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musickatia [10]
3 years ago
14

The bulk of geologic evidence indicates that the Earth's crust has been formed by unique, rapid ————- processes

Physics
1 answer:
jeka57 [31]3 years ago
8 0
Answer : It is Catastrophic!
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At what speed relative to the ship does she recoil toward the spaceship?
Andre45 [30]
She recoil at a high speed than a low
7 0
3 years ago
Calculate the mass of an object with a density of 102.5 g/mL and volume of 375 mL.
AveGali [126]

Answer:

38,437.5

Explanation:

Density(d)= 102.5g/ml

Volume (v)=375ml

Mass(m) = ?

D =m/v

102.5= m/375

102.5*375=m

38,437.5=m

therefore Mass = 38,437.5g/ml.

6 0
4 years ago
A black, totally absorbing piece of cardboard of area a = 2.1 cm2 intercepts light with an intensity of 8.8 w/m2 from a camera s
PolarNik [594]

Answer:

The value of radiation pressure is 2.933 \times 10^{-8} Pa

Explanation:

Given:

Intensity I = 8.8 \frac{W}{m^{2} }

Area of piece A = 2.1 \times 10^{-4} m^{2}

From the formula of radiation pressure in terms of intensity,

   P = \frac{I}{c}

Where P = radiation pressure, c = speed of light

We know value of speed of light,

 c = 3 \times 10^{8} \frac{m}{s}

Put all values in above equation,

  P = \frac{8.8}{3 \times 10^{8} }

  P = 2.933 \times 10^{-8} Pa

Therefore, the value of radiation pressure is 2.933 \times 10^{-8} Pa

8 0
4 years ago
The rest deltoid row is a back exercise true or false
bixtya [17]
False because your deltoids are in your shoulders not your back
3 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
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