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3 years ago

We reduce friction in machines? why

Physics

2 answers:

Alexxandr [17]3 years ago

6 0

Answer:

Because it causes a lot of wear and tear in machine parts that move against each other. It erodes the surfaces and destroys their symmetries

Explanation:

kvv77 [185]3 years ago

5 0

Answer:

friction reduces the efficiency of machines, thus we must reduce the friction force that is acting upon it.

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Please answer ASAP!!!

NARA [144]

The third (left hand corner) since the x and y are both negative.

4 0

3 years ago

What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a

Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

$A = W + \dfrac{W}{2}$

$A =\dfrac{3W}{2}$

$A =\dfrac{3mg}{2}$

When the car is at the bottom, the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

$N = m g + \dfrac{mv^2}{r}$

$\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}$

$\dfrac{mg}{2} = \dfrac{mv^2}{r}$

$v = \sqrt{\dfrac{rg}{2}}$

$v = \sqrt{\dfrac{40\times 9.8}{2}}$

v = 14 m/s

8 0

2 years ago

Help plzzzzzzz i need thissssssssss

frutty [35]

Answer:

The final graph

Explanation:

The graph that curves downwards is negative acceleration. While the position decreases the slop increases.

8 0

2 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :