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Sauron [17]
2 years ago
6

An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

speed of 44.4 km/s, just enough for the spacecraft to escape the Earth's gravity and leave the solar system. Ignoring air resistance and the Earth's rotation, what is the speed of the spacecraft when it is more than halfway to the star? Assume we are ignoring the effect of the Sun on the spacecraft.
Physics
1 answer:
Anna [14]2 years ago
6 0

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2

Mass gets cancelled

-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}

v_e=\sqrt{\frac{2Gm}{R}} = Escape velocity of Earth = 11.2 km/s

v_i = Velocity of projectile = 44.4 km/s

v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

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V125BC [204]

Answer: choices a and b

Explanation:

Telescope can be defined as am optical instrument which is designed to observe the distant objects clear and nearer. It comprises of arrangement of lenses which allow the rays of light to be collected. The collected light is focused and the image so produced is magnified in the form of an image. The telescopes are prepared and manufactured on mountains top as this will help in preventing the distortion of light obtain from the star due to the fluctuation of air mass in the atmosphere. The atmospheric distortion affects the resolution, and affects the vision. The atmospheric pressure is low at the mountain tops so it will help in better observation of the sky.

6 0
2 years ago
Light travels in a straight line at a constant speed of 3,0 x 108 m/s for 4,1
zepelin [54]

Answer:

As the velocity of light is constant so the acceleration of the light is equal to zero.

a= dv/dt

Explanation:

4 0
2 years ago
Using the formula F = M* A. What is the acceleration of a .5 kg
Katyanochek1 [597]

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7 0
2 years ago
The magnetic flux through each turn of a 110-turn coil is given by ΦB = 9.75 ✕ 10−3 sin(ωt), where ω is the angular speed of the
Xelga [282]

Answer:

Explanation:

Given that a coil has a turns of

N = 110 turns

And the flux is given as function of t

ΦB = 9.75 ✕ 10^-3 sin(ωt),

Given that, at an instant the angular velocity is 8.70 ✕ 10² rev/min

ω = 8.70 ✕ 10² rev/min

Converting this to rad/sec

1 rev = 2πrad

Then,

ω = 8.7 × 10² × 2π / 60

ω = 91.11 rad/s

Now, we want to find the induced EMF as a function of time

EMF is given as

ε = —NdΦB/dt

ΦB = 9.75 ✕ 10^-3 sin(ωt),

dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)

So,

ε = —NdΦB/dt

ε = —110 × 9.75 × 10^-3•ω Cos(ωt)

Since ω = 91.11 rad/s

ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)

ε = —97.71 Cos(91.11t)

The EMF as a function of time is

ε = —97.71 Cos(91.11t)

Extra

The maximum EMF will be when Cos(91.11t) = -1

Then, maximum emf = 97.71V

8 0
3 years ago
Answer this question correctly for a pat on the back.
ss7ja [257]
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2 years ago
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