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OlgaM077 [116]
3 years ago
6

A dockworker loading crates on a ship finds that a 21-kg crate, initially at rest on a horizontal surface, requires a 73-N horiz

ontal force to set it in motion. However, after the crate is in motion, a horizontal force of 55 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor. static friction kinetic friction
Physics
1 answer:
Nataliya [291]3 years ago
5 0

1) Static friction coefficient: 0.355

The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.

The maximum static frictional force is given by

F_s = \mu_s mg

where

\mu_s is the static coefficient of friction

m = 21 kg is the mass of the crate

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

The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have

F_s=73 N

Using this information into the previous equation, we can find the coefficient of static friction:

\mu_s = \frac{F}{mg}=\frac{73 N}{(21 kg)(9.8 m/s^2)}=0.355

2) Kinetic friction coefficient: 0.267

Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is

F_k = \mu_k mg (1)

where

\mu_k is the coefficient of kinetic friction

There is a horizontal force of

F = 55 N

pushing the crate. Moreover, the speed of the crate is constant: this means that the acceleration is zero, a = 0.

So we can write Newton's second law as

F-F_k = ma = 0

And by substituting (1), we can find the value of the coefficient of kinetic friction:

F-\mu_k mg = 0\\\mu_k = \frac{F}{mg}=\frac{55 N}{(21 kg)(9.8 m/s^2)}=0.267

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The speed of an object can be determined from the distance vs time graph.

You know that speed = distance/time

in the graph, distance/time = slope of the curve.

So SPEED IS GIVEN BY THE SLOPE of the curve in the graph.

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● If the distance vs time curve is a straight line, with some non-zero slope; That means speed is nonzero and constant. So the object is in uniform motion.

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4 0
2 years ago
Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis
ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × 10^{-9} m

wavelength λ  = 700 nm =  700 × 10^{-9} m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ    ...........1

here m is 1 for single slit and d is = \frac{1}{900*10^3 m}

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 380 × 10^{-9}

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 700 × 10^{-9}

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

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3 years ago
Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s
Vitek1552 [10]
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
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                                                        mn=mass of neutron=<span>1.67 10^-27 kg
                                                        </span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5=  - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
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the final answer is </span><span>Eb /nucleon </span><span>=  -5.625 x10^35 eV</span>
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c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930

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Answer:

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