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zysi [14]
4 years ago
15

What two factors affect the rate of acceleration in an object

Physics
2 answers:
Leto [7]4 years ago
8 0
Newton's second law states that F=ma so that means force =mass×acceleration force and mass
jonny [76]4 years ago
4 0
Force and mass affect acceleration
F=ma
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What is kinetic energy and what is net force formula
Over [174]

Answer:

Here's the equation for net force: F = ma. The work done on the plane, which becomes its kinetic energy, equals the following: Net force F equals mass times acceleration. Assume that you're pushing in the same direction that the plane is going; in this case, cos 0 degrees = 1, so.

Explanation:

In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes

Hope this help also looking it up helps ;)

6 0
3 years ago
With Christmas lights wired in series, if one light goes out, what happens to the rest?
daser333 [38]

Answer:

The ones that are after the light that went out are also out.

Explanation:

3 0
3 years ago
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Suppose that at room temperature, a certain aluminum bar is 1.0000 m long. The bar gets longer when its temperature is raised. T
Deffense [45]

Answer:

0.00034 m

Explanation:

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4 0
4 years ago
Kelvin, Celsius, and Fahrenheit are three types of<br> scales
Contact [7]

Answer:

yes, they are.

Explanation:

8 0
3 years ago
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A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

Speed, v_{1} = 5 m/s

m_{2} = 1 kg

v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
4 years ago
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