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Black_prince [1.1K]
2 years ago
14

What is the velocity of the 100 kg cart at point b?

Physics
1 answer:
Gre4nikov [31]2 years ago
8 0
I believe the answer is c
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What is the tangential velocity of a record player which makes 11 revolutions in 20 seconds? Please help
Deffense [45]

Answer:

oh umm it think its TV=11x*20s

Explanation:

4 0
3 years ago
A storm is moving in quickly. The weather station reported that it is traveling west at 40 miles per hour. This information desc
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This information describes the storm's velocity.
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3 years ago
If air resistance can be neglected, how does the acceleration of a ball that has been tossed straight upward compare with its ac
timama [110]
They are the same. If this is all happening on Earth, then the ball's acceleration is 9.8 m/s^2 in either case. That's the acceleration of gravity around here.
6 0
3 years ago
7. A sound wave begins traveling through a thin metal rod at one end with a speed that is 15 times the speed of sound in air. If
san4es73 [151]

Answer:

   L = 44,096 m

Explanation:

The speed of the sound wave is constant therefore we can use the relations of uniform kinematics

             v = x / t

the speed of the wave in the bar is

            v = 15 v or

            v = 15 343

             v = 5145 m / s

The sound at the bar goes the distance

             L = v t

Sound in the air travels the same distance

             L = v_air (t + 0.12)

as the two recognize the same dissonance,

             v t = v_air (t +0.12)

             t (v- v_air) = 0.12 v_air

              t = 0.12 v_air / (v -v_air)

l

et's calculate

             t = 0.12 343 / (5145 - 343)

             t = 8.57 10-3 s

The length of the bar is

              L = 5145 8.57 10-3

              L = 44,096 m

6 0
3 years ago
A 0.30-kg flying pig toy is attached to the ceiling with a string. When the pig’s wings flap, it moves at a constant speed of �
Lesechka [4]

Answer:

58.5 deg

Explanation:

v = mass of the pig toy = 0.30 kg

v = speed of the toy = 2 m/s

r = radius of the circle = 0.25 m

\theta = Angle of the string with the vertical = ?

T = Tension force in the string

Using equilibrium of force in vertical direction

T Cos\theta = mg                                     eq-1

Along the horizontal direction, the component of tension force provides the necessary centripetal force, hence

T Sin\theta = \frac{mv^{2}}{r}                                  eq-2

Dividing eq-2 by eq-1

\frac{T Sin\theta}{T Cos\theta} = \frac{\frac{mv^{2}}{r}}{mg}

tan\theta= \frac{v^{2}}{rg}

tan\theta= \frac{(2)^{2}}{(0.25)(9.8)}

tan\theta = 1.63

\theta = 58.5 deg

6 0
3 years ago
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