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Anna007 [38]
3 years ago
12

Need help on this place

Physics
1 answer:
serious [3.7K]3 years ago
6 0
Which question? 10-13?
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A sound wave traveling eastward through air
Wewaii [24]
A sound wave traveling eastward through air<span> causes the air molecules to Vibrate east and west. </span>
5 0
3 years ago
Read 2 more answers
What do you measure when you measure an object’s mass?
Effectus [21]

Answer:

how much space it takes up in the world

Explanation:

1) Mass is a measurement of the amount of matter something contains, while Weight is the measurement of the pull of gravity on an object. 2) Mass is measured by using a balance comparing a known amount of matter to an unknown amount of matter. Weight is measured on a scale.

3 0
3 years ago
A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc
Nookie1986 [14]

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

<h3>Fբ = 75 N</h3>

  • Next, we shall determine the net force acting on the crate

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

<h3>Fₙ = 25 N</h3>

  • Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

<h3>a = 0.5 m/s²</h3>

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

8 0
3 years ago
How long does it take for a 3.5 kW electric water heater to heat 40 kg of water? from 20 ° C to 75 ° C? The specific heat capaci
iren [92.7K]

Answer:

2633.7 s

Explanation:

From the question,

Heat lost by the water heater = Heat gained by the water

Applying,

P = cm(t₂-t₁)/t.................. Equation 1

Where P = power of the heat, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature, t = time

make t the subject of the equation

t = cm(t₂-t₁)/P.............. Equation 2

From the question,

Given: c = 4190 J/kgK, P = 3.5 kW = 3500 W, m = 40 kg, t₁ = 20°C, t₂ = 75°C

Substitute these values into equation 2

t = 4190×40(75-20)/3500

t = 9218000/3500

t = 2633.7 s

4 0
3 years ago
A music fan at a swimming pool is listening to a radio on a diving platform. The radio is playing a constant- frequency tone whe
joja [24]

Answer:

The Doppler Effect is given by the following relation;

f' = \left (\dfrac{v + v_0}{v - v_s} \right) \times f

Where;

f' = The frequency the observer hears

f = Actual frequency of the wave

v = The velocity of the sound wave

v_o = The velocity of the observer

v_s = The velocity of the source

Where the observer is stationary, we have;

(i) When the source is moving in the direction of the observer

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(ii) When the source is receding from the observer, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

Therefore;

(a) A person left behind on the platform

For a person left behind on the platform, we have that the radio source is receding, therefore, we have;

f' = \left (\dfrac{v }{v + v_s} \right) \times f

(1) Given that (v + v_s) > v, therefore, v < (v + v_s), f' < f, the frequency heard by the person left on the platform, f', is smaller (lower) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively smaller

(b) A person down below floating on a rubber raft

For the the person down below on the rubber raft, the radio source is advancing

Therefore, the radio source is moving towards the person at rest down on the rubber raft, therefore, we have;

f' = \left (\dfrac{v }{v - v_s} \right) \times f

(1) Given that (v - v_s) < v, therefore, f' > f, the frequency heard by the person down below floating on the rubber raft, f', is greater (higher) than the frequency produced by the radio

(2) The frequency is not constant as the speed of the source is increasing while it under the acceleration due to gravity

(3) During the fall, the speed of the source continuously increases under the effect of gravitational attraction and therefore the frequency heard by the person on the platform becomes progressively greater (higher)

Explanation:

7 0
3 years ago
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