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Anna11 [10]
3 years ago
12

Select an example of a vertical motion with a positive velocity and a negative acceleration. Select the correct example and expl

anation.a)A tossed ball on its way down. As the ball moves down it is slowing down, hence the acceleration is opposite the bvelocity (negative).b) A tossed ball on its way upward. As the ball moves upward it is slowing down, hence the acceleration is pointing in the same direction as the velocity (negative).c) A tossed ball on its way upward. As the ball moves upward it is slowing down, hence the acceleration is opposite the velocity (negative).d) A tossed ball on its way down. As the ball moves down it is slowing down, hence the acceleration is pointing in the same direction as the velocity (negative).
Physics
1 answer:
miss Akunina [59]3 years ago
8 0

Answer:

c)

Explanation:

When a ball is tossed upward, it rises due to the initial velocity which points upward (so it is positive if we choose the upward direction as positive), but once released, the only force (neglecting air resistance) acting on the ball is gravity.

This force is due to the mass of the Earth, and produces an acceleration in the same direction than the force, which is always downward, so in this case is negative.

As it has opposite direction to the velocity, causes the ball to slow down until it reach to a point when momentarily comes to an stop, and then changes direction falling with negative velocity, so it speeds up as it has the same sign than acceleration, that it is always present.

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A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg.
Svetllana [295]

Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' =  19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.

5 0
3 years ago
Question 2 (1 point)
finlep [7]
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4 0
3 years ago
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In a collision, a 25.0 kg mass moving at 3.0 m/s transfers all of its momentum to a 5.0 kg mass.
nadezda [96]

Answer:

Explanation:

The momentum of the 25 kg mass is

p=mv

p=25kg*3m/s= 75kg*m/s

If this whole momentum of the object is transferred to the 5.0 kg object then according to the law of conservation of momentum, the momentum of the 25.0 kg object must be transferred to the 5.0 kg object:

75kg*m/s = 5.0kg*v

v=\dfrac{75}{5}

\boxed{v=15m/s}

8 0
3 years ago
An object of mass 4kg is moving along a horizontal plane. If the coefficient of kinetic friction is 0.2 find the friction force
lana66690 [7]

Answer:

The friction force acting on the object is 7.84 N

Explanation:

Given;

mass of object, m = 4 kg

coefficient of kinetic friction, μk = 0.2

The friction force acting on the object is calculated as;

F = μkN

F = μkmg

where;

F is the frictional force

m is the mass of the object

g is the acceleration due to gravity

F = 0.2 x 4 x 9.8

F = 7.84 N

Therefore, the friction force acting on the object is 7.84 N

5 0
3 years ago
A train moving at a constant speed on a surface inclined upward at 10.0° with the horizontal travels a distance of 400 meters in
Amiraneli [1.4K]
If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:

Vv=80*sin(10)=80*0.1736=13.888 m/s. 

So the vertical component of the velocity of the train is Vv=13.888 m/s.
7 0
3 years ago
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