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2 years ago

How are velocity and acceleration related

Physics

1 answer:

VashaNatasha [74]2 years ago

8 0

Answer:

Acceleration is the rate of change of velocity. If an object is changing its velocity, its speed, or changing its direction, then it is said to be accelerating.

Explanation:

when you change directions or speed up you are accelerating (or decelerating) and when you change directions or speed its called velocity.

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Why is it that I feel alone

lesya [120]

Answer: quarantine

Explanation:

5 0

3 years ago

Read 2 more answers

The Moon orbits Earth in an average of p = 27.3 days at an average distance of a =384,000 kilometers. Using Newton’s version of

Korvikt [17]

Answer:

The mass of the earth, $M=6.023\times 10^{24}\ kg$

Explanation:

It is given that,

Time taken by the moon to orbit the earth, $T=27.3\ days=2358720\ m$

Distance between moon and the earth, $r=384000\ km=384\times 10^6\ m$

We need to find the mass of the Earth using Kepler's third law of motion as :

$T^2=\dfrac{4\pi^2}{GM}r^3$

$M=\dfrac{4\pi^2r^3}{T^2G}$

$M=\dfrac{4\pi^2\times (384\times 10^6)^3}{(2358720)^2\times 6.67\times 10^{-11}}$

$M=6.023\times 10^{24}\ kg$

So, the mass of the earth is $6.023\times 10^{24}\ kg$ . Hence, this is the required solution.

7 0

3 years ago

Two children are pulling on opposite sides of a blanket. The brother is pulling with a force of 3 N. The sister is pulling with

Citrus2011 [14]

Let F1=Force exerted by the brother (+F1)
F1= Force exerted by the sister (-F2)

Fnet=(+F1) + (-F2)
Fnet= (+F1) + (-F2)
Fnet=F1 - F2
Fnet= (+3N)+(-5N)
Fnet= -2N

-F

towards the sister (-F) (greater force applied)

7 0

3 years ago

Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5

schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0

3 years ago

A 0.5 kg cheeseburger is lobbed at a particularly unhappy customer with a force of 10 N.

aleksley [76]

The acceleration that the cheeseburger experienced is 20 m/s^2.

6 0

3 years ago