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yan [13]
1 year ago
5

how much work did the movers do (horizontally) pushing a 41.0- kg crate 10.3 m across a rough floor without acceleration, if the

effective coefficient of friction was 0.60?
Physics
1 answer:
11111nata11111 [884]1 year ago
7 0

W=2485.65 J ,work did the movers do friction force a 41.0- kg crate 10.3 m across a rough floor without acceleration

<h3>What is a basic friction force?</h3>

Two surfaces that come into contact and slide against one another produce a force known as frictional force. Several aspects that influence the frictional force include: The surface texture as well as the amount of force attracting them together have the most effects on these forces.

<h3>What outcomes does friction produce?</h3>

It generates heat, which is useful for warming our bodies or specific areas of any object. Power is also lost as a result. It makes noise during every operation. We are able to walk, run, dance, etc. due to friction.

To know  more about friction force visit:

brainly.com/question/1389727

#SPJ4

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Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t
LiRa [457]

Answer:

The minimum coefficient of friction is 0.544

Solution:

As per the question:

Radius of the curve, R = 48 m

Speed of the car, v = 16 m/s

To calculate the minimum coefficient of static friction:

The centrifugal force on the box is in the outward direction and is given by:

F_{c} = \frac{mv^{2}}{R}  

f_{s} = \mu_{s}mg

where

\mu_{s} = coefficient of static friction

The net force on the box is zero, since, the box is stationary and is given by:

F_{net} = f_{s} - F_{c}  

0 = f_{s} - F_{c}  

\mu_{s}mg = \frac{mv^{2}}{R}  

\mu_{s} = \frac{v^{2}}{gR}  

\mu_{s} = \frac{16^{2}}{9.8\times 48} = 0.544  

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State one difference between solar cells and solar heating systems?
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A thin beam of light of wavelength 625 nm goes through a thin slit and falls on a screen 3.00 m past the slit. You observe that
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Answer:

a = 2.275 10⁻⁴ m

Explanation:

This is a diffraction problem that is described by the equation

         a sin θ = m λ

         

The first dark minimum occurs for m = 1

         a = λ  / sin θ

The angle can be found by trigonometry,

       tan θ = y / x

       θ = tan⁻¹ y / x

Let's reduce the magnitudes to the SI system

        y = 8.24 mm = 8.24 10⁻³ m

        λ  = 625 nm = 625 10⁻⁹ m

       θ = tan⁻¹ 8.24 10⁻³ / 3.00

       θ = 0.002747 rad

We calculate

       a = 625 10⁻⁹ / sin 0.002747

       a = 2.275 10⁻⁴ m

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