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Anvisha [2.4K]
3 years ago
7

What does the star release, and in what form do we see it ?

Physics
1 answer:
Dimas [21]3 years ago
3 0

Stars are made of very hot gas. This gas is mostly hydrogen and helium, which are the two lightest elements. Stars shine by burning hydrogen into helium in their cores, and later in their lives create heavier elements.

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The greater the amplitude of a wave a. The quieter the sound b. The higher the frequency of the sound c. The dimmer the light in
Stolb23 [73]

Answer:

The greater the amplitude the greater the energy.

(Think of a water wave - which carries greater energy a 1 ft wave or

a 10 ft wave)

8 0
3 years ago
Calculate the recoil speed of a 1.4 kg rifle shooting 0.006 kg bullets with muzzle speed of 800 m/a.3.43 m/s
Angelina_Jolie [31]

Answer:

a) 3.43 m/s

Explanation:

Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.

The total momentum before the bullet is shot is zero, because they are both at rest, so:

p_i = 0

Instead the total momentum of the system after the shot is:

p_f = mv+MV

where:

m = 0.006 kg is the mass of the bullet

M = 1.4 kg is the mass of the rifle

v = 800 m/s is the velocity of the bullet

V is the recoil velocity of the rifle

The total momentum is conserved, therefore we can write:

p_i = p_f

Which means:

0=mv+MV

Solving for V, we can find the recoil velocity of the rifle:

V=-\frac{mv}{M}=-\frac{(0.006)(800)}{1.4}=-3.43 m/s

where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is

a) 3.43 m/s

5 0
3 years ago
What happens when a negatively charged object A is brought near a neutral object B?
castortr0y [4]

Answer: Option B

Explanation : When a negatively charged object A gets in contact with the neutral object B, the negative charge of object will induce the opposite charges on object B. Hence, there will be a positive charge on object B

8 0
3 years ago
A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it
Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given:  </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.  

<u>Solution  </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next  

Ef=Ei+W                                                 (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).  

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form  

(Kf + mc^2) = (KJ+ mc^2)                       (2)  

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.  

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W                              (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by  

W = FΔr                                                      (4)  

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut  

W = F Δr= (190N)(6 m) = 1140 J  

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate  

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,  

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W'                                 (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by  

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate  

1/2mvf'^2=1/2mvi'^2+W'

  vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

7 0
3 years ago
Determine the magnitude and direction of the force on a 200 m power line carrying a current of 5.0-A due west in a magnetic fiel
bogdanovich [222]

Answer:

Explanation:

l = 200 m

i = 5 A west

B = 6 micro Tesla direction 30 degree north of east

angle between length vector and the magnetic field vector = 180 - 30 = 150 degree

Write the length and the magnetic field in the vector form

\overrightarrow{l}=- 200 \widehat{i}metre

\overrightarrow{B}= 6\times 10^{-6}\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )Tesla

\overrightarrow{B}=\left ( 5.2\widehat{i}+3\widehat{j} \right )\times 10^{-6}Tesla

\overrightarrow{F} = i \overrightarrow{l}\times \overrightarrow{B}

\overrightarrow{F} = 5\times \left ( -200\widehat{i} \right )\times \left ( 5.2\widehat{i}+3\widehat{j} \right )\times 10^{-6}

\overrightarrow{F} =- 3\times 10^{-3}\widehat{k}Newton

Thus, the magnitude of force is 3 x 10^-3 newton and it is directed towards negative z axis direction.

7 0
3 years ago
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