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3 years ago

Electromagnetic Energy Example One Activity:

Physics

2 answers:

stich3 [128]3 years ago

6 0

One type is microwaves

Microwaves are used for radio and radar communications
Also used to cook food

diamong [38]3 years ago

6 0

Electromagnetic Energy Example One

Activity: Cellphones

Type of electromagnetic energy: Radio waves

Description of use: You use cellphones to make calls to other people

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Select the correct answer.

Serhud [2]

Answer:

b hope this helps

Explanation:

5 0

3 years ago

Read 2 more answers

Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B, moving in t

MA_775_DIABLO [31]

First, let's express the movement of Car A and B in terms of their position over time (relative to car B)
For car A: y=20x-200 Car A moves 20 meters every second x, and starts 200 meters behind car B
For Car B: y= 15x Car B moves 15 meters every second and starts at our basis point

Set the two equations equal to one another to find the time x at which they meet:
20x - 200 = 15x
200 = 5x
x= 40
At time x=40 seconds, the cars meet. How far will Car A have traveled at this time?
Car A moves 20 meters every second:
20 x 40 = 800 meters

8 0

3 years ago

If there was no frictional force acting on the ball in question 2 what would happen to the player? Use inertia and Newton’s 1st

laila [671]

Newwton's law of inertia states that an object will not be able to move unless force is applied to it

6 0

3 years ago

||||| A 1.5 kg block and a 2.5 kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless

amid [387]

Answer

The lighter block will have the Positive Acceleration that is +2.45 m/s square.

A (1.5 Kg) = + 2.45 m/s square

Explanation:

To solve this there is a hard way to do this and there is an easy way to do this. The hard way is to solve Newton's second law for each block individually and then combine them and you get two equations with two unknowns. you try your best to solve the algebra without losing any sins but lets be honest it usually goes wrong.

So the easy way to do this the way to get the magnitude of the acceleration of the blocks. That is to say that i want to know the magnitude at which 2.5 Kg block accelerates or 1.5 Kg block accelerates when the the blocks were released.

Take the net external force that tries to make system go and divide it by total mass of the the system.

A (of System) = F(net external force) / m (total mass of system)

This is the quick way to know the magnitude of acceleration of the objects in the system. But this is only possible if the system is moved in same magnitude of acceleration that is 2.5 Kg block will move downward and 1.5 Kg block will move upward with the same magnitude. So here in this case we have friction-less pulley and the blocks will move with the same magnitude of acceleration.

To find the external forces

External forces are the forces which exerted on the objects in our system from the objects outside of our system. So one external force is the force of gravity. Both 2.5 Kg block and the force of gravity will be in downward direction.

Force of gravity on 2.5 Kg block

F = + (2.5 x 9.8) = 24.5

After releasing the rope the 2.5 Kg block will drive the system and accelerates in downward direction so It will be a positive force.

Force of gravity on 1.5 Kg block

F = - (1.5 x 9.8) = 14.7

The force of gravity on 1.5 Kg block will be negative because it will accelerate in upward or opposite direction of the force of gravity. Because the whole system is moving in one direction but the force of gravity on 1.5 Kg block is opposing the acceleration of the system.

Now divide the the Net external forces by total mass of the blocks that is

A (of System) = F(net external force) / m (total mass of system)

A = (+ 24.5 - 14.7) / 2.5 + 1.5

A = 9.8 / 4

A = 2.45 meter per second square

So the Acceleration of 2.5 Kg block will be Negative that is -2.45 m/s square. Since block is accelerating down and we usually treat down as negative.

A (1.5 Kg) = - 2.45 m/s square

So the Acceleration of 1.5 Kg block will be Positive that is +2.45 m/s square. Since block is accelerating up and we usually treat up as positive.

A (1.5 Kg) = + 2.45 m/s square

4 0

4 years ago

A 50-cm-long spring is suspended from the ceiling. A 230g mass is connected to the end and held at rest with the spring unstretc

alukav5142 [94]

Answer:

k = 25.07 N/m

Amplitude = 9 cm

f = 1.66 Hz

Explanation:

Given:

- The original length of the spring L_o = 50 cm

- The mass hanged m = 230 g

- The amount of stretch given 2x = 18 cm @lowest point.

Find:

a. What is the spring constant? (K=)

b. What is the amplitude of the oscillation?

c. What is the frequency of the oscillation?

Solution:

- Make a FBD of the hanging mass, There are two external forces acting on it that is the force of gravity due to its weight and the springs restoring force when its stretched to its lowest point. After hanging the mass on the spring a new equilibrium position is achieved which also causes the spring to stretch. We can apply the Equilibrium conditions at this point in vertical direction as:

k*x - m*g = 0

k = m*g / x

Where, x is the extension of the spring or mean stretch. which 0.5*amplitude (Lowest point). x = 9 cm

k = 0.23*9.81 / 0.09

k = 25.07 N/m

Answer: For part a we have the stiffness of the spring k = 25.07 N/m

- The amplitude of the oscillating motion is the half the amount of total stretch or the amount the spring extends above or below the mean position.

Amplitude = x = 9 cm

- The frequency of any oscillatory motion which can be modeled by SHM can be expressed as:

f = 1 / 2*p* sqrt ( k / m )

- Plug the values in:

f = 1 / 2*pi* sqrt (25.07 / 0.23 )

f = 1.66 Hz

Answer: For part c the frequency of oscillation is f = 1.66 Hz

8 0

3 years ago