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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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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

2 years ago

A lacrosse ball that is thrown straight upwards reaches a maximum height of 4.5 m. At what initial velocity was it thrown? (note

shtirl [24]

Answer:

The initial velocity was 9.39 m/s

Explanation:

Lets explain how to solve the problem

The ball is thrown straight upward with initial velocity u

The ball reaches a maximum height of 4.5 m

At the maximum height velocity v = 0

The acceleration of gravity is -9.8 m/s²

We need to find the initial velocity

The best rule to find the initial velocity is v² = u² + 2ah, where v is

the final velocity, u is the initial velocity, a is the acceleration of

gravity and h is the height

⇒ v = 0 , h = 4.5 m , a = -9.8 m/s²

⇒ 0 = u² + 2(-9.8)(4.5)

⇒ 0 = u² - 88.2

Add 88.2 to both sides

⇒ 88.2 = u²

Take square root for both sides

⇒ u = 9.39 m/s

The initial velocity was 9.39 m/s

5 0

3 years ago

SOMEONE PLEASE HELP ME ASAP PLEASE!!!! SIGNIFICANT DIGITS!!!!!! THE ANSWER IS NOT −16.422

Aleks04 [339]

Explanation:

What exactly are u looking for?

7 0

2 years ago

Read 2 more answers

A 76 N crate is hung from a spring

lutik1710 [3]

Answer: 0.169 (3 s.f.)

Explanation:

Force = 76 N

Spring constant = 450 N/m

Extension/displacement = x

Hooke's law states that: F = kx

Therefore, 76 = 450 X x

76/450 = x

0.169 (3 s.f.) = x

4 0

2 years ago

Would it be m/s or kg?

Nesterboy [21]

Answer:

m.s

Explanation:

3 0

2 years ago