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12345 [234]
3 years ago
5

When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting

on the box is
1) 0N
2) Between 0N and 12N
3) Exactly 12 N
4) Greater than 12N​
Physics
1 answer:
Mrrafil [7]3 years ago
3 0

Answer:

12N

Explanation:

when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied

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20. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to
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Given:\\V_1=4m^3\\T_1=290K\\p_1=475kPa\\\\V_2=6.5m^3\\T_2=277K\\\\Find:\\p_2=?\\\\Solution:\\\\ \frac{pV}{T} =const.\\\\ \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \\\\\frac{p_1V_1T_2}{T_1}=p_2V_2\\\\\frac{p_1V_1T_2}{T_1V_2}=p_2\\\\p_2=p_1 \frac{V_1}{V_2}  \frac{T_2}{T_1} \\\\\\p_2=475kPa\cdot  \frac{4m^3}{6.5m^3} \cdot  \frac{277K}{290K} \approx 279.2kPa\\\\Correct\;is\;answer\;\;C
7 0
3 years ago
The slope of a velocity versus time graph gives
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4 0
3 years ago
How much work is required to stop an electron (m = 9.11 \times 10^ - 31 kg) which is moving with a speed of 2.10
Karo-lina-s [1.5K]

Answer:

-2.00876\times 10^{-18}\ J

Explanation:

v = Speed of electron = 2.1\times 10^6\ m/s (generally the order of magnitude is 6)

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Work done would be done by

W=K_i-K_f\\\Rightarrow W=0-\dfrac{1}{2}mv^2\\\Rightarrow W=0-\dfrac{1}{2}\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\\\Rightarrow W=-2.00876\times 10^{-18}\ J

The work required to stop the electron is -2.00876\times 10^{-18}\ J

6 0
3 years ago
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