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Marina86 [1]
2 years ago
12

Energy is the ability to do work. True or false

Physics
1 answer:
Blababa [14]2 years ago
4 0
Work can be defined as the energy transferred from a body to its sorroundings, the energy spent to move a body, or the energy you need to alter a charged particle, so no energy, no work; thus, the statement is true.
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The fast train known as the TGV (Train à Grande Vitesse) that runs south from Paris, France, has a scheduled average speed of 21
andrezito [222]

Answer:

Explanation:

Given

average speed of train(v_{avg})=216 kmph\approx 60 m/s

Maximum acceleration=0.05g

Now centripetal acceleration is

a_c=\frac{v^2}{r}

0.05\times 9.8=\frac{60^2}{r}

r=7346.93 m

(b)Radius of curvature=900 m

therefore a_c=\frac{v^2}{r}

v=\sqrt{a_cr}

v=\sqrt{0.05\times 9.8\times 900}

v=\sqrt{441}=21 m/s

8 0
3 years ago
The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th
MaRussiya [10]

Answer:

a). P=78.4W

b). P=392kPa

c.) It must be at the bottom

Explanation:

Given:

Volume flow V_f=2.0x10^{-4}m^3/s

Well depp h=40.m

a.

The power output of the pum

W=F*d

F=m*g

m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg

W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2

W=78.4J

P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W

b.

The pressure of difference the pum

ΔP=p*g*y'

ΔP=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa

P=392kPa

c.

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way  

4 0
3 years ago
Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.
amm1812
-- Equations  #2  and  #6  are both the same equation,
and are both correct.

-- If you divide each side by  'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by  'frequency', you get Equation #3,
which is also correct. 
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.
7 0
3 years ago
Read 2 more answers
A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2
stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

                     s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2

For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

We also have

                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

7 0
3 years ago
I select true helllpppp me
yanalaym [24]

Answer:

your right answer is true

hope it helps you

3 0
2 years ago
Read 2 more answers
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