What are the answers ?

Which is always true in a closed system?

Tanya [424]

Answer:

The correct option is momentum is conserved

Explanation:

A closed system is a system that is independent/free of external factors/force and does not exchange matter with its surrounding. Since a close system is free of external factors/force; <em>acceleration is constant in it, mass is conserved in it and there will be changes in velocities of objects in the closed system</em>.

This question actually seeks to test our knowledge of the law of momentum. The law of conservation of momentum states that the momentum of a closed system is conserved.

4 0

3 years ago

Identify which portions of the electromagnetic spectrum are used in each of the devices listed. a. a microwave oven b. a televis

dlinn [17]

The portions of the electromagnetic spectrum are used in the television set. It is used in the picture tube in the old television set.

<h3>What is the electromagnetic spectrum ?</h3>

The electromagnetic radiation distribution in accordance with the frequency or wavelength is known as the electromagnetic spectrum.

The portions of the electromagnetic spectrum are used in the television set. It is used in the picture tube in the old television set.

Hence, option b is correct.

To learn more about the electromagnetic spectrum, refer to the link;

brainly.com/question/15576247

#SPJ1

6 0

2 years ago

Two bumper cars in an amusement park ride collide elastically as one approaches the other directly fromthe rear. Car A has a mas

zubka84 [21]

Answer:

a) The velocity of car B after the collision is 4.45 m/s.

The velocity of car A after the collision is 3.65 m/s.

b) The change of momentum of car A is - 370.45 kg · m/s

The change of momentum of car B is 370.45 kg · m/s

Explanation:

Hi there!

Since the cars collide elastically, the momentum and kinetic energy of the system do not change after the collision.

The momentum of the system is calculated adding the momenta of each car:

initial momentum = final momentum

mA · vA + mB · vB = mA · vA´ + mB · vB´

Where:

mA = mass of car A

vA = initial velocity of car A

mB = mass of car B

vB = initial velocity of car B

vA´= final velocity of car A

vB´ = final velocity of car B

Let´s replace with the data we have and solve the equation for vA´:

mA · vA + mB · vB = mA · vA´ + mB · vB´

435 kg · 4.50 m/s + 495 kg · 3.70 m/s = 435 kg · vA´ + 495 kg · vB´

3789 kg · m/s = 435 kg · vA´ + 495 kg · vB´

3789 kg · m/s - 495 kg · vB´ = 435 kg · vA´

(3789 kg · m/s - 495 kg · vB´)/435 kg = vA´

Let´s write this expression without units for a bit more clarity:

vA´= (3789 - 495 vB´)/435

The kinetic energy of the system is also conserved, then, the initial kinetic energy is equal to the final kinetic energy:

initial kinetic energy of the system = final kinetic energy of the system

1/2 · mA · vA² + 1/2 · mB · vB² = 1/2 · mA · (vA´)² + 1/2 · mB · (vB´)²

Replacing with the data:

initial kinetic energy = 1/2 · 435 kg · (4.50 m/s)² + 1/2 · 495 kg · (3.70)²

initial kinetic energy = 7792.65 kg · m²/s²

7792.65 kg · m²/s² = 1/2 · 435 kg · (vA´)² + 1/2 · 495 kg · (vB´)²

multiply by 2 both sides of the equation:

15585.3 kg · m²/s² = 435 kg · (vA´)² + 495 kg · (vB´)²

Let´s replace vA´ = (3789 - 495 vB´)/435

I will omit units for clarity in the calculation:

15585.3 = 435 · (vA´)² + 495 · (vB´)²

15585.3 = 435 · (3789 - 495 vB´)²/ 435² + 495 (vB´)²

15585.3 = (3789² - 3751110 vB´ + 245025 vB²) / 435 + 495 (vB´)²

multiply both sides of the equation by 435:

6779605.5 = 3789² - 3751110 vB´ + 245025 vB² + 215325 vB´²

0 = -6779605.5 + 3789² - 3751110 vB´ + 460350 vB´²

0 = 7576915.5 - 3751110 vB´ + 460350 vB´²

Solving the quadratic equation:

vB´ = 4.45 m/s

vB´ = 3.70 m/s (the initial velocity)

a) The velocity of car B after the collision is 4.45 m/s

The velocity of car A will be teh following:

vA´= (3789 - 495 vB´)/435

vA´= (3789 - 495 (4.45 m/s))/435

vA´ = 3.65 m/s

The velocity of car A after the collision is 3.65 m/s

b) The change of momentum of each car is calculated as the difference between its final momentum and its initial momentum:

ΔpA = final momentum of car A - initial momentum of car A

ΔpA = mA · vA´ - mA · vA

ΔpA = mA (vA´ - vA)

ΔpA = 435 kg (3.648387097 m/s - 4.50 m/s) (I have used the value of vA´ without rounding).

ΔpA = - 370.45 kg · m/s

The change of momentum of car A is - 370.45 kg · m/s

ΔpB = mB (vB´ - vB)

ΔpB = 495 kg (4.448387097 m/s - 3.70 m/s) (I have used the value of vB´ without rounding).

ΔpB = 370.45 kg · m/s

The change of momentum of car B is 370.45 kg · m/s

I have used the values of the final velocities without rounding so we can notice that the change of momentum of both cars is equal but of opposite sign.

7 0

3 years ago

5. George walks to a friend's house. He walks 750 meters North, then realizes he walked too far.

dedylja [7]

Answer:

Average speed: approximately $76.9\; {\rm m\cdot s^{-1}}$ .

Average velocity: approximately $38.5\; {\rm m \cdot s^{-1}}$ (to the north.)

Explanation:

Consider an object that travelled along a certain path. Distance travelled would be equal to the length of the entire path.

In contrast, the magnitude of displacement is equal to distance between where the object started and where it stopped.

In this question, the path George took required him to travel $750\; {\rm m} + 250\; {\rm m} = 1000\; {\rm m}$ in total. Hence, the distance George travelled would be $1000\; {\rm m}$ . However, since George stopped at a point $(750\; {\rm m} - 250\; {\rm m}) = 500\; {\rm m}$ to the north of where he started, his displacement would be only $500\; {\rm m}$ to the north.

Divide total distance by total time to find the average speed.

Divide total displacement by total time to find average velocity.

The total time of travel in this question is $13\; {\rm s}$ .. Therefore:

$\begin{aligned}\text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &= \frac{1000\; {\rm m}}{13\; {\rm s}} \\ &\approx 76.9\; {\rm m\cdot s^{-1}}\end{aligned}$ .

$\begin{aligned}\text{average velocity} &= \frac{\text{total displacement}}{\text{total time}} \\ &= \frac{500\; {\rm m}}{13\; {\rm s}} && \genfrac{}{}{0px}{}{(\text{to the north})}{}\\ &\approx 38.5\; {\rm m\cdot s^{-1}} && (\text{to the north})\end{aligned}$ .

3 0

2 years ago

Overload refers to: A. Performing a weight-lifting exercise with the resistance (load) held overhead B. Using a demand (load) ab

andrey2020 [161]

Answer:

E. All of the answers are correct

Explanation:

Overload principle in fitness training is associated with a gradual development of an athlete's abilities by progressively increasing the athlete's load and training.

In order to do this, the athlete's limit must be surpassed albeit gradually at first then picked up later over time.

6 0

3 years ago