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saw5 [17]
2 years ago
6

You hear the engine roaring on a race car at the starting line. Predict the changes in the sound as the race starts and the car

approaches the camera. A) The sound of the engine will get louder and the pitch lower. Eliminate B) The sound of the engine will get louder and the pitch higher. C) The sound is determined by the car and it will not change in any way. D) The sound of the engine will get louder but the pitch will not change.
Physics
2 answers:
gogolik [260]2 years ago
4 0

B. The sound of the engine will get louder and the pitch higher.

Mariana [72]2 years ago
3 0

Answer: The correct option is B.

Explanation:

Doppler effect is the phenomenon in which there is an apparent change in the frequency when there is relative motion between source and listener.

When the source and the listener are approaching each other then the frequency increases. The loudness depends on the amplitude. The energy of the wave is directly proportional to the square of the amplitude.

In the given problem, the engine is roaring on a race car at the starting line. There is change in the sound as the race starts and the car approaches the camera.

Pitch depends on the frequency.

Therefore, the sound of the wave will get louder and the pitch will get higher.

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Anna needs to move a box of paperback books across the room. If she applies a force of 20 newtons to the box, what is the magnit
vlabodo [156]
Newton's third law tells us that for every force there is an equal and opposite force.  This means that if Anna exerts a force of 20 Newtons on the box, the box exerts a force of 20 Newtons on Anna.
5 0
2 years ago
Read 2 more answers
The velocity versus time graph of particle A is tangent to the velocity versus time graph for particle B at point O. What is the
lara [203]
As velocities are tangent, the value of both Particle A and Particle B would be same for that point O (Intersecting point)

a = v / t
Here, v = 7, t = 6
So, a = 7/6
a = 1.17 
As the graph is decreasing, value of acceleration would be negative.
So, a = -1.17 m/s²

In short, Your Answer would be Option C

Hope this helps!
7 0
2 years ago
Joule’s law is a linear relationship, that is, the more heat you provide, the greater the temperature change. However, in the pr
Nitella [24]

Answer:

hydrogen bridge

Explanation:

Joule's relationship to heat and temperature is true for all materials where we assume that interatomic forces are linear, when atoms separate these forces decrease. There is a point where the separation between atoms is enough that thermal agitation can separate the molecules and there is a change of state, generally from solid to liquid and from liquid to vapor. When these changes of state are occurring all the energy supplied is used to break the links, so the temperature does not change.

In the specific case of water, there is a bond called a hydrogen bridge that breaks around 4ºC, therefore, at this temperature there is a deviation from the curve since this link is being broken, this does not lead to a change of macroscopic state.

For the other temperatures the water behaves like the other bodies.

7 0
2 years ago
A food department is kept at â12°c by a refrigerator in an environment at 30°c. the total heat gain to the food department is
slava [35]

As per energy conservation in the reversible engine we can say

Q_2 + W = Q_1

here we know that

Q_2 = 3300 kJ/h

Q_1 = 4800 kJ/h

now from above equation

3300 + W = 4800

W = 1500 kJ/h

now we can convert it into kW

W = 1500\times \frac{kJ}{3600s}

W = 0.42 kW

so above is the power input to the refrigerator

now to find COP we know that

COP = \frac{Q_2}{W}

COP = \frac{3300}{1500} = 2.2

so COP of refrigerator is 2.2

3 0
2 years ago
A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t
ipn [44]

Answer:

ball clears the net

Explanation:

v_{o} = initial speed of launch of the ball = 20 ms^{-1}

\theta = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

v_{ox} = initial velocity = v_{o} Cos\theta = 20 Cos5 = 19.92 ms^{-1}

t = time of travel

X = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}       Eq-1

Consider the motion of the ball along the vertical direction

v_{oy} = initial velocity = v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}

t = time of travel

Y_{o} = Initial position of the ball at the time of launch = 2 m

Y = Final position of the ball at time "t"

a_{y} = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

Y = Y_{o} + v_{oy} t + (0.5) a_{y} t^{2}\\Y = 2 + (20 Sin5) (\frac{7}{20 Cos5}) + (0.5) (- 9.8) (\frac{7}{20 Cos5})^{2}\\Y = 2.00758 m\\

Since Y > 1 m

hence the ball clears the net

7 0
2 years ago
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