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4 years ago

What acceleration results from exerting a 125N force on a 0.65kg ball?

1 answer:

5 0

first you do your pyramid f is on top and ma is
on bottom were m=mass and a=acceleration
were in this case you do f÷m=force÷mass so again in this case 125÷0.65=192.3 to be more accurate 192.3076923077

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Think about what subsequently happens to the ketchup, which is initially at rest, and use newton's first law to explain why this

Vika [28.1K]

Newton's first law of ketchup is if the bottle of ketchup is smacked upward, the contents (the ketchup) will tend to remain in place, and it will be closer to the opening in the bottle.

<h3>Further explanation </h3>

If we are trying to get ketchup out of the bottle, the best way to do it is to turn the bottle upside down and give the bottle a upward smack which force the bottle rapidly upward.

Newton first law said that every object tries to continue in its state of rest or uniform motion unless an external force acted upon it (In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force). Newton first law describe the relationship between a body and the forces acting upon it, and its motion in response to those forces.

Inertia explained by Sir Isaac Newton in his first law of motion (Newton law of motion). The law states that an object at rest stays at rest and an object continues its state of motion until an external force acts on it. The examples are one's body movement to the side when a car makes a sharp turn

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<h3>Answer details</h3>

Grade: 9

Subject: physics

Chapter: newton's first law

Keywords: newton's first law

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3 years ago

Read 2 more answers

A 6,000 kg train car is moving to the right at 10 m/s and connects to a 4,000-kg train car that wasn't moving. What is the veloc

vredina [299]

1) The final velocity of the two trains is 6 m/s to the right

2) It is an inelastic collision

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two trains must be conserved before and after the collision.

Therefore we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 6,000 kg$ is the mass of the first train

$u_1 = 10 m/s$ is the initial velocity of the first train

$m_2 = 4,000 kg$ is the mass of the second train

$u_2 = 0$ is the initial velocity of the second train (initially at rest)

$v$ is the final combined velocity of the two trains

Solving the equation for v, we find the final velocity of the two trains:

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(6000)(10)+0}{6000+4000}=6 m/s$

There are two types of collisions:

Elastic collision: in an elastic collision, both the total momentum and the total kinetic energy of the system are conserved
Inelastic collision: in an inelastic collision, only the total momentum is conserved, while the total kinetic energy is not (part of it is converted into thermal energy due to internal frictions)

To verify what type of collision is this, we can compare the total kinetic energy before and after the collision:

Before:

$K=\frac{1}{2}m_1 u_1^2 = \frac{1}{2}(6000)(10)^2=300,000 J$

After:

$K=\frac{1}{2}(m_1 +m_2)v^2 = \frac{1}{2}(6000+4000)(6)^2=180,000 J$

As we can see, the kinetic energy is not conserved, so this is an inelastic collision.

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3 years ago

A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo

Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

$W = \Delta U = U_2 - U_1$

where the potential energy is defined as follows

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Let's first calculate the distance 'r' for both positions.

$r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m$

Now, we can calculate the potential energies for both positions.

$U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J$