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hammer [34]
1 year ago
9

Calculate the force needed to give a car of mass 800 kg an acceleration of 2.0 ms−2. plss quick

Physics
1 answer:
erik [133]1 year ago
4 0

The force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.

<h3>How to calculate force?</h3>

The force needed to push an object can be calculated by multiplying the mass of the object by its acceleration as follows:

Force = mass × acceleration

According to this question, a car of mass 800 kg has an acceleration of 2.0 ms−². The force is calculated as follows:

Force = 800kg × 2m/s²

Force = 1600N

Therefore, the force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.

Learn more about force at: brainly.com/question/13191643

#SPJ1

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Two particles are located on the x axis. particle 1 has a mass m and is at the origin. particle 2 has a mass 2m and is at x = +l
wlad13 [49]

The solution would be like this for this specific problem:

<span>
The force on m is:</span>

<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 1

The force on 2m is:</span>

<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 2

From (1), you’ll get M = 2mx^2 / L^2 and from (2) you get M = m(L - x)^2 / L^2 

Since the Ms are the same, then 

2mx^2 / L^2 = m(L - x)^2 / L^2 

2x^2 = (L - x)^2 

xsqrt2 = L - x 

x(1 + sqrt2) = L 

x = L / (sqrt2 + 1) From here, we rationalize. 

x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1) 

x = L(sqrt2 - 1) / (2 - 1) 


x = L(sqrt2 - 1) </span>

 

= 0.414L

 

<span>Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.</span>

8 0
3 years ago
A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c
mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

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