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3 years ago

How hot is the sun? First answer gets brainliest. :)

Physics

2 answers:

Andreyy893 years ago

5 0

Inner (Core) Temperature = 1.5 crore degree celsius
Surface temp. = 6 k degree celsius

Hope this helps!

AleksAgata [21]3 years ago

3 0

<span>5,778 K is how hot the sun is. ( ty will be receiving my brainliest now :) </span>

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4 years ago

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ElenaW [278]

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3 years ago

What happens when you stick your head in a microwave?

victus00 [196]

You'll probably die

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3 years ago

A 2.5. Gram arrow enters a 100 g apple with a speed of 115 m/s. If the apple was originally at rest, then what speed will it hav

stich3 [128]

Answer:

The speed of the apple will be 2.81 m/s when the arrow enters it.

Explanation:

We can find the speed of the apple by conservation of linear momentum:

$p_{i} = p_{f}$

$m_{ap}v_{ap_{i}} + m_{a}v_{a_{i}} = m_{ap}v_{ap_{f}} + m_{a}v_{a_{f}}$

Where:

$m_{ap}$ is the mass of the apple = 100 g = 0.1 kg

$m_{a}$ is the mass of the arrow = 2.5 g = 0.0025 kg

$v_{ap_{i}}$ and $v_{ap_{f}}$ is the initial and final speed of the apple respectively

$v_{a_{i}}$ and $v_{a_{f}}$ is the initial and final speed of the arrow respectively

Since the apple was originally at rest ( $v_{ap_{i}}$ = 0) and knowing that $v_{a_{f}}$ = $v_{ap_{f}}$ when the arrow enters into the apple, we have:

$0 + 0.0025 kg*115 m/s = v(0.0025 kg + 0.1 kg)$

$v = 2.81 m/s$

Therefore, the speed of the apple will be 2.81 m/s when the arrow enters it.

I hope it helps you!

5 0

3 years ago

6. A 15 kg box is given an initial push so that it slides across the floor and comes to a stop.

Illusion [34]

Answer:

(a) Frictional force = 44.145 N

(b) Acceleration = -2.943 m/s²

(c) Distance covered = 1.529 m

Explanation:

Given:

Mass of the box is, $m=15\ kg$

Coefficient of friction is, $\mu =0.30$

Acceleration due to gravity is, $g=9.81\ m/s^2$

Normal force acting on the box is, $N=mg=(15)(9.81) = 147.15\ N$

(a)

Frictional force is given as:

$f=\mu N=0.30\times 147.15=44.145\ N$

Therefore, the frictional force is 44.145 N.

(b)

The acceleration of the box is given using Newton's second law as:

$a=-\frac{f}{m}=-\frac{44.145}{15}=-2.943\ m/s^2$

Therefore, the acceleration of the box is -2.943 m/s².

(c)

Initial velocity is, $u=3.0\ m/s$

Final velocity is, $v=0\ m/s$

Acceleration of the box is, $a=-2.943\ m/s^2$

The displacement of the box can be determined using equation of motion as:

$v^2=u^2+2ad\\0=3^2+2\times (-2.943)d\\0=9-5.886d\\5.886d=9\\d=\frac{9}{5.886}=1.529\ m$

Therefore, the displacement of the box is 1.529 m.

4 0

4 years ago