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2 years ago

Why the brakes of a car gets hotter than the brakes of a bicycle?

Physics

1 answer:

anygoal [31]2 years ago

4 0

Answer:

Frictional force

Explanation:

The brakes of a car gets hotter than the brakes of a bicycle because more frictional force is expended in applying the brake on a car than a bicycle.

To brake a car, the concept of friction is usually adopted. Two surfaces that increases friction when they come in contact are used in braking.

Frictional force is a force that resists the motion of a body. This forces helps to have grasp of motion.

A car requires more resistance to motion to break it because it have more more mass and other component motions parameters.

A bicycle is lighter and will require little friction.

Some of the frictional force applied is converted to heat energy. The amount of this energy expended is proportional to the frictional force.

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3 years ago

Calculate the electric charge of the glass ball

-Dominant- [34]

Answer:

As per Coulomb's law we know that force between two charges is given as

F = \frac{kq_1q_2}{r^2}F=

here we know that

q_1 = 2.5 \times 10^{-6} Cq

=2.5×10

−6

q_2 = -5.0 \times 10^{-6} Cq

=−5.0×10

−6

r = 0.0050 mr=0.0050m

now from above formula we will have

F = \frac{(9 \times 10^9)(2.5 \times 10^{-6})(5 \times 10^{-6})}{(0.0050)^2}F=

(0.0050)

(9×10

)(2.5×10

−6

)(5×10

−6

)

F = 4500 NF=4500N

so they will attract towards each other as they are opposite in nature with force F = 4500 N

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2 years ago

In a double-slit interference experiment, the wavelength is λ = 432 nm, the slit separation is d = 0.100 mm, and the screen is d

andrew-mc [135]

In the double-slit interference experiment, the distance of the nth-maximum from the center of the screen is given by
$y= \frac{n \lambda D}{d}$
where
$\lambda$ is the wavelength
D is the distance between the screen and the slits
d is the distance between the slits

In our problem,
$\lambda=432 nm= 432 \cdot 10^{-9} m$
$D=42.0 cm=0.42 m$
$d=0.100 mm=0.1 \cdot 10^{-3} m$

By applying the previous formula, we can calculate the distance of the 4th maximum from the center of the screen:
$y_4 = \frac{(4)(432 \cdot 10^{-9} m)(0.42 m)}{(0.1 \cdot 10^{-3}m)}=7.25 \cdot 10^{-3} m$

Similarly, the distance of the 8th- maximum is
$y_8 = \frac{(8)(432 \cdot 10^{-9} m)(0.42 m)}{(0.1 \cdot 10^{-3}m)}=14.5 \cdot 10^{-3} m$

Therefore, the distance between the two maxima is
$\Delta y=y_8- y_4 = 14.5 \cdot 10^{-3} m- 7.25 \cdot 10^{-3} m =7.25 \cdot 10^{-3} m = 7.25 mm$

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3 years ago