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suter [353]
3 years ago
13

What forces act upon an object dropped in a vacuum

Physics
1 answer:
Vlada [557]3 years ago
8 0
Gravity is the only one, since there's no air resistance.
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Can someone explain how they got their answer or how I get the change in number? :(
enyata [817]

Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.

They then subtract the start time of the second quarter from the end time of the first quarter.

I hope this helps, good luck! :D

5 0
2 years ago
a 8000kg tow truck pulls a 3000kg car horizontally into a garage. if the motor of the tow truck exerts a forward thrust on the t
Evgen [1.6K]
Equation 1 :
m1 :   m1a =T 
Equation 2:m2 : m2a= F - T 

Adding 1 and 2a=0.073

Placing in  equation 1

we get T = 218.18N
8 0
3 years ago
Recall the topography of the Sierra Nevada Mountains east of California's central valley. Which of the following locations would
marysya [2.9K]

Answer:

3. western slope

Explanation:

If we investigate the physical features (such as natural and artificial features) of the Sierra Nevada Mountains located at the California's central valley eastern part, the western slope has the highest value of the precipitation level yearly. This is due to the location of the slope and the relevant conditions such as temperature and humidity.

7 0
3 years ago
A gas is placed in a storage tank at a pressure of 49.2 atm at 39.0C . As a safety device, a small metal plug in the tank is mad
Amiraneli [1.4K]

Answer:

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=49.2 atm\\T_1=39.0^oC = 312.15 K\\P_2=?\\T_2=198^oC=471.15 K

Putting values in above equation, we get:

\frac{49.2atm }{312.15 K}=\frac{P_2}{471.15 K}\\\\P_2=74.26 atm

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.

4 0
3 years ago
A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a
UNO [17]

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

6 0
3 years ago
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