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DIA [1.3K]
3 years ago
5

If the net force acting on an object doubles, how will the object's acceleration be affected?

Physics
1 answer:
Savatey [412]3 years ago
7 0

Answer:

It goes slower

Explanation:

If the net force is doubled it gains weight

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A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.
vekshin1

Answer:

A)  If the paintball stops completely the magnitude of the change in the paintball’s momentum is  p=0.273kg*m/s

B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, the change in the paintball’s momentum is  p=0.546kg*m/s

C) A paintball bouncing off your skin in the opposite direction with the same speed hurts more than a paintball exploding upon your skin because of the strength exerted is twice than if it explodes.

Explanation:

Hi

A) We use the formula of momentum p=mv, so we have p=0.0032kg*85.3m/s=0.273kg*m/s

B) We use the same formula above, then due we have a change of direction at the same speed, therefore the change in the momentum is the double so

p=2*0.0032kg*85.3m/s=0.546kg*m/s.

C) The average strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. therefore

F=\frac{\Delta p}{\Delta t}, as we don't have any data about the impact time but we know momentum is twice, time does no matter and strength is twice too.

4 0
4 years ago
PLEASE HELP ME VERY IMPORTANT
saul85 [17]

Answer: let's see, so it's a 1/4 chance of getting it right, hope your odds are good, have a wonderful day

Explanation:

:D

6 0
3 years ago
A 2.43ug particle moves at 1.97 x 108 m/s. What is its momentum?
Ira Lisetskai [31]

Answer:

0.48 kgm/s

Explanation:

m = mass of the particle = 2.43 μg = 2.43 x 10⁻⁶ x 10⁻³ kg = 2.43 x 10⁻⁹ kg

v  = velocity of the particle = 1.97 x 10⁸ m/s

p = momentum of the particle

momentum of the particle is given as

p = m v

inserting the values

p = (2.43\times 10^{-9})(1.97\times 10^{8})

p = 0.48 kgm/s

4 0
4 years ago
What is the difference between kinetic and potential energy and how do they work?
Iteru [2.4K]

To explain, I will use the equations for kinetic and potential energy:

PE = mgh\\KE = \frac{1}{2}mv^{2}

<h3>Potential energy </h3>

Potential energy is the potential an object has to move due to gravity.  An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>.  If gravity = 0 or height = 0, there is no potential energy.  Example:

An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2).  What is the object's gravitational potential energy?  <u>(answer: 5*5*9.8 = 245 J</u>)

(gravitational potential energy is potential energy)

<h3>Kinetic energy</h3>

Kinetic energy is the energy of an object has while in motion.  An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary).  An example:

An object of 5 kg is moving at 5 m/s.  What is the object's kinetic energy?  (<u>answer: 5*5 = 25 J</u>)

<h3>Kinetic and Potential Energy</h3>

Sometimes, an object can have both kinetic and potential energy.  If an object is moving (kinetic energy) and is above the ground (potential), it will have both.  To find the total (mechanical) energy, you can add the kinetic and potential energies together.  An example:

An object of 5 kg is moving on a 5 meter table at 10 m/s.  What is the objects mechanical (total) energy?  (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)

7 0
3 years ago
A muscle inserts 1.5 cm from the joint axis and exerts 300 N of force at an angle of pull of 60 degrees. How much torque is prod
Amanda [17]

Answer:

torque = 3.897 N-m

Explanation:

given data

force = 300 N

angle = 60 degree

distance = 15 cm

to find out

torque

solution

we will apply here torque formula that is given below

torque = force × sinθ × distance    ...................1

put here all these value in equation 1

we get torque

torque  = force × sinθ × distance

torque  = 300 × sin60 × 1.5 ×10^{-2}

torque  = 300 × 0.8660 × 1.5 ×10^{-2}

torque  = 259.80 × 1.5 ×10^{-2}

torque  = 389.711 ×10^{-2}

torque = 3.897 N-m

8 0
3 years ago
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