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2 years ago

What are the three examples that show relation between pressure and area in daily life.

Physics

1 answer:

makvit [3.9K]2 years ago

4 0

Answer: The area of the edge of a knife's blade is extremely small.

Syringes are used to take blood for blood tests.

When air is sucked out of a drinking straw, the air pressure inside if decreases and the atmospheric pressure outside forces the liquid to go inside the straw.

Explanation:

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A father uses a rope to pull a child on a sled at a constant speed. of the ones listed which forces are present?

Amanda [17]

All the forces may be present.

The child on the sled has a weight- which is due to the force of gravity on the child and the sled.

The sled and the child exert a force on the ground equal to the combined weight of the sled and the child. The ground exerts a normal force on the sled.

The force used by the father to pull the sled is the applied force.

The sled slides on the ground and as a result, the force of friction exists between the ground and the sled, directed opposite to the direction of the motion of the sled.

The father pulls the sled using a rope. As a result, the rope is under Tension.

As the sled moves it also experiences a force of air resistance, which is dependent on the sled's speed.

However, since the father pulls the sled along with constant speed, the sum of all the forces acting on the sled is zero.

Since there is no movement in the upward or downward directions, the weight of the child and the sled is equal to the normal force acted upon the sled by the ground.

The force applied by the father on the rope is equal to the tension in the rope.

Friction and air resistance act opposite to the direction of motion of the sled. for the sled to move at constant speed, the tension in the rope must be equal to the sum of the forces due to friction and air resistance.

4 0

3 years ago

Most towns use a water tower to store water and provide pressure in the pipes that deliver water to customers. The figure below

MatroZZZ [7]

The effective height of the water for Smith's house will be 24.61m.

<h3>How to calculate the height?</h3>

Based on the information given, the volume of the water in sphere will be:

= 4/3πr³ = (5.80 × 10^5)/1000

= 4.18r³ = 580

r³ = 138.7

r = 5.18m

The effective height of the water will be:

= 18.0 + 2(5.18)

= 28.36

The gauge pressure at Faucet of Jones house will be:

= pgh

= 1000(9.8)(28.36)

= 277.9kPa

The effective height of the water for Smith's house will be:

= 18.0 + 2(5.18) - 3.75

= 24.61m

The gauge pressure at Faucet of Jones house will be:

= 1000 × 9.8 × 24.61

= 241.2kPa

Learn more about height on:

brainly.com/question/983412

#SPJ1

8 0

2 years ago

In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e

Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

$\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}$

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

$\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}$

$\eta=1-\frac{(Q_1-Q_3)}{Q_1}}$

$\eta=1-\frac{(Q_3)}{Q_1}}$

also,

$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}$

$\frac{T_3}{T_1}=\frac{Q_3}{Q_1}$

thus,

$\eta=1-\frac{(T_3)}{T_1}}$

thus,

$\eta=1-\frac{(240\ K)}{500\ K}}$

$\eta=0.52$

Efficiency = 52%

8 0

3 years ago

HELP PLS ILL GIVE U BRAINLIST

jeyben [28]

B. insulator

hope this was helpful

6 0

3 years ago

Read 2 more answers

What is the minimum force require to move a 5kg wooden crate on a wooden floor?

kolbaska11 [484]

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with µ. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ F = p - f = 0

• net vertical:

∑ F = n - w = 0

where

p = magnitude of the pushing force

f = mag. of friction

n = mag. of the normal force

w = weight of the crate

The second equation gives

n = w = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of µ, so

f = µ (49 N) = 49µ N

To overcome static friction, the push has to exceed this in magnitude, so that

p > 49µ N

For instance, if p = 0.25, then p would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

7 0

3 years ago

What are the three examples that show relation between pressure and area in daily life.​

What are the three examples that show relation between pressure and area in daily life.