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jok3333 [9.3K]
2 years ago
10

You push a heavy crate down a ramp at a constant velocity. Only four forces act on the crate. Which force does the greatest magn

itude (it does not matter positive or negative) of work on the crate
Physics
1 answer:
Brilliant_brown [7]2 years ago
6 0

The friction force does the greatest magnitude of work on the crate

Consider all four forces. The normal force does no work at all, since there is no motion in the direction of that force, perpendicular to the ramp. The force of gravity is smaller than the force of friction, since you still need to push the crate to get constant velocity. The force of you pushing is also smaller than the force of friction, since you are moving down a ramp, and are therefore assisted by gravity. Therefore the force doing greatest magnitude of work is the force of friction. Note that, even though the frictional work is negative, it still has the greatest magnitude

Learn more about friction force here:

brainly.com/question/4618599

#SPJ4

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A difference In density between gases creates buoyancy which is the ability of an object to float. 0°C helium has a density of .
rjkz [21]

Answer:

mass = 0.18 [kg]

Explanation:

This is a classic problem where we can apply the definition of density which is equal to mass over volume.

density = \frac{mass}{volume} \\\\where:\\volume = 1 [m^3]\\density = 0.18[kg/m^3]

mass = 0.18*1

mass = 0.18 [kg]

6 0
3 years ago
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
4 years ago
What is the speed of a 200-kilogram car that is driving with 2000 joules of kinetic energy? (SHOW ALL WORK)
Katyanochek1 [597]

Answer:

v ≈ 4.47

Explanation:

The Formula needed = <u>KE = </u>\frac{1}{2}<u> m v²</u>

<u></u>

Substitute with numbers known:

2000J = \frac{1}{2} × 200kg × v²

Simplify:

÷100       ÷100      (Divide by 100 on both sides)

2000J = 100 × v²

\frac{2000J}{100} =  v²

20 = v²

√         √             (Square root on both sides)

√20 = √v²

4.472135955 = v (Round to whatever the question asks)

v ≈ 4.47       (I rounded to 2 decimal places or 3 significant figures, as that is what it usually is)

3 0
3 years ago
The weld bead in SMAW is formed by the?
SVETLANKA909090 [29]
The answer is C I just took this quiz.
7 0
3 years ago
What will be the answer??
AnnZ [28]
5-a).  Acceleration is a vector defined as the rate of change of velocity.
Its magnitude has units of [length/time²].  The SI unit is meter/second².
Its direction is the direction in which velocity is increasing. 

5-b).  The graph says that the object's speed is not changing.
When we look at any time, from zero to almost 50 minutes, the
object's speed is the same . . . 60 m/s .  This will make it easy.

There are 60 seconds in a minute, so 30 minutes = 1,800 seconds.
In every one of those seconds, the object covered 60 meters.
It travelled a total of (60 m/s)·(1,800 s) = 108,000 meters (108 km) .
8 0
4 years ago
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