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3 years ago

8

If you quadruple the temperature of a black body, by what factor will the total energy radiated per second per square meter incr

ease?

1 answer:

Anastaziya [24]3 years ago

8 0

Radiant heat transfer is proportional to the 4-th power of absolute temperature.
Therefore if the temperature is quadrupled, the radiant heat energy will increase by a factor of
4⁴ = 256

Answer: 256

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How does the force of gravity and the force of earth contribute to africa's poverty?

ivanzaharov [21]

Answer:

The force of gravity is not the same as being on the earth. when your on the earth there no gravitational pull its all up to the air

Explanation:

No explanation

8 0

3 years ago

Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg objec

yKpoI14uk [10]

Answer:

m = 3.91 kg

Explanation:

Given that,

Mass of the object, m = 3.74 kg

Stretching in the spring, x = 0.0161 m

The frequency of vibration, f = 3.84 Hz

When the object is suspended, the gravitational force is balanced by the spring force as :

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{3.74\times 9.8}{0.0161}$

k = 2276.52 N/m

The frequency of vibration is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$m=\dfrac{k}{4\pi^2f^2}$

$m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}$

m = 3.91 kg

So, the mass of the object is 3.91 kg. Hence, this is the required solution.

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3 years ago

Which law states that energy is neither created nor destroyed? a. Law of Conservation of Mass

Brrunno [24]

Answer:

C. Law of Conservation

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2 years ago

Read 2 more answers

A ball is thrown straight upward and rises to a maximum

Leviafan [203]

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

$v_f^2 - v_i^2 = 2 a d$

here we know that

$v_f = 0$

also we have

$a = -9.81 m/s^2$

so we have

$0 - v_i^2 = 2(-9.81)(16)$

$v_i = 17.72 m/s$

Now we need to find the height where its speed becomes half of initial value

so we have

$v_f = 0.5 v_i$

now we have

$v_f^2 - v_i^2 = 2 a d$

$(0.5v_i)^2 - v_i^2 = 2(-9.81)h$

$-0.75v_i^2 = -19.62 h$

$0.75(17.72)^2 = 19.62 h$

$h = 12 m$

3 0

3 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago