All categories

3 years ago

A 10 kg box is 1.3 m above the ground. How much potential energy does it have? (g on Earth of 9.8 m/s?

Physics

1 answer:

Volgvan3 years ago

8 0

Potential energy = mgh
Potential energy = 10 x 9.8 x 1.3
Potential energy = 127.4 J

You might be interested in

a 0.0215m diameter coin rolls up a 20 degree inclined plane. the coin starts with an initial angular speed of 55.2rad/s and roll

algol13

Answer:

$h = 0.0362\,m$

Explanation:

Given the absence of non-conservative force, the motion of the coin is modelled after the Principle of Energy Conservation solely.

$U_{g,A} + K_{A} = U_{g,B} + K_{B}$

$U_{g,B} - U_{g,A} = K_{A} - K_{B}$

$m\cdot g \cdot h = \frac{1}{2}\cdot I \cdot \omega_{o}^{2}$

The moment of inertia of the coin is:

$I = \frac{1}{2}\cdot m \cdot r^{2}$

After some algebraic handling, an expression for the maximum vertical height is derived:

$m\cdot g \cdot h = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega_{o}^{2}$

$h = \frac{r^{2}\cdot \omega_{o}^{2}}{g}$

$h = \frac{(0.0108\,m)^{2}\cdot (55.2\,\frac{rad}{s} )^{2}}{9.807\,\frac{m}{s^{2}} }$

$h = 0.0362\,m$

6 0

3 years ago

What are the two factors that determine the amount of gravitational force between two objects?

melisa1 [442]

Answer:

The mass of the objects

Distance between the objects

Explanation:

The two factors that determines the amount of gravitational force between two objects is the mass of the objects and the distance between them.

Gravitational force of attraction is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between them. This is the postulate of the newton's law of universal gravitation.

Mathematically:

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

m is the mass

r is the distance

G is the universal gravitation constant

6 0

3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$