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Advocard [28]
3 years ago
6

The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as meas

ured from the planet's surface). What type of relationship exists between the altitude and the atmospheric density, and what would the atmospheric density be at an altitude of 1,291 kilometers?

Physics
1 answer:
mixer [17]3 years ago
6 0
I know that the relationship between altitude and atmospheric density is that the higher the altitude, the lesser the density, and the lower the altitude the higher the density. Lower density float to the top, and higher density is 'heavy' so it comes down
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ivann1987 [24]

Answer:

This will require 266.9 of heat energy.

Explanation:

To calculate the energy required to raise the temperature of any given substance, here's what you require:The mass of the material, m The temperature change that occurs,  ΔT The specific heat capacity of the material,

c

(which you can look up). This is the amount of heat required to raise 1 gram of that substance by 1°C.

Here is a source of values of

c for different substances:

Once you have all that, this is the equation:

Q=m×c×ΔT(Q is usually used to symbolize that heat required in a case like this.)For water, the value of c is 4.186g°C So, Q=750×4.186×85=266=858=266.858

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Two balls are thrown against a wall. Ball 1 has a much higher speed than ball 2.
Sunny_sXe [5.5K]

Let both the balls have the same mass equals to m.

Let v_1 and v_2 be the speed of the ball1 and the ball2 respectively, such that

v_1>v_2\;\cdots(i)

Assuming that both the balls are at the same level with respect to the ground, so let h be the height from the ground.

The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, v, is \frac 12 m v^2

and the potential energy is due to the change in height is mgh [where g is the acceleration due to gravity]

So, the total energy of ball1,

=\frac 12 m v_1^2 + mgh\;\cdots(ii)

and the total energy of ball1,

=\frac 12 m v_2^2 + mgh\;\cdots(iii).

Here, the potential energy for both the balls are the same, but the kinetic energy of the ball1 is higher the ball2 as the ball1 have the higher speed, refer equation (i)

So, \frac 12 m v_1^2 >\frac 12 m v_2^2

Now, from equations (ii) and (iii)

The total energy of ball1 hi higher than the total energy of ball2.

6 0
3 years ago
An open-topped freight car with mass 24,000 kg is coasting without friction along a level track. It is raining very hard, and th
skelet666 [1.2K]

Answer:

(a) v = 3..6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

Explanation:

from the question we have the following:

mass of the car (Mc) = 24,000 kg

initial velocity of the car (u) = 4 m/s

mass of water (Mw) = 3000 kg

final velocity of the car (v) = ?

(a) we can calculate the final momentum of the car by applying the conservation of momentum where

initial momentum = final momentum

Mc x U = (Mc + Mw) x V

24000 x 4 = (24000 + 3000) x v

96,000 = 27000v

v =3.6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

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3 years ago
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