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andrew-mc [135]
3 years ago
14

How much force does it take to bring a 1,050 N car from rest to a velocity of 42 m/s in 13 seconds?

Physics
1 answer:
frozen [14]3 years ago
5 0

Answer:

F = 339.23 N

Explanation:

Weight of a car, W = 1050 N

Initial velocity, u = 0

Final velocity, v = 42 m/s

Time, t = 13 s

The weight of an object is given by :

W = mg

g is the acceleration due to gravity

m=\dfrac{W}{g}\\\\m=\dfrac{1050}{10}\\\\m=105\ kg

The force acting on car is :

F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{105\times (42-0)}{13}\\\\F=339.23\ N

So, the force acting on the car is 339.23 N.

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What is the eccentricity of an ellipse with a foci distance of 50,000,000 km and
inysia [295]

Answer:

25,000,000 Km ;)

Explanation:

5 0
3 years ago
(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

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5 0
2 years ago
Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
Which is likely to be more common in our Galaxy: white dwarfs or black holes? Why?
Nookie1986 [14]

Answer:

White dwarfs are likely to be much more common. The number of stars decreases with increasing mass, and only the most massive stars are likely to complete their lives as black holes. There are many more stars of the masses appropriate for evolution to a white dwarf.

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2 years ago
Why does an energy transfer not always result in phase change?
mash [69]

Answer:

These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. "If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance

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3 years ago
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