What was the change in internal energy (chemical energy plus thermal energy) of the person pulling the block?

Mars has a mass of about 6.58 × 1023 kg, and its moon Phobos has a mass of about 9.3 × 1015 kg. If the magnitude of the gravitat

NISA [10]

This looks complicated, but it's actually not too tough.

The formula for the gravitational force between two objects is

Force = G (one mass) (other mass) / (distance²) .

The question GAVE us all of those numbers except the distance.
All we have to do is pluggum in, massage it around, and find
the distance.

Force = 4.18 x 10¹⁵ N
G = 6.673 x 10⁻¹¹ N·m²/kg²
One mass = 6.58 x 10²³ kg
Other mass = 9.3 x 10¹⁵ kg .

The only tricky thing about this is gonna be the arithmetic ...
keeping all the exponents straight.

Take the formula for the gravitational force and plug in
everything we know:

Force = (G) · (one mass) · (other mass) / (distance²)

4.18x10¹⁵N = (6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) / (distance²).

Multiply each side by (distance²):

(distance²)·(4.18x10¹⁵N) = (6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg)

Divide each side by (4.18 x 10¹⁵ N) :

(distance²)=(6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) / (4.18x10¹⁵N)

That's the end of the Physics and Algebra. The only thing left is Arithmetic.
We have to simplify that whole ugly thing on the right side of the equation,
and then take the square root of each side.

When I crunch down the right side of that equation, I get

(distance²) = 9.769 x 10¹³ m²

and when I take the square root of each side, I get

distance = 9.884 x 10⁶ meters . **

You should check my Arithmetic. **
(Pause occasionally to let your calculator cool off.)

BY THE WAY ...
That "distance" in the equation for gravitational force is the distance
between the CENTERS of the two objects.
This doesn't make much difference for Phobos, because Phobos isn't
much bigger than a big sweet potato. But it does make a difference for
Mars.
The 'distance' we find with all of this nonsense is NOT the distance
between Phobos and the surface of Mars. It's the distance between
Phobos and the CENTER of Mars, so it includes the planet's radius.

** Consulting online resources between Floogle and Flickerpedia,
I found that the orbital distance of Phobos from Mars varies between
9,234 km and 9,517 km. Add the planet's radius to these, and I'm
beginning to feel confidence in the results of my back-of-the-napkin
calculation. But you should still check my Arithmetic.

5 0

3 years ago

My Notes An electron is released from rest on the axis of a uniform positively charged ring, 0.500 m from the ring's center. If

dsp73

Answer:

Velocity of the electron = v = 1.2\times 10^8\ m/s.

Explanation:

Given,

Mass of the electron = $m_e\ =\ 9\times 10^{-31}\ kg$
Charge on the electron = $q_e\ =\ 1.62\times 10^{-19}\ C$
Charge density of the ring = $\rho\ =\ +1.00\times 10^{-6}\ C/m$
Radius of the ring = R = 0.70 m
Distance between the electron ant the center or the ring = x = 0.5 m

Now total charge on the ring = $Q\ =\ \rho\times 2\pi R$

Potential energy due to the charged ring to the point on the x-axis is

$P.E.\ =\ \dfrac{KQq_e}{\sqrt{R^2\ +\ x^2}}\\$

Let v be the velocity of the electron at the center of the ring.

Total kinetic energy of the electron = $\dfrac{1}{2}m_ev^2\\$

Now, From the conservation of energy,

the total potential energy of the electron at initially is converted to the total kinetic energy of the electron at the center of the ring,

$\therefore P.E.\ =\ K.E.\\\Rightarrow \dfrac{KQ}{\sqrt{R^2\ +\ x^2}}\ =\ \dfrac{1}{2}m_ev^2\\\Rightarrow v\ =\ sqrt{\dfrac{2 KQq_e}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2Kq_e\rho \times 2\pi R}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2\times 9\times 10^9\times 1.0\times 10^{-6}\times 2\times 3.14\times 0.7\times 1.6\times 10^{-19}}{9\times 10^{-31}\times \sqrt{0.7^2\ +\ 0.5^2}}}\\\Rightarrow v\ =\ 1.2\times 10^8\ m/s.$

Hence the velocity of the electron on the center of the charged ring is $1.2\times 10^8\ m/s.$

5 0

3 years ago

First extinguish a match or candle by blasting it violently. Why?

Andreas93 [3]

Answer:

The wax vapor on burning candles could reignite the flame.

Explanation:

7 0

3 years ago

A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 18.0t and y =

In-s [12.5K]

Position: x = 18t y = 4t - 4.9t²

First derivative: x' = 18 y' = 4 - 9.8t

Second derivative: x'' = 0 y'' = - 9.8

Position vector: P = (18t) i + (4t - 4.9t²) j

Velocity vector: V = (18) i + (4 - 9.8t) j

Acceleration vector A = (- 9.8) j

6 0

3 years ago

Read 2 more answers

A man of mass m 1 5 70.0 kg is skating at v1 5 8.00 m/s behind his wife of mass m 2 5 50.0 kg, who is skating at v2 5 4.00 m/s.

ehidna [41]

Answer:

A. Kindly find attached free body diagram for your reference (smiles I guess I will make a terrible artist)

B. The collision is inelastic because both the husband and the wife moved together with same velocity as he grabs her on the waist

C. The general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf

Say mass of husband is m1

Mass of the wife is m2

Velocity of the husband is v1

Velocity of the wife is v2

According to the conservation of momentum principle momentum before impact m1v1+m2v2 =momentum after impact Common velocity after impact (m1+m2)vf

The momentum equation is

m1v1+m2v2= (m1+m2)vf

D. To solve for vf we need to make it subject of formula

vf= {(m1v1) +(m2v2)}/(m1+m2)

E. Substituting our given data

vf=

{(1570*58)+(2550*54)}/(1570+2558)

vf=91060+137700/4120

vf=228760/4120

vf=55.52m/s

Their speed after collision is 55.52m/s

7 0

3 years ago