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3 years ago

8

(True/False) According to Newton’s third law of motion, it is true that an object moving at a constant speed in a straight path

will continue to do so until a net force acts upon it.

1 answer:

algol133 years ago

7 0

The statement itself is true. But to be completely and compulsively correct, that's Newton's FIRST law of motion, not the third.

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Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac

Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

$l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h$ #The speed toward each other.

$\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h$

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0

2 years ago

2500000000000 in scientific notation

Alex_Xolod [135]

Your answer is 2.5 x 10^(11).

Please take the time to rate and thank me if you feel I helped you! Thanks, it means a lot to me. :)

4 0

2 years ago

Read 2 more answers

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

Muscular Strength exercises focus on high______and low___

Advocard [28]

Answer:

d d d d dd d d d d d dd d d d d dd d

4 0

2 years ago

A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in

chubhunter [2.5K]

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

m v + Mu = m v' + M u'

0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

5 0

2 years ago