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3 years ago

Light from the sun reaches Earth in 8.3min. The speed of light is 3.00 x 10^8 m/s. How far is the Earth from the sun?

Physics

2 answers:

ruslelena [56]3 years ago

8 0

Speed= distance/ time
distance= speed×time
distance= 3×10^8 × 8.3× 60
1494×10^8

strojnjashka [21]3 years ago

5 0

The answer is this, but i don't know how to simplify it. 3x^100000000<span />

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How much heat transfer (in kcal) is required to raise the temperature of a 0.850 kg aluminum pot containing 2.00 kg of water fro

hram777 [196]

To solve this problem we will apply the concept related to the heat transferred to a body to reach a certain temperature. This concept is shaped by the energy ratio of a body which is the product of the mass, its specific heat and the change in temperature. For the specific case, it will be the sum of the heat transferred to the Water, the Aluminum and the loss due to latency due to vaporization in the water. That is to say,

$\Delta Q = m_{Al} C_p \Delta T +m_wC_w \Delta T +m_w L_v$

Here,

$m_{Al}$ = Mass of Aluminum

$C_p$ = Specific Heat of Aluminum

$C_w$ = Specific Heat of Water

$m_w =$ Mass of water

$L_v =$ Latent of Vaporization

Replacing,

$\Delta Q = (0.85)(900)(100-45)+(2)(2000)(100-45)+(0.7)(2258000)$

Converting,

$\Delta Q = 1842675J (\frac{0.000239006kCal}{1J})$

$\Delta Q = 440.409kCal$

Therefore is required 440.409kCal

8 0

3 years ago

A disk shaped grindstone of mass 3.0 kg and radius 8 cm is spinning at 600 rpm. After the power is shut off the frictional torqu

seropon [69]

Answer:

$\theta=50\ revolution$

Explanation:

It is given that,

Mass of the grindstone, m = 3 kg

Radius of the grindstone, r = 8 cm = 0.08 m

Initial speed of the grindstone, $\omega_i=600\ rpm=62.83\ rad/s$

Finally it shuts off, $\omega_f=0$

Time taken, t = 10 s

Let $\alpha$ is the angular acceleration of the grindstone. Using the formula of rotational kinematics as :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{0-62.83}{10}$

$\alpha =-6.283\ rad/s^2$

Let $\theta$ is the number of revolutions of the grindstone after the power is shut off. Now using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

$\theta=\dfrac{\omega_f^2-\omega_i^2}{2\alpha }$

$\theta=\dfrac{-62.83^2}{2\times -6.283}$

$\theta=314.15\ radian$

$\theta=49.99\ revolution$

$\theta=50\ revolution$

So, the number of revolutions of the grindstone after the power is shut off is 50.

7 0

3 years ago

A sailcraft is stalled on a windless day. A fan is attached to the craft and blows air into the sail which bounces backward upon

mylen [45]

Answer:

Impulse = change in momentum w bounce

There are 2 impulses acting. Recoil of the fan going the negative direction and the impulse of the air bouncing off the sail. The greater impulse will bounce so the direction will be to the right moving the craft.

7 0

2 years ago

Explain what nuclear fusion is in the role it plays in Stars. DO NOT LOOK UP ANSWERS

Aloiza [94]

Answer:

Stars are powered by nuclear fusion in their cores, mostly converting hydrogen into helium. The production of new elements via nuclear reactions is called nucleosynthesis. A star's mass determines what other type of nucleosynthesis occurs in its core (or during explosive changes in its life cycle).

8 0

3 years ago

A 1.2 kg block of wood hangs motionless from strings. A 50 gram bullet, traveling horzontally, strikes the block and becomes emb

Firdavs [7]

Answer:

speed of the bullet before it hit the block is 200 m/s

Explanation:

given data

mass of block m1 = 1.2 kg

mass of bullet m2 = 50 gram = 0.05 kg

combine speed V= 8.0 m/s

to find out

speed of the bullet before it hit the block

solution

we will apply here conservation of momentum that is

m1 × v1 + m2 × v2 = M × V .............1

here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet

put all value in equation 1

m1 × v1 + m2 × v2 = M × V

1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8

solve it we get

v2 = 200 m/s

so speed of the bullet before it hit the block is 200 m/s

8 0

3 years ago