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brilliants [131]
3 years ago
15

What is the acceleration of an object in free fall near Earth's surface?

Physics
2 answers:
Vitek1552 [10]3 years ago
8 0

Answer:

D 9.8 m/s^2

Explanation:

The force of gravitational gravity on earth is 9.8 m/s^2

Licemer1 [7]3 years ago
8 0
D) 9.8 m/(s^2) is the approximate acceleration of a free falling object near Earth’s surface
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A weather balloon has a volume of 90.0 l when it is released from sea level. What
frozen [14]

The final atmospheric pressure is 5.19\cdot 10^4 Pa

Explanation:

Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,

pV=const.

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

p_1 V_1 = p_2 V_2

where in our problem we have:

p_1= 1.03\cdot 10^5 Pa is the initial pressure (the atmospheric pressure at sea level)

V_1 = 90.0 L is the initial volume

p_2 is the final pressure

V_2 = 175.0 L is the final volume

Solving the equation for p2, we find the final pressure:

p_2 = \frac{p_1 V_1}{V_2}=\frac{(1.01\cdot 10^5)(90.0)}{175.0}=5.19\cdot 10^4 Pa

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

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#LearnwithBrainly

3 0
3 years ago
Two vectors are being added, one at an angle of 20.0 , and the other at 80.0. The only thing you know about the magnitudes is th
postnew [5]

Answer:a

Explanation: they are all positive

4 0
2 years ago
How many significant figures are in 50.5?
Art [367]
3 as a single number is considered a sf
7 0
3 years ago
A fire engine is moving south at 35 m/s while blowing its siren at a frequency of 400 Hz.
vodomira [7]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as

f_d= f_s \frac{(v+v_d)}{(v-v_s)}

Here,

f_d=frequency received by detector

f_s=frequency of wave emitted by source

v_d=velocity of detector

v_s=velocity of source

v=velocity of sound wave

Replacing we have that,

f_d = 400(\frac{(343+18)}{(343-35)})

f_d=422 Hz

Therefore the frequencty that will hear the passengers is 422Hz

8 0
3 years ago
A loaded barge has a mass of 1 500 000 kg and is traveling at 3 m/s. If a tugboat applies an opposing force of 12 000 N for 10 s
yan [13]

Answer:

Explanation:

Initial momentum is 1.5e6(3) = 4.5e6 kg•m/s

An impulse results in a change of momentum

The tug applied impulse is 12000(10) = 120000 N•s or 0.12e6 kg•m/s

The remaining momentum is 4.5e6 - 0.12e6 =  4.38e6 kg•m/s

The barge velocity is now 4.38e6 / 1.5e6 = 2.92 m/s

The tug applies 0.012e6 N•s of impulse each second.

The initial barge momentum will be zero in

t = 4.5e6 / 0.012e6 = 375 s or 6 minutes and 15 seconds

To stop the barge in one minute(60 s), the tug would have to apply

4.5e6 / 60 = 75000 N•s /s or 75 000 N

5 0
2 years ago
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