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ivann1987 [24]
3 years ago
11

A 5.0 kg rock is dropped from the top of a building. The speed of the rock after it has fallen for 2.2 seconds is?

Physics
1 answer:
PSYCHO15rus [73]3 years ago
5 0
5x2.2/3 Because that's how you calculate velocity.

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D. voltage source. ttyl

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The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able
vovangra [49]

Answer:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

q_e=-1.6\cdot 10^{-19}\ C

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is 5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C

Let's test the possible charges listed in the question:

-8.0 \times 10 ^{-19 }. We have just found it's a possible charge of a particle

-3.2 \times 10 ^{-19 }. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

-1.2 \times 10 ^{-19 } this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 } cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge -4.8 \times 10 ^{-19 }\ C is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C

5 0
2 years ago
amping equipment weighing 6.0 kN is pulled across a frozen lake by means of ahorizontal rope. The coecient of kinetic friction i
Harman [31]

Answer:

Work done = 422.45 kJ

Explanation:

given,                                  

weight of equipment = 6 kN      

coefficient of kinetic friction = 0.05          

distance up to which it is pulled = 1000 m

constant acceleration = 0.2 m/s²                    

Work done by the camper = ?                

actual acceleration acting a'      

m a = m a' - μ mg            

a' = a + μ g                      

a' = 0.2 + 0.05 x 9.8                

a' = 0.69 m/s²                              

Work done = Force x distance

F = m a'                                                            

F = \dfrac{6000}{9.8} \times 0.69

F = 422.44897 N                            

Work done = F x d                          

Work done = 422.44897 x 1000

Work done = 422449 J                  

Work done = 422.45 kJ            

4 0
3 years ago
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