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3 years ago

True or False: The arrows on a motion map should point in the directions of motion.

Physics

1 answer:

slavikrds [6]3 years ago

3 0

Answer:

true,true,false

Explanation:

its false because if it is equal it would show an arrow pointing left and a 20 and the same for the right

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A professional golfer hits a golf ball of mass 46 g with her 5-iron, and the ball first strikes the ground 155 m away. The ball

RideAnS [48]

Answer:

$C=2.32\times 10^{-4}\ Ns^2/m^2$

Explanation:

It is given that,

Mass of the golf ball, m = 46 g = 0.046 kg

Terminal speed of the ball, v = 44 m/s

The drag force, $F_r=Cv^2$

Where, C is the drag coefficient. At terminal speed, the weight of the ball is balanced by the drag force.

$Cv^2=mg$

$C=\dfrac{mg}{v^2}$

$C=\dfrac{0.046\times 9.8}{(44)^2}$

$C=2.32\times 10^{-4}\ Ns^2/m^2$

Hence, this is the required solution.

4 0

3 years ago

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago

A race car drives one lap around a race track that is 500 meters in length.

pychu [463]

Explanation:

Check out the picture I drew for a minute before reading this...

B. Distance [the red line] is a scalar quantity reflecting how far an object has traveled. Displacement [the green line] is a vector quantity reflecting how far an object has moved from a point. The key difference is that distance can be any sort of path while displacement is always a vector (or a straight line) between a starting point and a finishing point. Sometimes distance and displacement are equal to one another. Sometimes you have a distance traveled, but zero displacement overall; which is what's going on in your question.

A. The distance that the racecar traveled is indeed 500m. But at the end of the lap, it is right back where it started. So overall, it has been displaced 0m.

3 0

3 years ago

Water is stored in 5 places on earth.Name each place an dnote the phrase that water is in liquid solid,gas

podryga [215]

Atmosphere - gas
sea and oceans - liquid and solid
land -liquid and solid
living things and plants - liquid

6 0

3 years ago

What is the acceleration of a 10kg pushed by a 5n force

Semenov [28]

Force equals mass times acceleration. Or:
F=ma
Plug it in:
5=10a
5/10=(10a)/10
.5m/s²=a

8 0

3 years ago