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Ad libitum [116K]
3 years ago
10

A black hole is a ________

Physics
1 answer:
masya89 [10]3 years ago
8 0

Answer:

Answer to the question:

Explanation:

A black hole is a finite region of space within which there is a mass concentration high and dense enough to generate a gravitational field such that no material particle, not even light, can escape it.

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A light bulb whose resistance is 240 ohms is connected to a 120 voltage source. What is the current through the bulb?
Sedaia [141]

Answer:

0.5A

Explanation:

Using R = \frac{V}{I},

R is the resistance (in Ohms)

V is the voltage (in V)

I is the current (in A)

240=\frac{120}{I}

I = 0.5A

7 0
2 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
HELP WILL GIVE BRAINLIEST IF CORRECT
anastassius [24]
It’s B i literally jus learned this
6 0
3 years ago
Read 2 more answers
Calculate the force between charges of 5.0 x 10^-8 c and 1.0 x 10^-7 if they are 5.0 feet apart
weqwewe [10]

Answer:

F = 19.375 x 10^-6  N

Explanation:

This problem can be solved by applying Coulomb's Law, which lets us determine the force between two electrically charged particles.

It is defined as

F = (ke * q1 * q2)/ r^2

Where,

ke =  is Coulomb's constant ≈ 9×10^9 N⋅m^2⋅C^−2

q1 = 5.0 x 10^-8 C

q2 = 1.0 x 10^-7 C

r = 5 ft = 1,524 m

F  = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)

F  = (9×10^9 N⋅m^2⋅C^−2)*(5.0 x 10^-8 C)*(1.0 x 10^-7 C)/ ((1,524 m)^2)

F = 19.375 x 10^-6  N

6 0
3 years ago
You are walking along a rugged coastline in British Columbia, Canada. As the tide goes out you realize there is a beautiful flat
kkurt [141]

Answer:

<em>Wave-Cut Platform</em>

Explanation:

A wave-cut platform, or shore platform <em>is the thin flat area frequently found at the bottom of a shore cliff or along a river, bay, or sea produced by erosion.</em>

When they become apparent as huge areas of flat rock, wave-cut platforms are often most noticeable at low tide.

7 0
3 years ago
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