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pantera1 [17]
3 years ago
11

The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward a

t the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.
Physics
1 answer:
tino4ka555 [31]3 years ago
3 0

Answer:

The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.

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I hand you two spheres. They have the same mass, the same radius, and the same exterior surface. I claim one is a solid sphere a
castortr0y [4]

Answer:

Explanation:

Between two spheres of equal masses having equal radius , the one which is hollow will have less moment of inertia and therefore less  radius of gyration ( k )

When they are allowed to roll over an inclined plane , they will have acceleration as follows

a = gsinθ / ( 1 + k²/r²)

For hollow sphere , k is less . therefore denominator is less and acceleration is more.

in other words , hollow sphere will have greater acceleration . So it will reach the bottom of an inclined surface in lesser time if allowed to roll over it .  

This is how we can identify a hollow sphere without breaking it open .

5 0
3 years ago
9. Electron travelling though two horizontal plates
VLD [36.1K]

Answer:

;

Explanation:

;

6 0
3 years ago
A small metal bead, labeled A, has a charge of 26 nC . It is touched to metal bead B, initially neutral, so that the two beads s
adell [148]

Answer:

q_A=25.953\ \rm nC

q_B=0.047\ \rm nC

Explanation:

Given:

  • Total initial charge on bead A=26 nC
  • The distance between them=5 cm
  • Magnitude o the force between them =4.4\times10^{-4}

Using coulombs law the force between two charged particle is \dfrac{q_Aq_B}{4\pi \epsilon_0 r^2}

where r is the radial distance between them

According to question we have

4.4\times10^{-4}=\dfrac{q_Aq_B\times9\times10^9}{0.05^2}\\\\4.4\times10^{-4}=\dfrac{q_A(q_A-26\times10^{-9})\times9\times10^9}{0.05^2}\\q_A=25.953\ \rm nC\\q_B=0.047\ \rm nC

Hence the charge on two metal beads is calculated.

5 0
3 years ago
A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.
Nataliya [291]

Answer:

Approximately 1.2 \times 10^{4}\; {\rm J} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: g = 9.81\; {\rm N \cdot kg^{-1}}.

The change in the gravitational potential energy ({\rm GPE}) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass m increased by \Delta h, the {\rm GPE} of that object would have increased by m\, g\, \Delta h.

In this question, the height of this object increased by \Delta h = 6.0\; {\rm m}. The mass of this object is m = 200\; {\rm kg}. Thus, the {\rm GPE} of this object would have increased by:

\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}.

(Note that 1\; {\rm N \cdot m} = 1\; {\rm J}.)

3 0
2 years ago
A charge of 5.4 C experiences a force of 25.0 in an electric field. What is the strength of electric field at that point ? If th
Airida [17]

The strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and  12.38 N.

<h3></h3><h3>What is the electric field strength?</h3>

The electric field strength is defined as the ratio of electric force to charge.

Given data;

q₁ = 5.4 C

F₁ is the electric force in case1

E is the electric field =?

F₂ is the electric force in case 2

q₂ is the charge 2

The strength of the electric field at that point is;

F₁=Eq₁

E₁=F/q₁

E₁=25.0 N / 5.4 C

E₁=4.587 N/C

The force would this charge experience at that point when the charge is 2.7 C;

F₂=Eq₂

F₂=4.587 N/C × 2.7 C

F₂ = 12.38 N

Hence the strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and  12.38 N.

To learn more about the electric field strength, refer to the link;

brainly.com/question/4264413

#SPJ1

8 0
2 years ago
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