Question 8 (1 point)

What is the total energy of a system with KE = 1368 J and a PE = 5782J? Express your answer in kJ.

mezya [45]

Answer:

7.150 KJ

Explanation:

The total energy of a system is the sum of the potential and kinetic energy in the system. According to this question, the kinetic energy (KE) = 1368 J and a potential energy (PE) = 5782J.

Total energy of the system will be PE + KE as follows;

1368J + 5782J

= 7150J

However, the question demanded that the answer be in kJ (kilojoules), which is higher than the joules.

Since 1kJ = 1000J

7150J = 7150/1000

= 7.150kJ

6 0

3 years ago

A potential difference of 10.0 volts exists between two points, A and B, within an electric field. What is the

Viefleur [7K]

Answer:

1. 5.0 x 10^2 C

Explanation:

V=W/Q

10 = 2.0 x 10^-2/Q

Q = 2.0 x 10^-2/ 10

Q = 5.0 x 10^2 C

7 0

4 years ago

A capacitor has plates of area 8.25 * 10 ^ - 5 m ^ 2 . To create a capacitance of 3.35*10^ -10 F , how far apart should the plat

sveticcg [70]

Answer:

A capacitor has plates of area 8.25 * 10 ^ - 5 m ^ 2 . To create a capacitance of 3.35*10^ -10 F , how far apart should the plates be ?

Explanation:

6 0

3 years ago

With what velocity Must a 0.53 kg softball be moving to be equal to the momentum of a 0.31 kg baseball moving at 21 m/s

vladimir2022 [97]

Answer:12.28m/s

Explanation:

momentum of baseball =mass of baseball x velocity of baseball

Momentum of baseball =0.31x21

Momentum of baseball =6.51kgm/s

For a softball to have same momentum with the baseball we can say :momentum of baseball =mass of softball x velocity of softball

6.51=0.53 x velocity of softball

Velocity of softball =6.51/0.53

Velocity of softball =12.28m/s

3 0

4 years ago

A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball

jonny [76]

(a) The impulse delivered to the ball by the racket is 5.24 kg.m/s

(b) The work that the racket does on the ball is -35.1 Joule

<h3>Further Explanation</h3>

<u>Given :</u>

mass of ball = m = 0.06 kg

initial velocity = v₁ = -50.4 m/s

final velocity = v₂ = 37.0 m/s

<u>Unknown :</u>

(a) Impulse = I = ?

(b) Work = W = ?

<u>Solution :</u>

<h2>Question (a) :</h2>

In this question , we could use the formula from Second Law of Newton :

$I = \Delta p$

$I = p_2 - p_1$

$I = m \times v_2 - m \times v_1$

$I = m \times (v_2 - v_1)$

$I = 0.06 \times (37.0 - (-50.4))$

$I = 0.06 \times (87.4)$

$I = 5.244~kg.m/s$

$\large { \boxed {I \approx 5.24~kg.m/s} }$

<h2>Question (b) :</h2>

$W = F \times d$

$W = (\frac{I}{\Delta t})(\frac {v_1 + v_2}{2} \Delta t)$

$W = \frac{I(v_1 + v_2)}{2}$

$W = \frac{5.244(-50.4 + 37)}{2}$

$W = \frac{5.244(-13.4)}{2}$

$W = -35.1348~Joule$

$\large { \boxed {W \approx -35.1~Joule}}$

<h3>Learn more</h3>

Newton's Law of Motion: brainly.com/question/10431582

Example of Newton's Law: brainly.com/question/498822

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Newton, Law, Impulse, Work

6 0

4 years ago