All categories

2 years ago

Two objects with a mass of 10,000-kg each and a distance of 1 meter between them.

Physics

1 answer:

Fofino [41]2 years ago

4 0

Answer:

d) What is the force if we doubled both the masses AND we doubled the distance

You might be interested in

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

2 years ago

Two football players collide head-on in midair while trying to catch a thrown football. The first player is 98.5 kg and has an i

romanna [79]

Answer:

$v_f=0.825m/s$

Explanation:

We must use conservation of linear momentum before and after the collision, $p_i=p_f$

Before the collision we have:

$p_i=p_1+p_2=m_1v_1+m_2v_2$

where these are the masses are initial velocities of both players.

After the collision we have:

$p_f=(m_1+m_2)v_f$

since they clong together, acting as one body.

This means we have:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Or:

$v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

Which for our values is:

$v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s$

7 0

3 years ago

How many stars has been observed by Hubble telescope till now ?

Hoochie [10]

Answer:

More Than 1.3 million.

Hope it helps :))

8 0

1 year ago

Scientist often use ______________ which are programs that combine what is known about atmospheric circulation

Serggg [28]

Answer:

General circulation model.

Explanation:

A general circulation model (GCM) is a type of climate model that employs a mathematical model of the general circulation of a planetary atmosphere or ocean. GCM uses the Navier–Stokes equations on a rotating sphere with thermodynamic terms for various energy sources (radiation, latent heat). These equations are the basic equations for computer programs used to simulate the Earth's atmosphere or oceans.

7 0

2 years ago

Read 2 more answers

Help Me Plss :,) I WILL MARK BRAINLIEST:

Elden [556K]

The correct answer is A. While the Big Bang is very possible it is still a theory. Even if you accept it, you have to see its flaws. Nothing is without flaw. There are many things it doesn't account for. Now, again this is not to say it is untrue just flawed. The best argument against it or anything similar would be that no one seen it happen.

6 0

2 years ago

Read 2 more answers