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Elena-2011 [213]
4 years ago
14

A crate slides down a ramp at a constant velocity. the ramp has an incline of 28 degrees above the horizontal the crate weighs 2

00 Newton’s
Physics
1 answer:
Sophie [7]4 years ago
5 0

First is the coefficient of static friction between the map

and the crate.

And the second one is at what angle does the crate begin to

slide if the mass is doubled.

1.    To look for the coefficient of static friction:

ΣF = 0

μ x m x g x cos θ = m * g * sin θ

μ = sin θ/ cos θ = tan θ

μ = tan 24 = 0.45

2.   At what angle… the solution is:

The static friction or μ does not depend

on mass, so the angle is still 24.

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Two 20kg spheres are placed with their
Maslowich

Answer:

F = 1.07 x 10⁻⁷ N

Explanation:

The gravitational force of attraction between two objects can be found by the use of Newton's Gravitational Law:

F = \frac{Gm_{1}m_{2}}{r^2}\\\\

where,

F = Gravitational Force of attraction = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = m₂ = mass of spheres = 20 kg

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Therefore,

F = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(20\ kg)(20\ kg)}{(0.5\ m)^2}\\\\

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5 0
3 years ago
The torque exerted by a crowbar on an object increases with increased _______.
Fofino [41]

Answer:

force and leverage distance

Explanation:

the formula for torque if = force x distance

(the distance above is the leverage distance on the crow bar)

therefore if there is an increase in either the torque or the leverage distance, or both, the torque exerted by the crow bar also increases.

for example

  • lets assume a force of 5 n is applied on the crow bar with a leverage distance of 2 m.

        the torque = 5 x 2 = 10 N.m

  • but if the force was increased to 7 N

        torque = 7 x 2 = 12 N.m

from the illustration above, we can see that the torque increased with an increase in force. There would also be an increase in torque if the distance were to be increased.

3 0
3 years ago
A level curve on a country road has a radius of 150 m. What is the maximum speed at which this curve can be safely negotiated on
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The maximum speed of the car is 24.3 m/s

Explanation:

For a car moving along an unbanked turn, the frictional force provides the centripetal force required to keep the car in circular motion. Therefore, we can write:

\mu mg = m\frac{v^2}{r}

where the term on the left is the frictional force while the term on the right is the centripetal force, and where

\mu=0.40 is the coefficient of friction between the tires and the road

m is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

v is the speed of the car

r = 150 m is the radius of the curve

Solving for v, we find the (maximum) speed at which the car can move along the turn:

v=\sqrt{\mu gr}=\sqrt{(0.40)(9.8)(150)}=24.3 m/s

For speed larger than this value, the frictional force is no longer enough to keep the car along the turn.

Learn more about circular motion:

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#LearnwithBrainly

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