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Class VI-VIII Tuition Fees

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₹ 2,000 to ₹ 5,000 per month

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Trending discussions on Class VI-VIII Tuition

Answered 5 hrs ago Tuition/Class VI-VIII Tuition

Ranjan Singh

Trainer

It's depend upon number of class and hours
Answers 145 Comments
Dislike Bookmark

Lesson Posted on 18/06/2017 Tuition/Class VI-VIII Tuition Tuition/Class IX-X Tuition Tuition/Class IX-X Tuition/Science +1 Tuition/Class VI-VIII Tuition/Science less

Numerical problems on force and pressure

Soumi Roy

I am an experienced, qualified tutor with over 5 years of experience in teaching maths and physics across...

Example 1: A force of 1200 N acts on the surface of area 10 cm2 normally. What would be the thrust and pressure on the surface?Given: Force F = 1200 N, Area A = 10 cm2 = 10 ×10-4 m2 = 10-3 m2Thrust = Normal pressure = F = 1200 NPressure P = F/A 1200N10−3m21200N10−3m2 ... read more

Example 1:

A force of 1200 N acts on the surface of area 10 cm2 normally. What would be the thrust and pressure on the surface?

Given:


Force F = 1200 N, Area A = 10 cm2 = 10 ×10-4 m2 = 10-3 m2

Thrust = Normal pressure = F = 1200 N

Pressure P = F/A 1200N103m21200N10−3m2
              
                      = 1.2 ×× 106 N/m2

Example 2:

The elephant weighs 20,000 N stands on one foot of area 1000 cm2. How much pressure would it exert on the ground?

Given:


Force applied by elephant F = 20,000 N, area A = 1000 cm2 = 1000/100×100m2 = 0.1 m2

The pressure is given by

Pressure P = F/A

      = 20,000N/0.1m2


        = 2,00,000 N/m2.

Example 3:

Calculate the pressure produced by a force of 800 N acting on an area of 2.0 m2.

Solution:

Pressure is defined as force per unit area or P = F / A
P = (800 N) / (2.0 m2
)
P = 400 N / m2
= 400 Pa

Example 4:

The pressure of a gas contained in a cylinder with a movable piston is 300 Pa. The area of the piston is 0.5 m2. Calculate the force that is exerted on the piston.

Solution

Pressure is defined as force per unit area or P = F / A
We multiply both sides of the equation by the area to solve for the force as
F = P A
F = (300 Pa) (0.5 m2
)
F = 150 (Pa) m2
= 150 (N / m2
) m2
F = 150 N

Example 5:

A swimming pool of width 9.0 m and length 24.0 m is filled with water to a depth of 3.0 m. Calculate pressure on the bottom of the pool due to the water.

Solution:

The pressure due to a column of fluid is calculated as the product of the height of the fluid times
the density of the fluid times the acceleration due to gravity. For water the density is d = 1000 kg /
m3
.
? P = d h g
? P = (1000 kg / m3
) (3.0 m) (9.8 m / s2
)
? P = 29400 kg m2 / m3
s2
To simplify the units we cancel one m term in the m2
term in the numerator and one m term in
the m3
term in the denominator to get
? P = 29400 kg m / s2
m2

We know that 1N = 1 kg m / s2
so we can write
? P = 29400 N / m2
= 29400 Pa

read less
Comments 6
Dislike Bookmark

Lesson Posted on 18/06/2017 Tuition/Class VI-VIII Tuition Tuition/Class IX-X Tuition Tuition/Class IX-X Tuition/Science +1 Tuition/Class VI-VIII Tuition/Science less

Numerical problems on force and pressure

Soumi Roy

I am an experienced, qualified tutor with over 5 years of experience in teaching maths and physics across...

Example 1: A force of 1200 N acts on the surface of area 10 cm2 normally. What would be the thrust and pressure on the surface?Given: Force F = 1200 N, Area A = 10 cm2 = 10 ×10-4 m2 = 10-3 m2Thrust = Normal pressure = F = 1200 NPressure P = F/A 1200N10−3m21200N10−3m2 ... read more

Example 1:

A force of 1200 N acts on the surface of area 10 cm2 normally. What would be the thrust and pressure on the surface?

Given:


Force F = 1200 N, Area A = 10 cm2 = 10 ×10-4 m2 = 10-3 m2

Thrust = Normal pressure = F = 1200 N

Pressure P = F/A 1200N103m21200N10−3m2
              
                      = 1.2 ×× 106 N/m2

Example 2:

The elephant weighs 20,000 N stands on one foot of area 1000 cm2. How much pressure would it exert on the ground?

Given:


Force applied by elephant F = 20,000 N, area A = 1000 cm2 = 1000/100×100m2 = 0.1 m2

The pressure is given by

Pressure P = F/A

      = 20,000N/0.1m2


        = 2,00,000 N/m2.

Example 3:

Calculate the pressure produced by a force of 800 N acting on an area of 2.0 m2.

Solution:

Pressure is defined as force per unit area or P = F / A
P = (800 N) / (2.0 m2
)
P = 400 N / m2
= 400 Pa

Example 4:

The pressure of a gas contained in a cylinder with a movable piston is 300 Pa. The area of the piston is 0.5 m2. Calculate the force that is exerted on the piston.

Solution

Pressure is defined as force per unit area or P = F / A
We multiply both sides of the equation by the area to solve for the force as
F = P A
F = (300 Pa) (0.5 m2
)
F = 150 (Pa) m2
= 150 (N / m2
) m2
F = 150 N

Example 5:

A swimming pool of width 9.0 m and length 24.0 m is filled with water to a depth of 3.0 m. Calculate pressure on the bottom of the pool due to the water.

Solution:

The pressure due to a column of fluid is calculated as the product of the height of the fluid times
the density of the fluid times the acceleration due to gravity. For water the density is d = 1000 kg /
m3
.
? P = d h g
? P = (1000 kg / m3
) (3.0 m) (9.8 m / s2
)
? P = 29400 kg m2 / m3
s2
To simplify the units we cancel one m term in the m2
term in the numerator and one m term in
the m3
term in the denominator to get
? P = 29400 kg m / s2
m2

We know that 1N = 1 kg m / s2
so we can write
? P = 29400 N / m2
= 29400 Pa

read less
Comments 6
Dislike Bookmark

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