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3 years ago

On which factors does friction depend ?

Physics

1 answer:

goldenfox [79]3 years ago

3 0

Answer:

It depends on 2 factors

the nature of the materials

the normal force.

Explanation:

It depends on 2 factors

the nature of the materials that are in friction and the treatment that they have followed. This factor is expressed by a numerical value called the coefficient of friction or friction.

the force exerted by one body on the other, that is, the normal force.

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Six artificial satellites circle a space station at a constant speed. The mass m of each satellite, distance L from the space st

damaskus [11]

Answer:

The answer to the question is;

Based on their acceleration the rank of the satellites from largest to smallest is.

B >→ A >→ E >→ C >→ F >→ D.

Explanation:

Acceleration is given by $\frac{v^2}{r}$

Therefore the acceleration for each of the satellite is given by

Satellite A) $\frac{160^2}{5000}$ = 5.12 m/s²

Satellite B) $\frac{160^2}{2500}$ = 10.24 m/s²

Satellite C) $\frac{80^2}{2500}$ = 2.56 m/s²

Satellite D) $\frac{40^2}{10000}$ = 0.16 m/s²

Satellite E) $\frac{120^2}{5000}$ = 2.88 m/s²

Satellite F) $\frac{80^2}{10000}$ = 0.64 m/s²

Therefore in order of decreasing acceleration, from largest to smallest we have

Satellite B) > Satellite A) >Satellite E) >Satellite C)>Satellite F)>Satellite D).

3 0

3 years ago

Read 2 more answers

A standard AA battery provides 1.5 V. How many pennies and nickels would you need to include in a voltaic pile to produce the sa

GrogVix [38]

Answer: In a battery, voltage determines how strongly electrons are pushed through a circuit, much like pressure determines how strongly water is pushed through a hose. Most AAA, AA, C, and D batteries are around 1.5 volts. Imagine the batteries shown in the diagram are rated at 1.5 volts and 500 milliamp-hours.

Explanation: Today "AA" is frequently used as a size designation, irrespective of the battery's electrochemical system. The main numbers used for the most common NiMH and NiCad battery

8 0

2 years ago

a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic

algol13

Answer:

Approximately $281.25\; \rm m$ . (Assuming that the drag on this ball is negligible, and that $g = 10\; \rm m \cdot s^{-2}$ .)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

Horizontal: no acceleration, velocity is constant (at $v(\text{horizontal})$ is constant throughout the descent.)
Vertical: constant downward acceleration at $g = 10\; \rm m \cdot s^{-2}$ , starting at $0\; \rm m \cdot s^{-1}$ .

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: $x(\text{horizontal}) = 30\; \rm m$ . Combine these two quantities to find the duration of this descent:

$\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}$ .

In other words, the ball in this question start at a vertical velocity of $u = 0\; \rm m \cdot s^{-1}$ , accelerated downwards at $g = 10\; \rm m \cdot s^{-2}$ , and reached the ground after $t = 7.5\; \rm s$ .

Apply the SUVAT equation $\displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t$ to find the vertical displacement of this ball.

$\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}$ .

In other words, the ball is $281.25\; \rm m$ below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be $281.25\; \rm m\!$ .