All categories

3 years ago

What is friction and explain

2 answers:

8 0

Friction is the force you get when you (for example) Rub something with another, it's a force that may generate heat and even some resistance. Another example is rubbing your hands together, they get hot, therefore friction is working, without friction you wouldn't be able to stop moving.

ASHA 777 [7]3 years ago

5 0

Friction is the resistance of mass when on another object.
If we avoid friction, we move faster.
But, in some cases, friction is also good. It is used in car brakes, when something rubs on the wheels to make the wheels stop rolling, or rotating.
Ice skates may be another thing. They provide a little friction on the blades, to be able to stop and to be able to move.

You might be interested in

A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan

PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d) V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

55 = 0 + 0 + 0.5*(9.81)*t^2

t = sqrt ( 55 * 2 / 9.81 )

t =3.349 s

- Use the second equation of motion in x direction:

x(f) = x(0) + V_x,i*t

150 = 0 + V_x,i*3.349

V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

V_y,f = V_y,i + g*t

V_y,f = 0 + 9.81*3.349

V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

V^2 = V_y,f^2 + V_x,i^2

V = sqrt (32.85^2 + 44.8^2)

V = 55.55 m/s

5 0

3 years ago

A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

8 0

3 years ago

Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface be

miv72 [106K]

consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

Y = v₀ t + (0.5) a t²

h = (0) (3.0) + (0.5) (9.8) (3.0)²

h = 44.1 m

4 0

3 years ago

The radius of Earth is . The average Earth-Sun distance is . How many Earths would fit between Earth and the Sun if they are sep

Verdich [7]

Answer:

The radius of the earth is 6,371 km.

The average Earth-Sun distance is 152.09 million km

How many Earths would fit between Earth and the Sun if they are separated by their average distance? Approximately 11,936 Earths.

I didn't really understand the last part, but if you don't get a better answer please mark me as brainliest.

6 0

3 years ago

A wire perpendicular to the screen carries a current in the direction shown.

____ [38]

A) Up is the direction of the magnetic field at point Z.

4 0

3 years ago

Read 2 more answers