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2 years ago

Two objects each with a mass of 5x10^15 kg have a gravitational

Physics

1 answer:

kipiarov [429]2 years ago

6 0

Answer:

Distance, r = 700.31 m

Explanation:

Mass of two objects, $m_1=m_2=5\times 10^{15}\ kg$

Force between the objects , $F=3.4\times 10^{15}\ N$

We need to find the distance between the two objects. The force of gravitational between two objects is given by :

$F=G\dfrac{m_1m_2}{r^2}\\\\r=\sqrt{\dfrac{Gm_1m_2}{F}} \\\\r=\sqrt{\dfrac{6.67\times 10^{-11}\times (5\times 10^{15})^2}{3.4\times 10^{15}}}\\\\=700.31\ m$

So, the distance between the objects is 700.31 m.

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You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

bezimeni [28]

Answer:

a) The resulting angular speed of platform is 1.38 rev/sec

b) The change in kinetic energy of the system is 53 J.

Explanation:

This question is incomplete. The complete question will be:

You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 8.8 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to 7.0 k g .m 2

a) What is the resulting angular speed of the platform? Answer in units of r e v / s .

b)What is the change in kinetic energy of the system? Answer in units of J.

<h3>ANSWER:</h3>

we know that:

Angular Momentum = L = Iω

From conservation of momentum:

Lo = Lf

(Io) (ωo) = (If) (ωf)

ωf = (Io) (ωo)/(If)

ωf = (8.8 kg.m²)(1.1 rev/s)/(7.0 kg.m²)

ωf = (1.38 rev/sec)(2π rad/ 1 rev) = 8.67 rad/sec

ωo = (1.1 rev/sec)(2π rad/ 1 rev) = 6.91 rad/sec

The kinetic energy for rotational motion is given as:

K.E = (1/2)Iω²

Thus, the change in kinetic energy will be:

ΔK.E = (K.E)f - (K.E)o

ΔK.E = (1/2)Ifωf² - (1/2)Ioωo²

ΔK.E = (1/2)(Ifωf² - Ioωo²)

ΔK.E = (1/2)[(7 kg.m²)(8.67 rad/sec)² - (8.8 kg.m²)(6.91 rad/sec)²

5 0

3 years ago

A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit

elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

= 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

$T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }$ second

Squaring the terms and solving it for 'g'

g = 4 π² $\frac{(R+h)^{3} }{R^{2}T^{2} }$ m/s²

Substituting the values in the above equation

g = 4 π² $\frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}$

g = 4.84 m/s²

Therefore, the weight

w = m x g newton

= 5832 Kg x 4.84 m/s²

= 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N

8 0

3 years ago

Consider the hydrogen atom. How does the energy difference between adjacent orbit radii change as the principal quantum number i

Kisachek [45]

Answer:

the energy difference between adjacent levels decreases as the quantum number increases

Explanation:

The energy levels of the hydrogen atom are given by the following formula:

$E=-E_0 \frac{1}{n^2}$

where

$E_0 = 13.6 eV$ is a constant

n is the level number

We can write therefore the energy difference between adjacent levels as

$\Delta E=-13.6 eV (\frac{1}{n^2}-\frac{1}{(n+1)^2})$

We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

$\Delta E=-13.6 eV(\frac{1}{1^2}-\frac{1}{2^2})=-13.6 eV(1-\frac{1}{4})=-13.6 eV(\frac{3}{4})=-10.2 eV$

While the difference between the levels n=2 and n=3 is

$\Delta E=-13.6 eV(\frac{1}{2^2}-\frac{1}{3^2})=-13.6 eV(\frac{1}{4}-\frac{1}{9})=-13.6 eV(\frac{5}{36})=-1.9 eV$

And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

5 0

3 years ago

Yosef applies an input force of 50 N to a crowbar. The crowbar applies a force of 750 N to the lid of a crate. What is the mecha

docker41 [41]

The mechanical advantage of the crowbar is 15.

5 0

3 years ago

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What are the three most abundant gases in the dry atmosphere?

Anastaziya [24]

Answer:

Nitrogen, Oxygen, Argon.

Explanation:

The three (3) most abundant gases in the dry atmosphere are"

- Nitrogen

- Oxygen

- Argon

These are not the only components of dry air. Dry atmosphere is made up of:

- 78.09% Nitrogen;

- 20.95% Oxygen;

- 0.93% Argon;

- 0.04% Carbon dioxide;

- Other gases

6 0

2 years ago

Read 2 more answers