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Nataly [62]
2 years ago
9

You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a

string. The load is not swinging, and the string is observed to make a constant angle of 45∘ with the horizontal. No forces other than tension and gravity are acting on the load. Which of the following statements are correct?
Check all that apply.
a. The train is an inertial frame of reference
b. The train is not an inertial frame of reference
c. The train may be instantaneously at rest
d. The train may be moving at a constant speed in a straight line
e. The train may be moving at a constant speed in a circle
f. The train must be speeding up
g. The train must be slowing down
h. The train must be accelerating
Physics
1 answer:
allochka39001 [22]2 years ago
6 0

Answer:

b.

c.

e.

h.

Explanation:

If we consider the ball,the tension is not balance by weight then the force acting will be non zero.It means that train car in not inertial frame of reference.And the same time the train car must accelerated with respected to the earth.It is not moving straight path.When train is in rest position but the string will be at the constant angle.

So the following option are correct:

b.

c.

e.

h.

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Degger [83]

Answer and Explanation:

curents i = 2.9 A

           i ' = 4.4 A

the magnitude (in T.m) of the path integral of B.dl around the window frame = μo * current enclosed

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Since from Ampere's law

where μ o = permeability of free space = 4π * 10 ^-7 H / m

plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)

                                 = 1.884 * 10^-6 Tm

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2 years ago
Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get
olga_2 [115]

Answer:

Part a)

\omega = 8.17 rad/s

Part b)

N = 7.74 rev

Part c)

\alpha = 0.69 rad/s^2

Part d)

\alpha = 0.48 rad/s^2

Part e)

t = 9.14 s

Explanation:

Part a)

Angular speed is given as

\omega = 2\pi f

\omega = 2\pi(\frac{78}{60})

\omega = 8.17 rad/s

Part b)

Since turn table is accelerating uniformly

so we will have

\theta = \frac{\omega_f + \omega_i}{2} t

\theta = \frac{8.17 + 0}{2}(11.9)

2N\pi = 48.6

N = 7.74 rev

Part c)

angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{t}

\alpha = \frac{8.17 - 0}{11.9}

\alpha = 0.69 rad/s^2

Part d)

When its angular speed changes to 120 rpm

then we will have

\omega_2 = 2\pi (\frac{120}{60})

\omega_2 = 12.56 rad/s

number of turns revolved is 15 times

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\omega_f^2 - \omega_i^2 = 2 \alpha \theta

12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)

\alpha = 0.48 rad/s^2

Part e)

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\omega_f - \omega_i = \alpha t

12.56 - 8.17 = 0.48 t

t = 9.14 s

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2 years ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

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l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

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2 years ago
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