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3 years ago

14

A 50kg skater at rest on a frictionless rink throws a 2kg ball, giving the ball a velocity of 10m/s. Which statement describes t

he skaters subsequent motion?

1 answer:

Andreas93 [3]3 years ago

3 0

0.4m/s in the opposite direction of the ball.

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How can u tell matched forces act on objects?

mamaluj [8]

Answer:If an object's speed changes, or if it changes the direction it's moving in,

then there must be forces acting on it. There is no other way for any of

these things to happen.

Once in a while, there may be a group of forces (two or more) acting on

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that happens, the object's speed will remain constant, and ... if the speed

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2 years ago

Student Exploration: Nuclear Decay. Has anyone done a Gizmos lab on this?

LUCKY_DIMON [66]

Answer:NOOPE

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2 years ago

You push your couch a distance of 3.9 m across the living room floor with a horizontal force of 220.0 n. the force of friction i

Mrac [35]

The general formula to calculate the work is:

$W=Fd \cos \theta$

where F is the force, d is the displacement of the couch, and $\theta$ is the angle between the direction of the force and the displacement. Let's apply this formula to the different parts of the problem.

(a) Work done by you: in this case, the force applied is parallel to the displacement of the couch, so $\theta=0^{\circ}$ and $\cos \theta=1$ , therefore the work is just equal to the product between the horizontal force you apply to push the couch and the distance the couch has been moved:

$W=Fd=(220.0 N)(3.9 m)=858 J$

(b) work done by the frictional force: the frictional force has opposite direction to the displacement, therefore $\theta=180^{\circ}$ and $\cos \theta=-1$ . Therefore, we must include a negative sign when we calculate the work done by the frictional force:

$W=-Fd=-(144.0 N)(3.9 m)=-561.6 J$

(c) The work done by gravity is zero. In fact, gravity (which points downwards) is perpendicular to the displacement of the couch (which is horizontal), therefore $\theta=90^{\circ}$ and $\cos \theta=0$ : this means

$W=0$ .

(d) Work done by the net force:

The net force is the difference between the horizontal force applied by you and the frictional force:

$F=220 N-144 N=76 N$

And the net force is in the same direction of the displacement, so $\theta=0^{\circ}$ and $\cos \theta=1$ and the work done is

$W=Fd=(76 N)(3.9 m)=296.4 J$

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3 years ago

I WILL GIVE YOU BRAINLIEST IF YOU ANSWER THOS QUESTION IN THE NEXT 5 MINUTES!!

schepotkina [342]

Answer:

the blue shopping cart.

Explanation:

The blue shopping cart doesnt have to worry about running someone over in the front. The red one does, so it slows down more.

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3 years ago

The minimum stopping distance of a car moving at 20.5 mi/h is 11.6 m. Under the same conditions (so that the maximum braking for

pshichka [43]

Answer:

d = 69 .57 meter

Explanation:

First case

Speed of car ( v ) = 20.5 mi/h = 9.164 M/S

distance ( d ) = 11.6 meter ( m = mass of the car )

Work done = 0.5 m v² = 0.5 * 9.164² * m J = 41.99 m J

Force = ( workdone /distance ) = ( 41.99 m / 11.6 ) = 3.619 m N

Second case

v = 50.2 mi/h = 22.44135 m/s

d = ?

Work done = 0.5 * 22.44² * m J = 251.7768 * m J

Since the braking force remains the same .

3.619 m = ( 251.7768 m / d )

d = 69 .57 meter

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3 years ago