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3 years ago

Find the pressure if a force of 2N is applied to an area of 0.00004m^2

Physics

1 answer:

BigorU [14]3 years ago

3 0

Answer:

5 x 10^4 N/m^2

Explanation:

Pressure, force and area are related witg the following equation;

Pressure = Force /Area

From the question, we obtained the following information;

Force = 2N

Area = 0.00004m^2

Pressure =?

Pressure = Force /Area

Pressure = 2/0.00004

Pressure = 5 x 10^4N/m^2

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Newton’s laws do not apply to small objects?

Soloha48 [4]

Answer:

Yes Newton's laws apply to small objects

EX: Newton s first law

when body at rest always want to be at rest

or body at motion always want to be at motion

unles an external force acts upon it

for example a eraser on the table will be at rest

if so e apply some force then it comes motion

so, Newton s law apply to small object s

7 0

2 years ago

HELP IM TIMED. ILL MARK BRAINLIEST PLSSSSSS.

Alex17521 [72]

Answer:

According to Newton's Second Law of Motion :

Where,

F = Force Applied

m = Mass of the object

a = Acceleration

Now, we will use this law to solve this question.

Given :

Acceleration or a = 15.3 m/s²

Force = 44 N

Mass = ?

Substitute, the given values in the formula.

F = ma

⇒ m = F/a

m = 44/15.3

5 0

2 years ago

The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to

LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power ,P = F .v

F=force

v=velocity

v= 100 km/h

$v=\dfrac{1000}{3600}\times 100\ m/s$

v=27.77 m/s

75 x 1000 = F x 27.77

$F= \dfrac{75000}{27.77}\ N$

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

3 0

3 years ago

How much work is done if a force of 20N is used to move an object 6 metres? <br><br> pls help

Dmitriy789 [7]

I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.

Work = 20 N * 6 m * 1

Work = 120 Nm

Work = 120 joules

Hope this helps!

8 0

2 years ago

Read 2 more answers

A rotating viscometer consists of two concentric cylinders –an inner cylinder of radius Rirotating at angular velocity (rotation

RSB [31]

Answer:

b) the result we got can be termed approximation because we are neglecting the shear stress acting on the two ends of the cylinder. Here we have considered only the share stress acting on the curved surface area only.

Explanation:

check attachment for solution to A

4 0

2 years ago