Your physics textbook is sliding to the right across the table. identify all forces acting on the object.

2 answers:

7 0

There are in total 4 forces acting on the object. Let's call m the mass of the object and g the gravitational acceleration. We have:

1) The force that is pushing the book, making it sliding along the table. This force acts horizontally, in the same direction of the displacement.

2) The frictional force between the textbook and the table, acting horizontally in the opposite direction of the displacement. This force is equal to
$F=\mu m g$
where $\mu$ is the coefficient of friction.

3) The weight of the object, acting vertically and downwards, equal to
$W=mg$

4) The reaction force of the table against the textbook, acting vertically and upwards, equal to the weight of the object.

OverLord2011 [107]3 years ago

6 0

Answer:

The Force of gravity, Normal Force, Kinetic Force

Explanation:

Gravity is the force downwards.

Normal force is the force upwards from the object.

Kinetic force is the force pointing left of the object, kinetic because the object is moving (it would be static force if the object were no moving)

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An insulator can do which of the following? Conduct charge through it. Become positively or negatively charged. Transfer protons

ser-zykov [4K]

The correct answer is:
<span>Become positively or negatively charged

In fact, an insulator is a material where charges cannot move freely. Therefore, it can be positively or negatively charge (for example, if it is rubbed against another object, the insulator can remain with an excess of charge), but it cannot transfer charge to other objects.</span>

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3 years ago

A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?

posledela

Answer:

$118\; \rm Hz$ .

Explanation:

The frequency $f$ of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of $v= 295\; \rm m\cdot s^{-1}$ . In other words, the wave would have traveled $295\; \rm m$ in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that $295\; \rm m$ ? The wavelength of this wave $\lambda = 2.50\; \rm m$ gives the length of one wave cycle. Therefore:

$\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118$ .

That is: there are $118$ wave cycles in $295\; \rm m$ of this wave.

On the other hand, Because that $295\; \rm m$ of this wave goes through that point in each second, that $118$ wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

$f = 118\; \rm s^{-1} = 118\; \rm Hz$ .