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nordsb [41]
2 years ago
13

Which of the following is a primary source?

Physics
1 answer:
Anarel [89]2 years ago
6 0

The answer is C :), I think.

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An electroscope is a device that has small gold foil strips suspended from a metal rod. A student brings a negatively charged ro
Sever21 [200]
I think it is neutral
4 0
3 years ago
A 1.5-kg ball is throw at 10 m/s. What is the balls momentum?
alexdok [17]

Answer : The momentum of ball is, 15 kg.m/s

Explanation :

Momentum : It is defined as the motion of a moving body. Or it is defined as the product of mass of velocity of an object.

Formula of momentum is:

where,  

p = momentum  = ?

m = mass  = 1.5 kg

v = velocity = 10 m/s

Now put all the given values in the above formula, we get:

Therefore, the momentum of ball is 15 kg.m/s

3 0
3 years ago
Qual o valor a ser pago, no final de 5 meses e 18 dias, correspondente a um empréstimo de $125.000,00, sabendo-se que a taxa de
liberstina [14]
25% ao semestre = 25%/6 ao mes.
Um mes tem em media 365/12 dias, ou 30,42 dias/mes.
18 dias = 18/30,42 mes

5 meses + 18 dias = 5 meses + 18/30,42 mes = 5,592 mes

a taxa de juros por 5 meses e 18 dias e 25%/6 * 5.592 = 23,3%

123,3% * $125.000 = $154.125,00
6 0
3 years ago
Read 2 more answers
An oscillator consists of a block of mass 0.628 kg connected to a spring. When set into oscillation with amplitude 27 cm, the os
oksian1 [2.3K]

Answer:

T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N

Explanation:

a.)

Period: It is already given in the question "oscillator repeats its motion every 0.372 s".

So T=0.372 s

b)

frequency= f = 1/ T

f = 1/ 0.372

f=2.7 Hz

c).

Angular frequency= w= 2πf

w= 2*π*2.7

w=16.9 rad/s

d)

Spring Constant:

As w=\sqrt{k/m}

⇒w²= k/m

⇒k= m*w²

⇒k= 0.628 * 16.9² N/m

⇒k=179.2 N/m

e)

The mass will have maximum speed when it passes through the mean position.

At mean position

Maximum elastic potential energy = Maximum kinetic energy

1/2 k A² = 1/2 m v²    ( A is amplitude of oscillation)

⇒ v=\sqrt{k A^2/m}

⇒ v= \sqrt{179.2 * 0.27/ 0.628}\

⇒ v= 8.78 m/s

f)

Maximum force will be exerted on the block when it is at maximum distance.

F= k* A   ( A is amplitude of oscillation)

F= 179.2 * 0.27 N

F= 48.4 N

5 0
3 years ago
A catapult with a spring constant of 10,000 N/m is used to launch a target from the deck of a ship. The spring is compressed a d
Mnenie [13.5K]

Answer:

(C) 40m/s

Explanation:

Given;

spring constant of the catapult, k = 10,000 N/m

compression of the spring, x = 0.5 m

mass of the launched object, m = 1.56 kg

Apply the principle of conservation of energy;

Elastic potential energy of the catapult = kinetic energy of the target launched.

¹/₂kx² = ¹/₂mv²

where;

v is the target's  velocity as it leaves the catapult

kx² = mv²

v² = kx² / m

v² = (10000 x 0.5²) / (1.56)

v² = 1602.56

v = √1602.56

v = 40.03 m/s

v ≅ 40 m/s

Therefore, the target's velocity as it leaves the spring is 40 m/s

6 0
2 years ago
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