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kvasek [131]
3 years ago
6

What type of kinetic energy does flowing water have?

Physics
1 answer:
mina [271]3 years ago
6 0

Answer:

none it's Hydroelectric energy

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At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
2 years ago
Are we the same age as the universe because matter cannot be created nor destroyed
wlad13 [49]
No. We aren't the same age as the universe.
4 0
3 years ago
Read 2 more answers
Find the energy released when there is a decrease of 0.3kg of material in a nuclear reaction
MrRissso [65]

Answer:

E = 2.7 x 10¹⁶ J

Explanation:

The release of energy associated with the mass can be calculated by Einstein's mass-energy relation, as follows:

E = mc^2

where,

E = Energy Released = ?

m = mass of material reduced = 0.3 kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

E = (0.3\ kg)(3\ x\ 10^8\ m/s)^2

<u>E = 2.7 x 10¹⁶ J</u>

4 0
2 years ago
Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1200 km/s , measured relative to the
Radda [10]

Answer:

The maximum electrical force is 2.512\times10^{-2}\ N.

Explanation:

Given that,

Speed of cyclotron = 1200 km/s

Initially the two protons are having kinetic energy given by

\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2

When they come to the closest distance the total kinetic energy is converts into potential energy given by

Using conservation of energy

mv^2=\dfrac{kq^2}{r}

r=\dfrac{kq^2}{mv^2}

Put the value into the formula

r=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{1.67\times10^{-27}\times(1200\times10^{3})^2}

r=9.57\times10^{-14}\ m

We need to calculate the maximum electrical force

Using formula of force

F=\dfrac{kq^2}{r^2}

F=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{(9.57\times10^{-14})^2}

F=2.512\times10^{-2}\ N

Hence, The maximum electrical force is 2.512\times10^{-2}\ N.

8 0
3 years ago
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T
PIT_PIT [208]

Answer:

160 m

Explanation:

The intensity, I, of the sound is inversely proportional to the square of the distance, r, from the source.

I\propto \dfrac{1}{r^2}

Hence,

I_1r_1^2 = I_2r_2^2

r_2 = r_1\sqrt{\dfrac{I_1}{I_2}}

From the question, I_2 is half of I_1

r_2 = r_1\sqrt{\dfrac{I_1}{0.5I_1}}

r_2 = r_1\sqrt{2}

r_2 = 113\text{ m}\sqrt{2} = 160 \text{ m}

3 0
3 years ago
Read 2 more answers
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