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3 years ago

11

How much work is done when a barbell is lifted 2.5 meters into the

1 answer:

ohaa [14]3 years ago

8 0

Answer:

the formula given out a Distance x force you get work done

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A device that converts high voltages to lower voltages is a(n) ________.

AysviL [449]

Answer: A) Transformer

Explanation: A Transformer is an apparatus, device or a component in a system that is used to convert high voltage into low voltage. It is used to either increase or decrease the voltage of an alternating current.

A transformer uses the basic principle of electro magnetic induction, having two or more coils, the voltage is changed from one coil to another but with thesame frequency as alternating current energy passes through them.

7 0

3 years ago

An object moving north with an initial velocity of 14 m/s accelerates 5 m/s^2 for 20 seconds.what is the final velocity of the o

bekas [8.4K]

Answer:

Option C = 114 m/s

Explanation:

a=v-u/t

By substituting,

5=v-14/20

100=v-14

Thus, v=100+14

v= 114 m/s

Hope it helps :)

5 0

3 years ago

Read 2 more answers

37 Questions - All Questions Are Required

ANTONII [103]

Acceleration = vf-vi /t
10-22/3=2.6m/s^2

5 0

3 years ago

Read 2 more answers

Can someone please give me the (Answers) to this? ... please ...<br><br> I need help….

IceJOKER [234]

#1.

<em>Car </em>1<em> weighs </em>300 kilograms<em> and is moving right at </em>3 meters per second (m/s)

v1 (before) = 3 m/s

v2 (before) = -1 m/s

v1 (after) = 0.5 m/s

#2.

Law of conservation of momentum

momentum before collorion = momentim after collosion

MV + mv = MV' + mv'

1500x25+ 1000x5

37500 + 15000

6 0

2 years ago

Dust particles in air have a typical mass of 5.0 x 10-16 kg. They undergo irregular motion due to collisions with air molecules.

Mars2501 [29]

Answer:

Rms speed of the particle will be $38.68\times 10^8m/sec$

Explanation:

We have given mass of the air particle $m=5\times 10^{-16}kg$

Gas constant R = 8.314 J/mol-K

Temperature is given T = $27^{\circ}C=273+27=300K$

We have to find the root mean square speed of the particle

Which is given by $v_{rms}=\sqrt{\frac{3RT}{m}}=\sqrt{\frac{3\times 8.314\times 300}{5\times 10^{-16}}}=38.68\times 10^8m/sec$

So rms speed of the particle will be $38.68\times 10^8m/sec$

5 0

3 years ago