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4 years ago

8

What is kinetic energy? Question 1 options: energy that is due to motion energy that is stored the mass of an object the rate at

which work is performed

1 answer:

mario62 [17]4 years ago

8 0

<span>Kinetic Energy means energy that a body possesses by virtue of being in motion. I hope this helps!!!</span>

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Write this large number in scientific notation. Determine the values of Vm) and (n) when the following mass of the Earth is writ

hichkok12 [17]

Answer : The answer is, 5.97, 24

Explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example :

5000 is written as $5.0\times 10^3$

889.9 is written as $8.899\times 10^{-2}$

In this examples, 5000 and 889.9 are written in the standard notation and $5.0\times 10^3$ and $8.899\times 10^{-2}$ are written in the scientific notation.

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

As we are given the 5,970,000,000,000,000,000,000,000 in standard notation.

Now converting this into scientific notation, we get:

$\Rightarrow 5,970,000,000,000,000,000,000,000=5.97\times 10^{24}$

As, the decimal point is shifting to left side, thus the power of 10 is positive.

Hence, the answer is, $5.97\times 10^{24}$

Now the answer is comparing to $m.\times 10^n$

So, m = 5.97 and n = 24

Thus, the answer is, 5.97, 24

5 0

3 years ago

As the scattering angle of the photon increases, what happens to the wavelength associated with the photon?

frez [133]

As the scattering angle of the photon increases, the wavelength associated with the photon increases.

<h3><u>Explanation:</u></h3>

The particle with quantum mechanical property is known as Compton wavelength. The wavelength of a photon increases during collision. When the scattering angle of the photon is 0 degree then the photon's wavelength increases by 0 and when the scattering angle is 180 degree then the wavelength of the photon will become double. This is known as Compton wavelength.

When a photon undergoes collision process, the photo loses its energy and this energy is transferred to the electrons. This causes energy of the photon to decrease and thus the frequency also decreases. Thus, the wavelength of the photon will increase.

6 0

3 years ago

What is momentum?????

Leokris [45]

Momentum is defined as the mass (m) times the velocity (v). If an object is steady so its velocity is zero resulting in zero momentum.

7 0

3 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Angelina_Jolie [31]

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

a)

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

$\alpha=\frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time elapsed

The angular velocity can be written as

$\omega=\frac{2\pi}{T}$

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

$\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})$

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

$T'=T+8.09\cdot 10^{-6} =0.01400809 s$

Therefore, the change in angular velocity after 1 year is

$\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s$

So, the angular acceleration of the pulsar is

$\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y$

b)

To solve this part, we can use the following equation of motion:

$\omega'=\omega + \alpha t$

where

$\omega'$ is the final angular velocity

$\omega$ is the initial angular velocity

$\alpha$ is the angular acceleration

t is the time

For the pulsar in this problem:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s$ is the initial angular velocity

$\omega'=0$ , since we want to find the time t after which the pulsar stops rotating

$\alpha = -0.259 rad/s/y$ is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

$t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y$

c)

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

$\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y$

which means that the period of the pulsar increases by

$\Delta T=8.09\cdot 10^{-6} s$

For every year:

$\Delta t=1 y$

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

$T=T_0+\frac{\Delta T}{\Delta t}\Delta t$

where $T_0$ is the original period and

$\Delta t=869 y$

is the time that has passed; solving for T0,

$T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s$

6 0

3 years ago

A bug of mass 2.2 g is sitting at the edge of a cd of radius 3.0 cm. if the cd is spinning at 280 rpm, what is the angular momen

m_a_m_a [10]

M = 2.2 g = 2.2 x 10⁻³ kg, the mass of the bug.
r = 3.0 cm = 0.03 m, the radial distance from the center.

The angular speed is
ω = 280 rpm
= (280 rev/min)*(2π rad/rev)*(1/60 min/s)
= 29.3215 rad/s

The moment of inertia of the bug is
I = mr²
= (2.2 x 10⁻³ kg)*(0.03 m)²
= 1.98 x 10⁻⁶ kg-m²

Calculate the angular momentum of the bug.
J = Iω
= (1.98 x 10⁻⁶ kg-m²)*(29.3215 rad/s)
= 5.806 x 10⁻⁵ (kg-m²)/s

Answer: 5.806 x 10⁻⁵ (kg-m²)/s

5 0

4 years ago