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3 years ago

An example of intentionally increasing friction?

Physics

2 answers:

eimsori [14]3 years ago

5 0

You can increase friction intentionally by rubbing two objects together.

balu736 [363]3 years ago

3 0

When a car gets stuck in a marshy pond, we place a wooden block so that the friction increases and the car is helped out of the marsh. In this situation, we intentionally increased friction. Of course there are many other such examples. Hope that helped.

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How does the mass of the bob affect the number of swings of a pendulum?

gladu [14]

Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period

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3 years ago

FILL IN THE BLANKS here on earth, the pull of gravity on a mass of 1 kg is ......... newtons

Aleks [24]

Answer:

9.8 Newton

Explanation:

At average gravity on earth (conventionally, g=9.80665m/s2),

a kilogram mass exerts a force of about 9.8 newton

I hope this answer helps you

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3 years ago

The term used to describe the ability of a conduit to conduct flow is:

MissTica

The answer is , Conductance

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4 years ago

Read 2 more answers

How long will it take a running horse to travel 260 m and attain a speed at 12 m/s from rest?

Greeley [361]

Answer:

Explanation:

s=vt

t=s/v

s=260 m

v=vf-vi=12-0=12 m/s

therefore

t=260/12=21.67 sec

8 0

3 years ago

A 20-foot ladder is leaning against the wall. If the base of the ladder is sliding away from the wall at the rate of 3 feet per

tamaranim1 [39]

Answer:

<u>6.87 ft/s</u> is the rate at which the top of ladder slides down.

Explanation:

Given:

Length of the ladder is, $L=20\ ft$

Let the top of ladder be at height of 'h' and the bottom of the ladder be at a distance of 'b' from the wall.

Now, from triangle ABC,

AB² + BC² = AC²

$h^2+b^2=L^2\\h^2+b^2=20^2\\h^2+b^2=400----1$

Differentiating the above equation with respect to time, 't'. This gives,

$\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2$

In the above equation the term $\frac{dh}{dt}$ is the rate at which top of ladder slides down and $\frac{db}{dt}$ is the rate at which bottom of ladder slides away.

Now, as per question, $h=8\ ft, \frac{db}{dt}=3\ ft/s$

Plug in $h=8$ in equation (1) and solve for $b$ . This gives,

$8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=\sqrt{336}=18.33\ ft$

Now, plug in all the given values in equation (2) and solve for $\frac{dh}{dt}$

$8\times \frac{dh}{dt}+18.33\times 3=0\\8\times \frac{dh}{dt}+54.99=0\\8\times \frac{dh}{dt}=-54.99\\ \frac{dh}{dt}=-\frac{54.99}{8}=-6.87\ ft/s$

Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.

8 0

4 years ago