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3 years ago

Help ?

1 answer:

8 0

I think it would be the one talking about if there’s water there would still be energy because water is used as a source of energy because there’s so much of it and it can be used again and again

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In a mixture what happens when the ingredients intermingled

Zinaida [17]

The ingredients do not react with or chemically bond to each other.

7 0

3 years ago

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Three science questions help please!

Nimfa-mama [501]

6. D
7. D
8. B
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4 0

4 years ago

The distance between two stations is 1995 Km. How much time will it take to cover the distance at an average speed of 19KM/hour

Flura [38]

105 hours or 4.375 days

5 0

3 years ago

The distance between slits in a double-slit experiment is decreased by a factor of 2. If the distance between fringes is small c

Elodia [21]

Answer:

the distance between adjacent fringes is increased by a factor o 2

Explanation:

To find how the distance between fringes is modified you can use the following formula for the calculation of the distance between fringes:

$\Delta y=\frac{\lambda D}{d}$

D: distance to the screen

d: distance between slits

λ: wavelength of the light

if d is decreased by a factor of 2, that is d'=1/2d, you have:

$\Delta y'=\frac{\lambda D}{d'}=\frac{\lambda D}{(1/2)d}=2\Delta y$

hence, the distance between adjacent fringes is increased by a factor o 2

4 0

3 years ago

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