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4 years ago

The amount of matter I’m an object is know as

Physics

1 answer:

Oduvanchick [21]4 years ago

8 0

The amount of matter I’m an object is know as <em>the object's mass</em> .

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Read the scenario below and identify the Independent Variable, Dependent Variable, the Control group, and at least two Constants

Softa [21]

Answer:

Explanation:

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3 years ago

Suppose that a spiral galaxy is located at the center of a spherically symmetric dark matter halo

Novosadov [1.4K]

To solve the problem it is necessary to use the concepts of Orbital Speed considering its density, and its angular displacement.

In general terms the Orbital speed is described as,

$V_{orbit} = \sqrt{\frac{G\rho 4\pi r^3}{3}}$

PART A) If the orbital speed of a star in this galaxy is constant at any radius, then,

$\frac{4\pi G\rho r}{3} = \frac{v^2}{r}$

$\frac{4\pi G\rho r}{1} = \frac{3v^2}{r}$

$\frac{\rho}{1} = \frac{3v^2}{r^2 4\pi G}$

$\rho = \frac{1}{r^2}$

PART B) This time we have $v=\omega t$ , where $\omega$ is the angular velocity (constant at this case)

$\frac{4\pi G\rho r}{3} = \frac{v^2}{r}$

$\frac{4\pi G\rho r}{3} = \frac{(\omega r)^2}{r}$

$\rho = \frac{3\omega r}{4\pi Gr}$

$\rho = \frac{3\omega^2}{4\pi G} \propto constant$

PART C) If the total mass interior to any radius r is a constant,

$\frac{4\pi G\rho r}{3} = \frac{v^2}{r}$

$\frac{GM}{r^2}=\frac{v^2}{r}$

$v = \sqrt{\frac{GM}{r}}$

$v= \sqrt{\frac{1}{r}}$

3 0

4 years ago

A pendulum consists of a 2.0 kg stone swinging on a4.0 m string of negligible mass. The stone has a speed of 8.0 m/swhen it pass

arlik [135]

Answer:

a) $v_{60^{o}} =4.98 m/s$

b) $\theta_{max}=79.34^{o}$

Explanation:

This problem can be solved by doing an energy analysis on the given situation. So the very first thing we can do in order to solve this is to draw a diagram of the situation. (see attached picture)

So, in an energy analysis, basically you will always have the same amount of energy in any position of the pendulum. (This is in ideal conditions) So in this case:

$K_{lowest}+U_{lowest}=K_{60^{o}}+U_{60^{0}}$

where K is the kinetic energy and U is the potential energy.

We know the potential energy at the lowest of its trajectory will be zero because it will have a relative height of zero. So the equation simplifies to:

$K_{lowest}=K_{60^{o}}+U_{60^{0}}$

So now, we can substitute the respective equations for kinetic and potential energy so we get:

$\frac{1}{2}mv_{lowest}^{2}=\frac{1}{2}mv_{60^{o}}^{2}+mgh_{60^{o}}$

we can divide both sides of the equation into the mass of the pendulum so we get:

$\frac{1}{2}v_{lowest}^{2}=\frac{1}{2}v_{60^{o}}^{2}+gh_{60^{o}}$

and we can multiply both sides of the equation by 2 to get:

$v_{lowest}^{2}=v_{60^{o}}^{2}+2gh_{60^{o}}$

so we can solve this for $v_{60^{o}}$ . So we get:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

so we just need to find the height of the stone when the pendulum is at a 60 degree angle from the vertical. We can do this with the cos function. First, we find the vertical distance from the axis of the pendulum to the height of the stone when the angle is 60°. We will call this distance y. So:

$cos \theta = \frac{y}{4m}$

so we solve for y to get:

$y = 4cos \theta$

so we substitute the angle to get:

y=4cos 60°

y=2 m

so now we can find the height of the stone when the angle is 60°

$h_{60^{o}}=4m-2m$

$h_{60^{o}}=2m$

So now we can substitute the data in the velocity equation we got before:

$v_{60^{o}}=\sqrt{v_{lowest}^{2}-2gh_{60^{0}}}$

$v_{60^{o}} = \sqrt{(8 m/s)^{2}-2(9.81 m/s^{2})(2m)}$

$v_{60^{o}}=4.98 m/s$

b) For part b, we can do an energy analysis again to figure out what the height of the stone is at its maximum height, so we get.

$K_{lowest}+U_{lowest}=K_{max}+U_{max}$

In this case, we know that $U_{lowest}$ will be zero and $K_{max}$ will be zero as well since at the maximum point, the velocity will be zero.

So this simplifies our equation.

$K_{lowest} =U_{max}$

And now we substitute for the respective kinetic energy and potential energy equations.

$\frac{1}{2}mv_{lowest}^{2}=mgh_{max}$

again, we can divide both sides of the equation into the mass, so we get:

$\frac{1}{2}v_{lowest}^{2}=gh_{max}$

and solve for the height:

$h_{max}=\frac{v_{lowest}^{2}}{2g}$

and substitute:

$h_{max}=\frac{(8m/s)^{2}}{2(9.81 m/s^{2})}$

to get:

$h_{max}=3.26m$

This way we can find the distance between the axis and the maximum height to determine the angle of the pendulum about the vertical.

y=4-3.26 = 0.74m

next, we can use the cos function to find the max angle with the vertical.

$cos \theta_{max}= \frac{0.74}{4}$

$\theta_{max}=cos^{-1}(\frac{0.74}{4})$

so we get:

$\theta_{max}=79.34^{o}$

5 0

3 years ago

This picture shows some fortified cereal in a bowl.

ikadub [295]

Cereals are usually fortified with three most common minerals and vitamins which are:
1- Iron
2- Vitamin D
3- Vitamin B-12

Comparing the above to the given choices, we will find that the best answer is:
b. iron

3 0

4 years ago

Energy is released during the fission of pu-239 atoms as a result of the

pav-90 [236]

<h2>Answer: process of converting matter into energy</h2><h2></h2>

Nuclear fission consists of dividing a heavy nucleus into two or more lighter or smaller nuclei, by means of the bombardment with neutrons to make it unstable. In this process that takes place in the atomic nucleus, neutrons, gamma rays and <u>large amounts of energy are emitted. </u>

Then, with this division a great release of energy occurs and the emission of two or three neutrons, other particles and gamma rays.

This means fission is a process in which energy is released by the separation of the components of the nucleous of the atom.

In other words:

<h2>Matter is converted to energy .</h2>

6 0

3 years ago