PLEASE HELP 15 POINTS!!!!

Physics

1 answer:

goldfiish [28.3K]3 years ago

8 0

Answer:

15 degrees Celsius

Explanation:

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What temperature can a solar flare reach

balandron [24]

It reaches 10 or 20 million degrees kelvin but it can get as high as 10 million degrees kelvin

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3 years ago

A 25 kg child runs at a speed of 5.0 m/s and jumps onto a stationary shopping cart and holds on for dear life. The cart has mass

makkiz [27]

Answer:

3.38 m/s

Explanation:

Mass of child = m₁ = 25

Initial speed of child = u₁ = 5 m/s

Initial speed of cart = u₂ = 0 m/s

Mass of cart = m₂ = 12 kg

Velocity of cart with child on top = v

This is a case of perfectly inelastic collision

$m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\frac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\frac{25\times 5+12\times 0}{25+12}\\\Rightarrow v=\frac{125}{37}\\\Rightarrow v=3.38\ m/s$

Velocity of cart with child on top is 3.38 m/s

7 0

3 years ago

Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac

Serjik [45]

The strength of the gravitational field is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
$r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m$

And now we can use the previous equation to calculate the field strength at that altitude:
$g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2} = 8.95 m/s^2$

And we can see this value is a bit less than the gravitational strength at the surface, which is $g_s = 9.81 m/s^2$ .