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3 years ago

What are the types of energies released when a candle burns?

Physics

2 answers:

iragen [17]3 years ago

7 0

Answer:

A) Thermal and Radiant

Explanation:

When candle is burnt then the energy that is released in the form of light as well as in the form of thermal energy.

Here the candle is formed by wax which is melted due to due to burning and this will convert the energy present in the form of chemical energy into thermal energy as well as in the form of radiant energy.

Thermal energy will increase the temperature of surrounding air while radiant energy is in the form of light which is an electromagnetic wave.

So correct answer will be

A) Thermal and Radiant

Phantasy [73]3 years ago

4 0

When a candle is burning the candle is releasing thermal and radiant energy

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Take a close look at the energy transfers and transformations shown in the above diagram. Which type of energy is transformed in

labwork [276]

Kinetic Energy

How to calculate

Kinetic energy is given by K.E.

$\\ \sf\longmapsto K.E=\dfrac{1}{2}mv^2$

Where

m is denoted to mass
v is denoted to velocity

4 0

2 years ago

As the time required to run up the stairs increases, the power developed by that person

weeeeeb [17]

Yes. Power will decrease.

'cause Power = Work / time
So, power is indirectly proportional to time so, when one increases other would decrease

Hope this helps!

8 0

2 years ago

Arrow_forward

garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0

2 years ago

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0

otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$