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balu736 [363]
3 years ago
9

I AM DDDUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUMMMMMMMMBBB

Physics
2 answers:
aivan3 [116]3 years ago
8 0

Answer:

same

Explanation:

umka21 [38]3 years ago
6 0
That’s relatable haha
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What is the acceleration due to gravity at an altitude of 116?
Schach [20]
Gravitational acceleration (Ga) is  inversely proportional to  k / Distance^2

so Ga * Distance^2 = K

On the surface of Earth acceleration due to gravity is about 9.8m/s^2 with an average distance to the earths core of about 6371 km (Wolfram alpha).

So k = 9.8 * 6371^2
I'm presuming that your distance of 116 is km

As 
Ga = k / distance^2 

Ga = ((9.8 * 6371^2) / (6371 + 116)^2 ) = 397778481.8 / 42081169

= 9.45 m/s^2 to 2sf
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4 years ago
Which is a possible measurement of the momentum of a moving object?
Delvig [45]

Momentum is a vector, although we don't hammer on that.  In order to completely describe a momentum, you need a magnitude AND a direction ... just like force and velocity.

So choice #2 is the magnitude of a momentum, without its direction.

<em>Choice #3</em> is  the full package, with both the magnitude and the direction.

Choice #1 has units of energy, and choice #4 has units of acceleration, so neither of those can be it.

8 0
4 years ago
Read 2 more answers
2 Points
Tom [10]
The answers are A and E!
6 0
3 years ago
Which volcanic hazard can block the sunlight and temporarily cool the Earth’s surface?
marissa [1.9K]
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4 0
4 years ago
A
True [87]

Answer:

\Delta T=3.615^{\circ}C is the drop in the water temperature.

Explanation:

Given:

  • mass of ice, m_i=14.7\ g=0.0147\ kg
  • mass of water, m_w=324\ g=0.324\ kg

Assuming the initial temperature of the ice to be 0° C.

<u>Apply the conservation of energy:</u>

  • Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.

<u>Now from the heat equation:</u>

Q_i=Q_w

m_i.L=m_w.c_w.\Delta T ......................(1)

where:

L= latent heat of fusion of ice =333.55\ J.g^{-1}

c_w= specific heat of water =4.186\ J.g^{-1}.^{\circ}C^{-1}

\Delta T= change in temperature

Putting values in eq. (1):

14.7 \times 333.55=324\times 4.186\times \Delta T

\Delta T=3.615^{\circ}C is the drop in the water temperature.

8 0
3 years ago
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