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ELEN [110]
3 years ago
11

6. Draw conclusions: Newton’s first law states that an object in motion will travel at a constant velocity unless acted upon by

an unbalanced force. How do these experiments show this?
Physics
1 answer:
elixir [45]3 years ago
6 0

Answer:

The experiments are not shown, so I will answer in a general way.

By the first Newton's law, an object will only change it's velocity if there is a net force different than zero acting on the object.

Then in the experiments (depending on the experiment), you can see different things.

If an object is not moving and you apply a force in it, the object will move.

If an object is moving and you apply a force in the opposite direction of it's motion, the motion will: decrease the speed, stop at all, or move in the opposite direction. Depending on the force that you apply.

An excellent experiment (but hard to do) is dropping an object from a really high place.

The gravitational force will pull down the object and the object will start to increase it's velocity.

But there is the air resistance, that opposes to this motion and increases with the speed of the object.

Then there is a given speed such that the air resistance force will be equal to the gravitational force, then we have a balanced force (the net force is zero) which means that the object will keep falling at a constant velocity.

You might be interested in
The kinetic energy and the potential energy of the cannonball is constantly ________ as it travels through the air.
enot [183]

Answer:

C. Constant

Explanation:

The total energy of the cannonball remains constant as it travels through the air.

4 0
3 years ago
A cord is wrapped around the rim of a solid uniform wheel 0.300 m in radius and of mass 8.00 kg. A steady horizontal pull of 34.
disa [49]

Answer:

A. α = 94.4 rad/s

B. a = 28.32 m/s

C. N = 34N

D. α = 94.4 rad/s

    a = 28.32 m/s

     N =  44.4  N

Explanation:

part A:

using:

∑T = Iα

where T is the torque, I is the moment of inertia and α is the angular momentum.

firt we will find the moment of inertia I as:

I = \frac{1}{2}MR^2

Where M is the mass and R is the radius of the wheel, then:

I = \frac{1}{2}(8 kg)(0.3 m)^2

I = 0.36 kg*m^2

Replacing on the initial equation and solving for α, we get::

∑T = Iα

Fr = Iα

34 N =  0.36α

α = 94.4 rad/s

part B

we need to use this equation :

a = αr

where a is the aceleration of the cord that has already been pulled off and r is the radius of the wheel, so replacing values, we get:

a = (94.4)(0.3 m)

a = 28.32 m/s

part C

Using the laws of newton, we know that:

N = T

where N is the force that the axle exerts on the wheel part and T is the tension of the cord

so:

N = 34N

part D

The anly answer that change is the answer of the part D, so, aplying laws of newton, it would be:

-Mg + N +T = 0

Then, solving for N, we get:

N = -T+Mg

N = -34 + (8 kg)(9.8)

N =  44.4  N

6 0
3 years ago
A 5 kg box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force of 17 new
Juliette [100K]

Answer:

The acceleration of the box is 2.

Explanation:

According to Newton's second law of motion, the acceleration of any object will be directly proportional to the net unbalanced force acting on the object and inversely proportional to the mass of the object.

Net force = Mass × Acceleration

So Acceleration = \frac{Net force}{Mass}

Since in this case, the box is experiencing a force from east of magnitude 27 N and resisting force of about 17 N from west. So the net force will be the difference of acting and reacting force.

Net force = 27-17 = 10 N.

Thus, Acceleration = \frac{10 N}{5 kg}

So 2 m/s^{2} is the acceleration of the box. Thus the magnitude of acceleration of the box is 2.

5 0
4 years ago
A ball of mass 0.075 kg is fired horizontally into a ballistic pendulum. The pendulum mass is 0.350 kg. The ball is caught in th
Tatiana [17]
1) In the initial situation, the total mechanical energy of the system is given only by the kinetic energy of the ball that is moving with speed v:
E_i =K= \frac{1}{2}m_b v^2
where m_b = 0.075 kg is the mass of the ball.

In the final situation, where the system (ball+pendulum) rises a vertical distance of h=0.145 m, the system is stationary (v=0) so the total mechanical energy of the system is the gravitational potential energy:
E_f = U = (m_b+m_p)gh
where m_p = 0.350 kg is the mass of the pendulum.

For the law of conservation of energy, E_i=E_f , so we can find the initial speed v of the ball:
\frac{1}{2}m_bv^2 = (m_b+m_p)gh
v= \sqrt{ 2 \frac{m_b+m_p}{m_b}gh } =4.0 m/s

2) The kinetic energy lost in the collision is the initial kinetic energy of the ball:
K= \frac{1}{2}m_bv^2= \frac{1}{2}(0.075 kg)(4.0 m/s)^2=0.6 J
8 0
3 years ago
A 3kg box has 45J of gravitational potential energy, how high is it off the ground?
frez [133]

Answer:

1.53m

Explanation:

Given parameters:

Mass of box  = 3kg

Gravitational potential energy  = 45J

Unknown

Height of the box  = ?

Solution:

To solve this problem;

 Gravitational potential energy = mgh

m is the mass

g is the acceleration due to gravity

h is the height

            45  = 3 x 9.8 x h

            h  = 1.53m

7 0
3 years ago
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