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stepan [7]
3 years ago
11

A car is stopped at a red light. When the light turns green, the car accelerates until it reaches a final velocity of 45 m/s. It

takes the car 12 s to reach this speed. How far does the car travel during this time.
Physics
1 answer:
sertanlavr [38]3 years ago
7 0

Speed at the beginning  =  zero

Speed after 12 seconds  =  45 m/s

Average speed during the 12 seconds  =  (1/2) · (0 + 45)  =  22.5 m/s

Distance covered in 12 seconds at an average speed of 22.5 m/s  :

Distance = (speed) x (time)

Distance = (22.5 m/s) x (12 sec)

<em>Distance   =  270 meters </em>  

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The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

<em>T</em> - 25 N = (8 kg) <em>a</em>

where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.

The hanging mass has a net force of

(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>

where <em>g</em> = 9.8 m/s².

Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :

(<em>T</em> - 25 N) + ((6 kg) <em>g</em> - <em>T </em>) = (14 kg) <em>a</em>

33.8 N = (14 kg) <em>a</em>

<em>a</em> = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

<em>T</em> - 25 N = (8 kg) (2.4 m/s²)

<em>T</em> ≈ 25 N + 19.31 N ≈ 44 N

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Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for
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2. For each of the listed parts of a power plant, make a selection to indicate in what
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Answer: Find the answer in the explanation

Explanation: Given the Roman numeral and the representation

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II. part of a nuclear power plant

III. part of a coal-fired power plant and part of a nuclear power plant

a.) Boiler : I

b.) Combustion chamber: I

c.) Condenser: I

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7 0
3 years ago
A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b
jonny [76]

Answer:

Part a)

I = 3.16 \times 10^{-5} W/m^2

Part b)

P_o = 0.162 Pa

Explanation:

Part a)

Level of sound = 75 dB

now we know that

L = 10 Log\frac{I}{I_0}

here we know that

I_0 = 10^{-12} W/m^2

now we have

75 = 10 Log(\frac{I}{10^{-12}})

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Part b)

Intensity of sound wave is given as

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here we know that

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so we have

I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c

I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}

now we know

\rho = 1.2 kg/m^3

c = 340 m/s

B = 1.4 \times 10^5 Pa

now we have

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P_o = 0.162 Pa

5 0
3 years ago
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