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nadya68 [22]
3 years ago
6

What is the frequency of a wave with a wavelength of 15 m and a wavespeed of 300 m/s?

Physics
1 answer:
STatiana [176]3 years ago
6 0

Answer: f=20 (i think)

Explanation:

all I did was divide 300 and 15.

300/15= 20

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When does a star reach equilibrium?
LenaWriter [7]
A protostar reaches equilibrium when gravity is balanced by the outward force of nuclear fusion. 
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3 years ago
A pan hangs from a 50 cm spring. When a 10 kg mass is placed in the pan, it stretches the spring 6 cm. What is a function rule l
MrRissso [65]

Answer:

Explanation:

A Spring stretches / compresses when force is applied on them and they are governed by the Hookes Law which states that the force required to stretch or compress a spring is directly proportional to the distance it is stretched.

F = -kx

F is the force applied and x is the elongation of the spring

k is the spring constant.

negative sign indicates the change in direction from equilibrium position.

In the given question, we dont have force but we know that the pan is hanging. We also know from the Newton's second law of motion that

F=mg

Inserting this into Hooke's Law

mg=-kx

computing it for x,

-x=mg/k

This is the model which will tell the length of the spring against change in the mass located in the pan.

3 0
3 years ago
What is the GPE of a 15,000 kg airplane sitting on the ground?
Vesna [10]

Answer:

C, it is not moving

it has no potential

7 0
2 years ago
A 71 kW radio station broadcasts its signal uniformly in all directions. - What is the average intensity of its signal at a dist
marshall27 [118]

Answer:

Explanation:

Energy of signal being radiated per second on all sides = 71 x 10³ J .

At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.

So energy crossing per unit area

= \frac{71\times10^3}{4 \times \pi\times(220)^2}

= 11.67 x 10⁻² Wm⁻²s⁻¹.

This is the intensity of the signal.

At 2200 m this intensity will further reduce by 100 times

So there it becomes equal to

11.67 x 10⁻⁴ Wm⁻² s⁻¹.

3 0
3 years ago
A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it
Savatey [412]

The charge of the object must be 1.11 \times e^{-5} \text { coulomb }

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

         Electric field, E=\frac{\text { Force }(F)}{q}

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

      q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }

6 0
3 years ago
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