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Trava [24]
4 years ago
15

How would I answer this equation? E= mc^2 for m

Physics
1 answer:
zubka84 [21]4 years ago
5 0

The E stands for energy. M stands for mass c squared stands for speed of light squared

energy= mass speed of light squared

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NEED HELP ASAP!!!!!!!!!!!!!
love history [14]

Answer: Decreased risk of heart attack

Explanation:

thats the answer because it actually

increase rish of heart attack

Hope this helps :)

5 0
3 years ago
Read 2 more answers
A train moves from rest to a speed of 30 m/s in 32.0 seconds. What is its acceleration?​
kramer

Answer:

0.94m/s (squared)

Explanation:

a=change in velocity/time

a=v-u/t

a=32-0=32/32

a=0.94m/s (squared)

6 0
3 years ago
The polar coordinates of the collar A are given as functions of time in seconds by r = 2+ 0.7 t2 ft and ????= 3.5t rad. What are
r-ruslan [8.4K]

Answer with explanation:

Part a)

v_{radial}=\frac{dr}{dt}=\frac{d(2+0.7t^{2})}{dt}\\\\v_{radial}=1.4t\\\\\therefore v_{radial}|_{t=4}=1.4\times 4=5.6ft/s\\\\v_{angular}=r|_{t=4}\times \frac{d\theta }{dt}=13.2\frac{3.5t}{dt}=46.2fts^{-1}\\\\\therefore v=\sqrt{v_{radial}^{2}+v_{angular}^{2}}\\\\v=46.53ft/s

Part b)

a_{radial}=\frac{d^{2}r}{dt^{2}}=\frac{d^{2}(2+0.7t^{2})}{dt^{2}}\\\\a_{radial}=1.4ft/s^{2}\\\\a_{angular}=r\times \frac{d^{2}\theta }{dt^{2}}\\\\a_{angular}=r\times \frac{d^{2}(3.5t) }{dt^{2}}\\\\\therefore a_{angular}=0\\\\\therefore Accleration=1.4ft/s^{2}

8 0
3 years ago
A pump is used to lift 100 KG of water from a wel 60 m deep,in 20 S If force of gravity on 1 KG is 10 N,find
Harlamova29_29 [7]

Explanation:

Given,

  • m = 100 kg
  • g = 10 N/kg¹
  • h = 60 m
  • t = 20 s

To Find:

a) Work done by the pump

b) Potential energy stored in the water

c)Power spent by the pump

d)Power rating of the pump.

Solution:

  • a) Work done by the pump

We know that,

\rm \: Work  \: done = Force * Distance  \: moved

  • f = 100 kg * 10N/kg
  • d = 60 m

\rm \: Work\; Done =(100 \: kg \times  \cfrac{10N}{kg} ) \times 60 \: m

\rm \: Work\; Done =1000 \times 60 \: joule

\boxed{\rm \: Work\; Done =60000 \: joule}

[The unit'll be joule since N×M = J]

  • b) Potential energy stored in the water

\rm \: P.E = m \cdot g \cdot  h

  • m = 100 kg
  • g = 10N/kg
  • h = 60

\rm \: P.E =100 \:kg \:  \times  \cfrac{10 \: N}{kg}  \times 60

\boxed{\rm \: P.E =60000  \: joule}

  • same condition here as well, N×M = J
  • c) Power of the Pump

\rm \: P = W/T

  • where P = Power; W = Work done & T = Time taken
  • As we got the value of work done on question (a),& ATQ time taken is 20 S.

\rm \: P =  \cfrac{60000 \: joule}{20 \: seconds}  =\boxed{\rm { 3000 \: Watts }}\: or \: \boxed{\rm 3 \: kW}

  • d) Power rating of the pump = 3 kW

Assumption: The pump is 100% efficient & works well.

6 0
2 years ago
Please reply to me.
inysia [295]
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➷ It would be C

<h3><u>✽</u></h3>

➶ Hope This Helps You!

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➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

6 0
4 years ago
Read 2 more answers
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