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2 years ago

15

Which of the following is a form of potential energy? O A. Sound energy O B. Elastic energy O C. Light energy O O O D. Kinetic e

nergy

1 answer:

olga55 [171]2 years ago

6 0

Answer:

O C. Light energy

Explanation:

it conducts energy in it and is an energy itself.

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On which principal is the thermometer based

olchik [2.2K]

Answer:

thermometer is based on the principle of thermal expansion.

4 0

3 years ago

A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this dista

cluponka [151]

Answer:

4.20214231077 m/s

29.429 m/s²

4316.29 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of player = 110 kg

Equations of linear motion

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 0.9}\\\Rightarrow u=4.20214231077\ m/s$

The velocity is 4.20214231077 m/s

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{4.20214231077^2-0^2}{2\times 0.3}\\\Rightarrow a=29.429\ m/s^2$

The acceleration is 29.429 m/s²

Force is given by

$F=m(a+g)\\\Rightarrow F=110(29.429+9.81)\\\Rightarrow F=4316.29\ N$

The force he exerts is 4316.29 N

6 0

3 years ago

Which statement about he diagram is accurate

xeze [42]

B is my answer

.......

6 0

3 years ago

How much force is needed to accelerate a 2500 kg car at a rate of 4m/s²?

Vika [28.1K]

Answer:

Momentum (p) is equal to the product of an object's mass (m) and its change in velocity (v).

Change in velocity (v) results in change in momentum (p), which is equal to impulse. Impulse (J) is also equal to an applied force (F) over a period of time (t).

Combining p=mv and J=F t together:

m (v final - v initial) = F t

(2,500 kg) (70 m/s - 30 m/s) = F (10 s)

Explanation:

Solving for F we get:

10,000 N of force

4 0

2 years ago

Read 2 more answers

(refer to photo attached) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the

mixas84 [53]

The electric field strength at a point 1.00 cm to the left of the middle is -2.0 x 10⁷ N/C.

The magnitude of the force is 94.4 N and direction of the force on it towards the right.

<h3>Electric field strength</h3>

The electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;

E = kq/r²

Electric field due to first charge

E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²

E1 = 1.35 x 10⁸ N/C

Electric field due to second charge

E2 = -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²

E2 = - 1.35 x 10⁸ N/C

Electric field due to third charge

E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²

E3 = -2.0 x 10⁷ N/C

<h3>Net electric field</h3>

E = E1 + E2 + E3

E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - -2.0 x 10⁷ N/C

E = -2.0 x 10⁷ N/C

<h3>Force on the charge −4.72 µC</h3>

F = Eq

F = - 2.0 x 10⁷ x -4.72 x 10⁻⁶

F = 94.4 N

Thus, the direction of the force will be towards the right.

Learn more about force on charge here: brainly.com/question/25923373

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6 0

1 year ago