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quester [9]
3 years ago
9

Is a burning wax candle nuclear energy?

Physics
1 answer:
sasho [114]3 years ago
6 0

no because nuclear energy come from kinetic not potential energy. burning a wax candle is an example of heat/thermal energy .

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What is surface tension??​
Nadya [2.5K]

Answer:

<h2>internal force per unit area on a solid or liquid surface is called surface tension...</h2>
6 0
3 years ago
A 95 kg fullback, running at 8.2 m/s, collides in midair with a 128 kg defensive tackle moving in the opposite direction. Both p
Sindrei [870]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below is the answers:

Fullback running 

<span>Mo = mass * velocity </span>
<span>Mo = 95kg * 8.2 m/s =779 kg*m/s (a </span>

<span>He got stopped Change in Mo = 779 kg*m/s (b </span>

<span>Both stopped ===> Tackle's mo = - Halfback's Mo = - 779 kg*m/s (c & d </span>

<span>- 779 = 128 * v </span>
<span>v= - 6.09 m/s (e</span>
6 0
3 years ago
Give examples of plants that can reproduce by vegetative reproduction.​
marishachu [46]

Answer:

Some examples of vegetative propagation are farmers creating repeated crops of apples, corn, mangoes or avocados through asexual plant reproduction rather than planting seeds. Vegetative propagation can be accomplished from side-shoots, slips, stems and sections of tubers, bulbs or rhizomes.

Explanation:

5 0
3 years ago
A girl pulls on a 10-kg wagon with a constant horizontal force of 30 n. if there are no other horizontal forces, what is the wag
olga2289 [7]
Force = mass * acceleration
F = ma

Given m = 10 kg, F = 30 N;

F = ma
30 = 10a

Solving for a:
a = 3 m/s^2

The acceleration is 3 meters per second squared.
8 0
3 years ago
Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig
HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

y=-\frac{1}{2} gt^2+v_0t+y_0

With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

1) v_0=y-y_0+\frac{g}{2}

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) E=\frac{1}{2}m{v_0}^2+mgy_0

If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

4) h=\frac{{v_0}^2}{2g}+y_0

Combining equation 1 and 4:

h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0

4 0
3 years ago
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