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2 years ago

A piston in a gas supply pump has an

Physics

2 answers:

eimsori [14]2 years ago

6 0

<h2>Option A is the correct answer.</h2>

Explanation:

Pressure is the ratio of normal force to the area.

$\texttt{Pressure = }\frac{\texttt{Normal force}}{\texttt{Area}}$

Force = Pressure x Area

Area = 500 cm² = 0.05 m²

Pressure = 4000 Pa

Force = 0.05 x 4000 = 200 N

We know that

Work done = Force x Displacement

W = F x d

Displacement, d = 30 cm = 0.3 m

W = 200 x 0.3 = 60 Joules.

Option A is the correct answer.

KonstantinChe [14]2 years ago

5 0

Answer:

A. 60 Joule

Explanation:

First find the force given the pressure and area.

F = AP

F = 500 cm² × (1 m / 100 cm)² × 4000 N/m²

F = 200 N

Now find the work:

W = Fd

W = 200 N × 30 cm × (1 m / 100 cm)

W = 60 J

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Recalculate 100 km/h to m/s

qwelly [4]

= 27.777

Explanation:

A kilometer has 1,000 meters, and an hour has 3,600 seconds, so 100 kilometers per hour is: 100 x 1,000 / 3,600 = 27.777... m/s.

4 0

3 years ago

What are the SI units of thermal conductivity?

kotykmax [81]

Answer:

The SI unit of thermal conductivity is watts per meter-kelvin (W/(m⋅K)).

Explanation:

hope this will help u

7 0

2 years ago

a person throws a ball in such a way that its speed is zero at one particular point in its path. How did the person throw the ba

Keith_Richards [23]

vertical! or in other words, up!

6 0

3 years ago

A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter

expeople1 [14]

Answer: $56.87m/s^{2}$

Explanation:

If we make an analysis of the net force $F_{net}$ of the rock that was thrown upwards, we will have the following:

$F_{net}=F_{up}-W$ (1)

Where:

$F_{up}=200N$ is the force with which the rock was thrown

$W$ is the weight of the rock

Being the weight the relation between the mass $m=3kg$ of the rock and the acceleration due gravity $g=9.79m/s^{2}$ :

$W=m.g=(3kg)(9.79m/s^{2})$ (2)

$W=29.37 N$ (3)

Substituting (3) in (1):

$F_{net}=200N-29.37 N$ (4)

$F_{net}=170.63 N$ (5) This is the net Force on the rock

On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:

$F_{net}=m.a$ (6)

Finding the acceleration $a$ :

$a=\frac{F_{net}}{m}$ (7)

$a=\frac{170.63 N}{3kg}$ (8)

Finally:

$a=56.87m/s^{2}$

3 0

2 years ago

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$

Substitute the given values

$u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56$

6 0

3 years ago