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aleksandr82 [10.1K]
3 years ago
5

The greater the amplitude of a wave a. The quieter the sound b. The higher the frequency of the sound c. The dimmer the light in

tensity d. The greater the amount of energy in the waveImmersive Reader (1 Point) a b c d
Physics
1 answer:
Stolb23 [73]3 years ago
8 0

Answer:

The greater the amplitude the greater the energy.

(Think of a water wave - which carries greater energy a 1 ft wave or

a 10 ft wave)

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The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?
postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

Initial\ Distance = 2cm ---- r_1

New\ Distance = 4cm --- r_2

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

F_1 = \frac{kq_1q_2}{r_1^2} --- Coulomb's law

Substitute 2 for r1

F_1 = \frac{kq_1q_2}{2^2}

F_1 = \frac{kq_1q_2}{4}

The new force (F2) is

F_2 = \frac{kq_1q_2}{r_2^2}

Substitute 4 for r2

F_2 = \frac{kq_1q_2}{4^2}

F_2 = \frac{kq_1q_2}{4*4}

F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}

Substitute F_1 = \frac{kq_1q_2}{4}

F_2 = \frac{1}{4}*F_1

F_2 = \frac{F_1}{4}

The new force is 1/4 of the previous force.

3 0
2 years ago
9- Under what circumstances would a vector have components that are equal in
valkas [14]

Explanation:

c. if the vector is oriented at 0° from the X -axis.

6 0
2 years ago
What is the answer to my question
Andrew [12]
Well that depends...what is your question?
7 0
3 years ago
Read 2 more answers
A force of 10N is required to stretch a spring from 20cm to 25cm. What is the spring constant in N/m2 Be careful of unit
kupik [55]

Answer:

C) 40 N/m

Explanation:

If we ASSUME that the spring is un-stretched at the zero cm position

k = F/Δx = 10/0.25 = 40 N/m

5 0
3 years ago
Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti
liq [111]

Answer:

The speed of water must be expelled at 6.06 m/s

Explanation:

Neglecting any drag effects of the surrounding water we can assume the linear momentum in this case is conserves, that is, the total initial momentum of the octopus and the water kept in it cavity should be equal to the total final linear momentum. That's known as conservation of momentum, mathematically expressed as:

p_f=p_i

with Pi the total initial momentum and Pf the final total momentum. The total momentum is the sum of the momentums of the individual objects, in our case the octopus and the mass of water that will be expelled:

p_{of}+p_{wf}=p_{oi}+p_{wi}

with Po the momentum of the octopus and Pw the momentum of expelled water. Linear momentum is defined as mass times velocity:

m_o*v_{of}+m_w*v_{wf}=m_o*v_{oi}+m_w*v_{wi}

Note that initially the octopus has the water in its cavity and both are at rest before it sees the predator so v_{oi}=v_{wi} = 0\frac{m}{s}:

m_o*v_{of}+m_w*v_{wf}=0

We should find the final velocity of water if the final velocity of the octopus is 2.70 m/s, solving for v_{wf}:

v_{wf}=-\frac{m_o*v_{of}}{m_w}=-\frac{(6.00-1.85)*(2.70)}{1.85}

v_{wf}=-6.06\frac{m}{s}

The minus sign indicates the velocity of the water is opposite the velocity of the octopus.

3 0
3 years ago
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