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sergij07 [2.7K]

3 years ago

15

What comes before Action-reaction?

1 answer:

Anarel [89]3 years ago

4 0

These two forces are called action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects.

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A physics professor is pushed up a ramp inclined upward at 30.0° above the horizontal as she sits in her desk chair, which slide

11111nata11111 [884]

Answer:

V = 3.17 m/s

Explanation:

Given

Mass of the professor m = 85.0 kg

Angle of the ramp θ = 30.0°

Length travelled L = 2.50 m

Force applied F = 600 N

Initial Speed u = 2.00 m/s

Solution

Work = Change in kinetic energy

$F_{net}d = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}\\\frac{2F_{net}d }{m} = v^{2} -u^{2}\\ v^{2} =\frac{2F_{net}d }{m} +u^{2}\\ v^{2} =\frac{2(600cos30 - 85\times 9.8 \times sin30) \times 2.5 }{85} +2.00^{2}\\ v^{2} = 10.066\\v = 3..17m/s$

7 0

3 years ago

Light traveling though air strikes the surface of the four different materials shown. Which material reflects light but does not

dangina [55]

The spoons .................

8 0

3 years ago

If it takes Ashley 10 seconds to run from the batter's box to first base at an average speed

abruzzese [7]

Answer:

65 meters

Explanation:

Time = 10 seconds

Speed = 6.5 m/s

Distance = ?

$Distance = speed\times time\\\\d = 6.5 \times 10 \\d = 65 m$

7 0

3 years ago

How is a light element like helium is formed in stars and how is a heavier element like gold formed by stars?

aliina [53]

Answer:

C

Explanation:

Because *censored*. Then the element *censored* your answer

7 0

2 years ago

A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f

meriva

Answer:

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

$I=\dfrac{2}{5}mr^2$

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for rolling without slipping v= ωr

Form energy conservation

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

v= ωr

$I=\dfrac{2}{5}mr^2$

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2$

$mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2$

$2mg(R-r)=mv^2+\dfrac{2}{5}mv^2$

$2g(R-r)=\dfrac{7}{5}v^2$

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

3 0

3 years ago