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3 years ago

7

Need 3 examples of unbalanced forces

2 answers:

enyata [817]3 years ago

5 0

a roller coaster going down the first drop.

jarptica [38.1K]3 years ago

4 0

1. pulling tug of war
2. monkey swinging from vine
3. arm wrestle

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Infer why the output force exerted by a rake must be less than input force?

adell [148]

<h3><u>Answer and Explanation</u>;</h3>

input force refers to the force exerted on a machine, also known as the effort, while the output force is the force machines produce or the Load. The ratio of output force to input force gives the mechanical advantage of a simple machine
<em><u>The output force exerted by the rake must be less than the input force because one has to use force while raking. The force used to move the rake is the input force. </u></em>
<em><u>The rake is not going to be able to convert all of the input force into output force, the force the rake applies to move the leaves, because of friction.</u></em>

5 0

3 years ago

A solid sphere rolls without slipping down an incline, starting from rest. At the same time, a box starts from rest at the same

NISA [10]

Answer:

The box arrives first.

Explanation:

Hope this helps!! :))

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brainly.com/question/20164917

4 0

2 years ago

Mr Johnson launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, how much time does it take

vazorg [7]

Answer:c

Explanation:

3 0

2 years ago

A wave has a wavelength of 50 m and a speed of 200 m/s. What is the frequency of this wave in hertz?

Kobotan [32]

Answer:

freq= 200/50= 4 Hz

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3 0

2 years ago

Three equal 1.55-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the

kati45 [8]

Answer:

0.12959085 J

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

$U=k\dfrac{q^2}{d}$

In this system with three charges which are equidistant from each other

$U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}$

$\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J$

The potential energy of the system is 0.12959085 J

6 0

2 years ago