All categories

4 years ago

A body is placed on a rough inclined plane. Why does the frictional force decrease with the increase of angle of inclination

1 answer:

kondor19780726 [428]4 years ago

7 0

The frictional force is directly proportional to the force that is perpendicular on the surface.

When the body is placed on a horizontal level with zero inclination, the only force acting on the body is the gravitational force which always pulls the body down. The gravitational force, in this case, is the perpendicular force to the surface. Accordingly, this entire force is used to generate friction

Now as the inclination of the surface increases, the gravitational force is no longer the perpendicular force of the body, its value decreases, which means only a part is used to generate frictional force. Consequently, frictional force decreases.

When the inclination reaches 90 degrees, the gravitational force does not act along the normal and accordingly, no friction force is generated.

You might be interested in

To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th

Kryger [21]

Answer:

$H = 10.05 m$

Explanation:

If the stone will reach the top position of flag pole at t = 0.5 s and t = 4.1 s

so here the total time of the motion above the top point of pole is given as

$\Delta t = 4.1 - 0.5 = 3.6 s$

now we have

$\Delta t = \frac{2v}{g}$

$3.6 = \frac{2v}{9.8}$

$v = 17.64 m/s$

so this is the speed at the top of flag pole

now we have

$v_f - v_i = at$

$17.64 - v_i = (-9.8)(0.5)$

$v_i = 22.5 m/s$

now the height of flag pole is given as

$H = \frac{v_f + v_i}{2}t$

$H = \frac{22.5 + 17.64}{2} (0.5)$

$H = 10.05 m$

5 0

3 years ago

A SPARK PLUG IN AN AUTOMOBILE ENGINE CONSISTS OF TWO

kirill [66]

Answer:

Electric potential, $V=3.52\times 10^{12}\ volts$

Explanation:

It is given that,

The magnitude of electric field, $E=4.7\times 10^7\ V/m$

Distance between automobile engines that consists of metal conductors, $d = 0.75\ mm = 7.5\times 10^4\ m$

Let V is the magnitude of the potential difference between the conductors. The relation between the electric field and the electric potential is given by:

$V=E\times d$

$V=4.7\times 10^7\ V/m \times 7.5\times 10^4\ m$

$V=3.52\times 10^{12}\ volts$

So, the magnitude of the potential difference between the conductors is $3.52\times 10^{12}\ volts$ . Hence, this is the required solution.

6 0

4 years ago

WILL GLADLY GIVE BRAINIEST FOR THIS EASY QUESTION

Alekssandra [29.7K]

The atomic number of Copper is 29. So immediately we know that there are 29 protons in the nucleus of each Copper atom, and 29 electrons in the cloud surrounding the nucleus of each [neutral] Copper atom.

Copper has two stable isotopes ... ⁶³Copper and ⁶⁵Copper. So if the Copper atom you're holding in your hand is not radioactive, then it has either (63-29)=34 neutrons or (65-29)=36 neutrons in its nucleus. (There's never a need to round the number of neutrons, because there's no such thing as a part of a neutron.)

4 0

3 years ago

Read 2 more answers

True or false solids possess a greater level of kinetic energy than gases

Kipish [7]

Answer:

False

Explanation:

Solid particles have the least amount of energy, and gas particles have the greatest amount of energy. The temperature of a substance is a measure of the average kinetic energy of the particles. A change in phase may occur when the energy of the particles is changed.

8 0

3 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago