maximum angular velocity is:

\omega_{max}=35 \cdot 2\pi = 219.8 rad/s

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

\alpha = \frac{\Delta \omega}{t}

where

\Delta \omega is the change in angular velocity

t is the time elapsed

Here we have:

\omega_0 = 0 is the initial angular velocity

\omega_{max}=219.8 rad/s is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

\alpha = \frac{219.8-0}{11}=20.0 rad/s^2

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the radius of the circle

Here we have

\alpha =20.0 rad/s^2

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

a_t=(20.0)(0.145)=2.9 m/s^2

d)

The magnitude of the radial (centripetal) acceleration is given by

a_c = \omega^2 r

where

\omega is the angular velocity

r is the radius of the circle

Here we have:

\omega_{max}=219.8 rad/s is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

a_c=(219.8)^2(0.145)=7005 m/s^2

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

\alpha=0, since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

a_t=(0)(0.145)=0 m/s^2

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

v=\omega r

where

v is the linear speed

\omega is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

\omega=219.8 rad/s

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

v=(219.8)(0.145)=31.9 m/s

7 0
2 years ago
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