When an unbalanced force acts on an object,

What is gravitational potential energy?

Kay [80]

Answer:

D). energy resulting from the attraction between two masses.

Explanation:

4 0

3 years ago

8. List three temperature scales. Then write the boiling and freezing

Tomtit [17]

Explanation:

For most temperature scales, the boiling point of water and the freezing point is used to calibrate it.

Three known temperature scales;

Kelvin scale
Celcius scale
Fahrenheit scale

Kelvin scale Celcius scale Fahrenheit scale

Freezing point 273K 0°C 32°F

Melting point 373K 100°C 212°F

5 0

2 years ago

Water is traveling into what sphere during infiltration?

o-na [289]

When precipitation hits the ground and hydrosphere

5 0

3 years ago

One car has two and a half times the mass of a second car, but only half as much kinetic energy. When both cars increase their s

Leokris [45]

Answer:

$v_1 = 7.96 m/s$

$v_2 = 17.8 m/s$

Explanation:

Let the mass of the other car is "m" and its kinetic energy is

$K = \frac{1}{2}mv^2$

now the mass of the first car is two and half times and its kinetic energy is half that of other car

so we will have

$\frac{1}{2}(2.5m)v_1^2 = \frac{1}{2}(\frac{1}{2}mv^2)$

$2.5 v_1^2 = 0.5 v^2$

$v_1 = 0.447 v$

now speed of both cars is increased by value of 9 m/s

so now we will have same kinetic energy for both cars

$\frac{1}{2}(2.5 m)(0.447v + 9)^2 = \frac{1}{2}m(v + 9)^2$

$2.5(0.447 v + 9)^2 = (v + 9)^2$

$1.58(0.447v + 9) = v + 9$

$0.293v = 5.22$

$v = 17.8 m/s$

so speed of first car is

$v_1 = 0.447 v = 7.96 m/s$

$v_2 = 17.8 m/s$

3 0

3 years ago

A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

centripetal force acting on satellite = $\dfrac{mv^2}{r}$

gravitational force = $\dfrac{GMm}{r^2}$

equating both the above equation

$\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$

$v = \sqrt{\dfrac{GM}{r}}$

$v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}$

v = 5578.5 m/s

b) $T= \dfrac{2\pi\ r}{v}$

$T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}$

T = 14416.92 s

$T = \dfrac{14416.92}{3600}\ hr$

T = 4 hr

c) gravitational force acting

$F = \dfrac{GMm}{r^2}$

$F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}$

F = 5202 N

4 0

3 years ago