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inysia [295]
3 years ago
8

If the frequency of sound wave is doubled, the wavelength:

Physics
1 answer:
Ede4ka [16]3 years ago
4 0

Answer:

decreases

Explanation:

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If a 2 kg object is falling at 3 m/s at what rate is gravity working on the object
777dan777 [17]

Answer:

+9.8m/s^2

Explanation:

The rate of gravity of the object is constant thriughout the surface of the earth.

For falling object, the rate of gravity is positive since the body is coming down (falling)

The rate of gravity is negative if the body is going up

The constant value for acceleration due to gravity is 9.8m.s^2

Since the object is falling, hence the acceleration due to gravity is positive.

Rate of gravity working on the object will be +9.8m/s^2

4 0
3 years ago
In a discussion person A is talking 1.2 dB louder than person B, and person C is talking 3.2 dB louder than person A. What is th
liberstina [14]

Answer: 3.84dB

Explanation:

Since person A is talking 1.2dB louder than B, we will have

A = 1.2B... (1)

Similarly, person C is talking 3.2 dB louder than person A, we have

C = 3.2A... (2)

From equation 1, B = A/1.2... (3)

To get the ratio of the sound intensity of person C to the sound intensity of person B, we will divide equation 2 by 3 to give

C/B = 3.2A/{A/1.2}

C/B = 3.2A×1.2/A

C/B = 3.2×1.2

C/B = 3.84dB

3 0
3 years ago
Put the waves in order from highest frequency to lowest frequency
S_A_V [24]
BCA for sure, b the lines are showing more movement
6 0
3 years ago
Read 2 more answers
(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
3 years ago
Read 2 more answers
A 45.2 kg softball player slides across dirt with Uk=0.340. What is her acceleration?
alisha [4.7K]

Answer:

the ball didnt hit my face so

Explanation:

8 0
3 years ago
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