All categories

3 years ago

1. Spring force is proportional to stretch distance, F=kd. Do you expect doubling the stretch distance double the force?

Physics

1 answer:

Solnce55 [7]3 years ago

3 0

Answer:

Yes

Explanation:

The spring force is given as:

F = kd

F is the spring force

K is the spring constant

d is the magnitude of the stretch

Since k is a constant, therefore, doubling the stretch distance will double the force.

Both stretch distance and force applied can be said to be directly proportional to one another.

You might be interested in

Will xenon react with nitrogen and why

Andreyy89

<span>No, xenon wont react with nitrogen.

This is because xenon is a noble gas and noble gases on</span>ly react with other elements under very unusual circumstances.<span>

I hope this has helped you.</span>

3 0

3 years ago

Which two statements about kinetic energy are true

gtnhenbr [62]

Answer:

A, C

Explanation:

It can be found in objects like a swing. It creates kinetic energy when it is coming back or going away, while moving.

8 0

2 years ago

Identify and explain the role of all the parts of the female reproductive system.

vekshin1

Answer:

Explanation:

Hopefully this helps you

5 0

3 years ago

A ____________ is the measure of how far the particles in the medium move away from their normal rest position.

Bezzdna [24]

Explanation:

Amplitude is your answer..

Mate..

hope you like it

5 0

3 years ago

The mass of the Sun is 2x1030 kg, and the mass of Mars is 6.4x1023 kg. The distance from the Sun to Mars is 2.3X1011 m. Calculat

user100 [1]

Answer: force from sun to mars = $1.62*10^{19} N$ , force from mars to sun = $-1.62*10^{19} N$

Explanation: The force of attraction or repulsion between 2 object ( in this case the sun and mars ) in space is given by newton's law of gravitation which is given mathematically below as

$F =\frac{Gm_{s} m_{m} }{r^{2} }$

where $G = 6.67 * 10^{-11} , m_{s} = 2*10^{30}kg m_{m} = 6.4 *10^{23} kg\\r = 2.3 *10^{11} m$

G = gravitational constant, $m_{m}$ = mass of mars, $m_{s}$ = mass of sun

$F = \frac{6.64 * 10 ^{-11}* 2 * 10^{30} *6.4*10^{23}} {2.3*10^{11} * 2.3* 10^{11} } =\frac{85.376 *10^{42} }{5.25 *10^{22} } \\\\F = 1.62 * 10 ^{19} N\\$

the mars also exerts the same magnitude of force above on the sun but in the opposite direction, thus the force mars exerts on the sun is $-1.62*10^{19} N$

8 0

3 years ago