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sesenic [268]
3 years ago
11

24 g of water is heated from 25 c to 35 c. how much heat energy was required

Physics
1 answer:
Viefleur [7K]3 years ago
4 0
If you were to rase to temperature 10c and the heat energy required to rase it 1 c is 4190 j/(kg C). 24g • 4190 j/(kg C) • (35-25)C 24 • 4190 • 10= 1,005,600 j or 1.5 MJ
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A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3
Advocard [28]

Answer:

r = 5.07 m

Explanation:

given,

velocity of the man , v = 3.43 m/s

centripetal acceleration, a = 2.32 m/s²

magnitude of position of = ?

using centripetal acceleration formula

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

2.32 =\dfrac{3.43^2}{r}

r =\dfrac{3.43^2}{2.32}

   r = 5.07 m

The magnitude of the position vector relative to rotational axis is equal to 5.07 m.

5 0
3 years ago
Note: The rope is 20 m long. Answer like this: (1.<br> 2._____ etc)
qaws [65]

Answer:

1. <u>Potential energy</u>, 2. <u>Potential and kinetic energy</u>, 3. <u>Potential and kinetic energy</u>, 4. <u>Potential and kinetic energy</u>, 5. <u>Potential energy</u>

Explanation:

We note that the total mechanical energy (M.E.) of the body is given as follows;

M.E. = K.E. + P.E. = Constant

Where;

K.E. = The kinetic energy of the body = (1/2)·m·v²

P.E. = The potential energy of the body = m·g·h

m = The mass of the person

v = The velocity with which the person is in motion

g = The acceleration due to gravity ≈ 9.81 m/s²

h = The height of the person above the ground

The length of the rope = 20 m

The initial height at location 1, h₁ = 40.0 m

At location 1, the velocity, v₁ = 0.00 m/s

The mechanical energy, M.E. = K.E.₁ + P.E.₁

∴  K.E.₁ = 0 and P.E.₁ = m ×9.81×40

M.E. = (1/2) ×m ×0² + m ×9.81×40

∴ M.E. = 0 + P.E.₁ the type of energy present at location 1 is only potential energy

At location 2, the velocity, v₂ = 10.0 m/s

The mechanical energy, M.E. = K.E.₂ + P.E.₂ = (1/2) ×m ×10² + m ×9.81×40

∴  K.E.₂ = 50·m and P.E.₂ = m ×9.81×35 = 343.35·m

M.E. = 50·m + 343.35·m the type of energy at location 2 is both kinetic energy, K.E. and potential energy, P.E.

At location 3, the velocity, v₃ = 20.0 m/s

The mechanical energy, M.E. = K.E.₃ + P.E.₃ = (1/2) ×m ×20² + m ×9.81×20

∴  K.E.₃ = 200·m and P.E.₃ = m ×9.81×20 = 196.2·m

M.E. = 200·m + 196.2·m the type of energy at location 3 is both kinetic energy, K.E. and potential energy, P.E.

At location 4, the velocity, v₄² = 350.0 m²/s², h₄ = 15.0 m

The mechanical energy, M.E. = K.E.₄ + P.E.₄ = (1/2) × m ×350 + m ×9.81×15

∴  K.E.₄ = 175·m and P.E.₄ = m×9.81×15 = 147.15·m

M.E. = 175·m + 147.15·m the type of energy at location 4 is both kinetic energy, K.E. and potential energy, P.E.

At location 5, the velocity, v₅ = 0 m/s, h₅ = 10.0 m

The mechanical energy, M.E. = K.E.₅ + P.E.₅ = (1/2) × m × 0 + m ×9.81×10

∴  K.E.₅ = 0·m and P.E.₅ = m×98.1 = 98.1·m

M.E. = 0·m + 98.1·m the type of energy at location 5 is only potential energy, P.E.

Therefore, we have;

\left|\begin{array}{ccc}Location&&Type(s) \ of \ Energy \ Presents\\1&&Potential \ Energy\\2&&Potential  \ and \ Kinetic \ Energy\\3&&Potential  \ and \ Kinetic \ Energy\\4&&Potential  \ and \ Kinetic \ Energy\\5&&Potential  \  Energy\end{array} \right |

5 0
2 years ago
A car initially at rest accelerates at 10m/s^2. The car’s speed after it has traveled 25 meters is most nearly... A.) 0.0m/s B.)
STALIN [3.7K]

The car traverses a distance x after time t according to

x=\dfrac12at^2

where a is its acceleration, 10 m/s^2. The time it takes for the car to travel 25 m is

25\,\mathrm m=\left(5\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=\sqrt 5\,\mathrm s

5 is pretty close to 4, so we can approximate the square root of 5 by 2. Then the car's velocity v after 2 s of travel is given by

v=\left(10\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)\approx20\dfrac{\rm m}{\rm s}

which makes C the most likely answer.

3 0
3 years ago
Identifying Advantages of Parallel Circuits
melisa1 [442]

Answer: If one bulb goes out the other bulbs stay lit.

If there is a break in one branch of the circuit, current can still flow through the other branches.

Explanation:

3 0
3 years ago
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Two animals that can be found at the Grand Canyon include the Abert’s and Kaibab squirrels. The Abert’s squirrel is found at the
Tom [10]
D. <span>When placed together in pairs for study, no pair of Abert's squirrel and Kaibab squirrel result in offspring.</span>
6 0
3 years ago
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