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Oksanka [162]
3 years ago
12

You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction

force felt by the desk?
Physics
1 answer:
const2013 [10]3 years ago
7 0

If the desk doesn't move, then it's not accelerating.

If it's not accelerating, then the net force on it is zero.

If the net force on it is zero, then any forces on it are balanced.

If there are only two forces on it and they're balanced, then they have equal strengths, and they point in opposite directions.

So the friction on the desk must be equal to your<em> 245N</em> .

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Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

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(A) The force acting on the object is given by :

F = kx

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F = ma

a is acceleration at that instant

a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2

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Gravitational attraction depends on the mass of the objects as well as their distance. The gravitational force between objects i
shepuryov [24]
<h2>Answer: Gravitational attraction will be the same</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}    (1)

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

Now, if we double both masses and the distance also doubles, this means:

m_{1} and m_{2} will be now 2m_{1} and 2m_{2}

r will be now 2r

Let's rewrite the equation (1) with this new values:

F=G\frac{(2m_{1})(2m_{2})}{(2r)^2}    (2)

Solving and simplifying:

F=4G\frac{m_{1}2m_{2}}{4r^2}    

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As we can see, equation (3) is the same as equation (1).

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7 0
3 years ago
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Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

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3 years ago
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Answer:

is the same

Explanation:

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