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Morgarella [4.7K]
3 years ago
10

WILL GIVE BRAINLIST: Does an objects weight affect how fast it falls?

Physics
1 answer:
maw [93]3 years ago
5 0

Answer:

it's depent on height and gravity

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A 1,000 kg ball traveling at 5 m/s would have
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15 because 5×5×5 is the same thing as 5×3 which equals to 15

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Unlike the idealized voltmeter, a real voltmeter has a resistance that is not infinitely large. part a a voltmeter with resistan
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The EMF of the battery includes the force to to drive across its internal resistance. the total resistance:  
R = internal resistance r + resistance connected rv 
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Now find the current:  
V 1= IR 
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find the voltage at the battery terminal (which is net of internal resistance) using  
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If riding a lawnmower engine exerts 19 hp in one minute to move the lawnmower how much work is done
ioda

Answer:

the work done by the lawnmower is 236.14 J.

Explanation:

Given;

power exerted by the lawnmower engine, P = 19 hp

time in which the power was exerted, t = 1 minute = 60 s.

1 hp = 745.7 watts

The work done by the lawnmower is calculated as follows;

Work = Energy = \frac{Power}{time} \\\\Work = \frac{(19 \times 745.7)}{60} = 236.14 \ J

Therefore, the work done by the lawnmower is 236.14 J.

6 0
3 years ago
The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an
9966 [12]

Answer:

-4.941*10^8J.

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

\Delta U = \Delta_{perogee}-\Delta_{Apogee}

Where,

U= \frac{-GmM_e}{r}

Replacing

\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}

\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})

Our values are given by,

m = 85.5Kg

M_e = 5.97*10^{24}Kg

r_A = 7330Km

r_p = 6610Km

G = 6.67*10^{-11}Nm^2/Kg^2

Replacing at the equation,

\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})

\Delta U = -4.941*10^8J

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was -4.941*10^8J.

4 0
3 years ago
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