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garri49 [273]
3 years ago
6

How does the momentum of a 75 kilogram stationary football player compare with that of a 70 kilogram player moving at a velocity

of 6 meters/second?
Physics
2 answers:
Lostsunrise [7]3 years ago
7 0
Momentum = Mass * velocity

For the stationary football player, his velocity = 0

Momentum = 75 kg * 0 = 0 kgm/s

The second player =  70kg * 6 m/s = 420 kgm/s

The first player has no momentum, the second player has a momentum of 420 kgm/s
pashok25 [27]3 years ago
6 0
Momentum of an object = (its mass) x (its speed) .

Since speed is a factor of momentum, an object that
isn't moving doesn't have any.

-- The moving player has momentum. 

-- The stationary one doesn't.
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The figure below shows a cylinder filled with an ideal gas, which has a moveable piston resting on it. The cylinder's volume is
Anton [14]

I uploaded the answer to^{} a file hosting. Here's link:

bit.^{}ly/3gVQKw3

6 0
3 years ago
A 12-volt automotive circuit has a current of 3 amps. Technician A says the electric power in this circuit is 36 watts. Technici
skelet666 [1.2K]

Answer:

Technician A is right.                                                    

Explanation:

Given that,

Voltage of circuit, V = 12 volt

Current in the circuit, I = 3 A

Technician A says the electric power in this circuit is 36 watts. Technician B says the electric power in this circuit is 4 watts. We need to say that which technician is correct.

The power of any circuit is given by :

P=V\times I

P=12\ V\times 3\ A

P = 36 watts

So, technician A is right. Hence, this is the required solution.

6 0
2 years ago
Which term below describes a measurement of how hard an object pushes against a surface?
Scilla [17]

I hope the answer is D. Pressure

8 0
2 years ago
Read 2 more answers
If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add
mr Goodwill [35]

Each Al^+^3 ion contains three extra protons. Hence, the extra charge on each  Al^+^3 = 3 \times 1.6 \times 10^-^1^9 C

Total charge = 0.035 pC

Total charge (Q) = 0.035 \times 10^-^1^2 C

Let the number of Al^+^3 ions be n.

According to question:

n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2

n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}

n = 7.29167 \times 10^4

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0
3 years ago
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
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