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Nataly [62]
3 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]3 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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Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
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Answer:

80 ft/s

Explanation:

Given:

Δy = 100 ft

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a = 32.2 ft/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 ft/s)² + 2 (32.2 ft/s²) (100 ft)

v = 80.2 ft/s

Rounded, the speed when it reaches the ground is 80 ft/s.

4 0
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Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperatu
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Complete Question

Use Stefan's law to find the intensity of the cosmic background radiation emitted by the fireball of the Big Bang at a temperature of 2.81 K. Remember that Stefan's Law gives the Power (Watts) and Intensity is Power per unit Area (W/m2).

Answer:

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Explanation:

From the question we are told that

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Now  According to Stefan's law

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Where  \sigma is the Stefan Boltzmann constant with value  \sigma  =  5.67*10^{-8} m^2 \cdot kg \cdot s^{-2} K^{-1}

  Now the intensity of the cosmic background radiation emitted according to the unit from the question is mathematically evaluated as

        I  =  \frac{P}{A}

=>      I  =  \frac{\sigma *  A  * T^4}{A}

=>      I  =  \sigma  *  T^4

substituting values

      I  = 5.67 *10^{-8}  *  (2.81)^4

       I  = 3.535 *10^{-6} \  W/m^2

       

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