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Veseljchak [2.6K]

2 years ago

9

how does gravity affect objects of different mass close to earth, and how does that effect change as an object moves farther fro

m earth?

1 answer:

Luda [366]2 years ago

3 0

Answer:

It makes it lighter when its closer and heavier when its farther way.

Explanation:

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To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?

Nimfa-mama [501]

The energy stored in a capacitor is given by:
$U= \frac{1}{2}CV^2$
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
$C=3.0 \mu F = 3.0 \cdot 10^{-6} F$
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
$V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V$

8 0

3 years ago

1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t

tiny-mole [99]

Explanation:

Charges, $q_1=8\ \mu C=8\times 10^{-6}\ C$

$q_2=-5\ \mu C=-5\times 10^{-6}\ C$

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

$F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N$

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0

2 years ago

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

2 years ago

The kinetic energy of a body of mass 15 kg is 30 joule. What is its momentum?

lys-0071 [83]

This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum. So here they are:

Kinetic energy = (1/2) · (mass) · (speed²)

Momentum = (mass) · (speed)

So, now ... We know that

==> mass = 15 kg, and

==> kinetic energy = 30 Joules

Take those pieces of info and pluggum into the formula for kinetic energy:

Kinetic energy = (1/2) · (mass) · (speed²)

30 Joules = (1/2) · (15 kg) · (speed²)

60 Joules = (15 kg) · (speed²)

4 m²/s² = speed²

Speed = 2 m/s

THAT's all you need ! Now you can find momentum:

Momentum = (mass) · (speed)

Momentum = (15 kg) · (2 m/s)

<em>Momentum = 30 kg·m/s</em>

<em>(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum. When I saw this, I wondered whether that's always true. So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)</em>

8 0

2 years ago

What is the equivalent resistance for a series circuit with three resistors : 5.0 ohms, 2.0 ohms, and 12.0 ohms

vesna_86 [32]

Answer:19ohms

Explanation:

equivalent resistance=5+2+12

equivalent resistance=19ohms

8 0

2 years ago