All categories

2 years ago

Why is physics important to everyone? Describe 5 applications of physics in everyday life.

Physics

1 answer:

Sedaia [141]2 years ago

4 0

1. Communication

In early ages people were using letters to communicate between them. Now days instantly we can call one person from anywhere, these are based on optical fiber communication and it is an application of physics.

2. Transportation

We can travel from place to anther place by aeroplane , these technologies are introduced after advancement in physics.

3. Construction

Earlier people used to stay in small houses, now days apartments with many number of storeys are built by considering all the forces coming on the structure.

4. Medical Department

The invention of many equipment increased the health of humans.

5. Weather forecasting

The prediction of weather considering all the available factors, prediction of natural disasters all are application of physics.

You might be interested in

Need physics help ASAP please!

Effectus [21]

Answer:

Explanation:

7 0

2 years ago

A driver with a 0.80-s reaction time applies the brakes, causing the car to have acceleration opposite the direction of motion.

jeka94

Answer:

a) During the reaction time, the car travels 21 m

b) After applying the brake, the car travels 48 m before coming to stop

Explanation:

The equation for the position of a straight movement with variable speed is as follows:

x = x0 + v0 t + 1/2 a t²

where

x: position at time t

v0: initial speed

a: acceleration

t: time

When the speed is constant (as before applying the brake), the equation would be:

x = x0 + v t

a)Before applying the brake, the car travels at constant speed. In 0.80 s the car will travel:

x = 0m + 26 m/s * 0.80 s = 21 m

b) After applying the brake, the car has an acceleration of -7.0 m/s². Using the equation for velocity, we can calculate how much time it takes the car to stop (v = 0):

v = v0 + a* t

0 = 26 m/s + (-7.0 m/s²) * t

-26 m/s / - 7.0 m/s² = t

t = 3.7 s

With this time, we can calculate how far the car traveled during the deacceleration.

x = x0 +v0 t + 1/2 a t²

x = 0m + 26 m/s * 3.7 s - 1/2 * 7.0m/s² * (3.7 s)² = 48 m

4 0

3 years ago

I need help, please answer

Burka [1]

This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.

7 0

3 years ago

ILI

Nikolay [14]

The original kinetic energy will be 0 J and the final kinetic energy will be 7500 J and the amount of work utilized will be similar to the final kinetic energy i.e., 7500 J.

Explanation:

As it is known that the kinetic energy is defined as the energy exhibited by the moving objects. So the kinetic energy is equal to the product of mass and square of the velocity attained by the car. Thus,

$\text {Kinetic energy}=\frac{1}{2} m v^{2}$

So the initial kinetic energy will be the energy exerted by the car at the initial state when the initial velocity is zero. Thus the initial kinetic energy will be zero.

The final kinetic energy is

$\text {Kinetic energy}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 600 \times 5 \times 5$ = 7500 J

As the work done is the energy required to start the car from zero velocity to 5 m/s velocity.

Work done = Final Kinetic energy - Initial Kinetic energy

Thus the work utilized for moving the car is

Work done = 7500 J - 0 J = 7500 J

Thus, the initial kinetic energy of the car is zero, the final kinetic energy is 7500 J and the work utilized by the car is also 7500 J.

7 0

2 years ago

Which type of line is shown on the map

emmainna [20.7K]

Answer:

Explanation:

completed the test

3 0

2 years ago