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rusak2 [61]
3 years ago
12

A racecar drives at a constant speed down a straight track. The car is in _?_

Physics
1 answer:
vagabundo [1.1K]3 years ago
7 0

Answer:

the answer is

the car is in motion

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FromTheMoon [43]
Yes. There is a substantial number of people ... members of Brainly as well as non-members ... students, puzzle solvers, and just average educated thinkers, who would be able to solve it.
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3 years ago
1. You get hit with a force of 45 N by a ball with a mass of 0.75 kg. How fast was the
dlinn [17]

Answer:

60 {ms}^{ - 2}

Explanation:

As we know,

=》Force = Mass × Acceleration

=》45 N = 0.75 × Acceleration

=》Acceleration = 45 ÷ 0.75

=》Acceleration = 60

hence, the Acceleration of the ball would be. 60 meters per second square

60m {s}^{ - 2}

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3 years ago
What term best describes the difference in colors of the birds below?
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4 years ago
Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary. The mass of the yo
patriot [66]

To solve this problem it is necessary to use the conservation equations of both kinetic, rotational and potential energy.

By definition we know that

KE + KR = PE

Where,

KE =Kinetic Energy

KR = Rotational Kinetic Energy

PE = Potential Energy

In this way

\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh

Where,

m = mass

v= Velocity

I = Moment of Inertia

\omega = Angular velocity

g = Gravity

h = Height

We know as well that \omega = v/r for velocity (v) and Radius (r)

Therefore replacing we have

\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh

[tex]h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2[/tex]

h= \frac{1}{2}v^2 ( \frac{1}{g} +\frac{I}{mg}\frac{1}{r^2} )

h= \frac{1}{2}0.75^2 ( \frac{1}{9.8} +\frac{2.9*10^{-5}}{(0.056)(9.8)}\frac{1}{(0.0064)^2} )

h = 0.3915m

Therefore the height must be 0.3915 for the yo-yo fall has a linear speed of 0.75m/s

6 0
3 years ago
Sledding down a hill, you are traveling at 10 m/s when you reach the bottom. You (inertia 80 kg ) then move across horizontal sn
RSB [31]

Answer:

V = 0.32 m /s

Explanation:

Momentum of the man on the sledge along with sledge

= (80+8) x 10 = 880 kg. m/s When the man jumps off the sledge , the velocity of the sledge will remain intact at 10 m/s

When the sledge hits the boulder and bounces back the momentum of the sledge - boulder system will remain conserved .

change in momentum of sledge = 8 x (10 +6 )

= 128  kg m/s

Change in the momentum of boulder

= 400 V -0

= 400V

400V = 128

V = 0.32 m /s

5 0
3 years ago
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