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AveGali [126]
3 years ago
7

A space transportation vehicle releases a 470-kg communications satellite while in a circular orbit 350 km above the surface of

the Earth. A rocket engine on the satellite boosts it into an orbit 2350 km above the surface of the Earth. How much energy does the engine have to provide for this boost?
Physics
1 answer:
natima [27]3 years ago
8 0

Answer:

E = 3.194 x 10⁹ J = 3.194 GJ

Explanation:

The formula for the absolute potential energy is:

U = - GMm/2r

where,

G = Gravitational Constant = 6.67 x 10⁺¹¹ N m²/kg²

M = mass of Earth = 5.972 x 10²⁴ kg

m = mass of satellite = 470 kg

r = distance between the center of Earth and satellite

Thus, the energy required from engine will be difference between the potential energies.

E = U₂ - U₁

E = - GMm/2r₂ - (- GMm/2r₁)

E = (GMm/2)(1/r₁ - 1/r₂)

where,

r₁ = Radius of Earth + 350 km = 6371 km + 350 km = 6721 km = 6.721 x 10⁶ m

r₂=Radius of Earth + 2350 km=6371 km + 2350 km= 8721 km = 8.721 x 10⁶ m

therefore,

E = [(6.67 x 10⁺¹¹ N m²/kg²)(5.972 x 10²⁴ kg)(470 kg)/2](1/6.721 x 10⁶ m - 1/8.721 x 10⁶ m)

<u>E = 3.194 x 10⁹ J = 3.194 GJ</u>

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Violet light (λ = 400 nm) passing through a diffraction grating for which the slit spacing is 6.0 μm forms a pattern on a screen
sergey [27]

Answer:

3 order dark fringe

Explanation:

y = Distance from central bright fringe = 204 mm

λ = Wavelength = 400 nm

L = Distance between screen and source = 1 m

d = Slit distance = 6 μm

tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{204}{1000}=0.2012^{\circ}

dsin\theta=m\lambda\\\Rightarrow m=\frac{dsin\theta}{\lambda}\\\Rightarrow m=\frac{6\times 10^{-6}sin0.2012}{400\times 10^{-9}}=2.9982\approx 3

Order of fringe is 3

So, it is a Dark order fringe

3 0
4 years ago
A small electronic device is rated at 0.16 W when connected to 115 V. What is the resistance of this device?
Ierofanga [76]

Answer:

150156.25 Ω

Explanation:

Resistance: This can be defined as the opposition to the flow of electric current in a circuit. The S.I unit of resistance is Ohm's (Ω)

The expression for resistance is given as

P = V²/R................ equation 1

Where P = power, V = Voltage, R = Resistance.

Making R the subject of the equation,

R = V²/P.................. Equation 2

Given: V = 115 V , P = 0.16 W.

Substitute into equation 2

R = 155²/0.16

R = 150156.25 Ω

Hence,

The resistance = 150156.25 Ω

8 0
3 years ago
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True because the kind of light we get is white light and everything we see is actually every colour except the color we are seeing because that is wwhat is reflected

6 0
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A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above
gizmo_the_mogwai [7]

Answer:

B.The force of friction between the block and surface will decrease.

Explanation:

The force of friction is given by

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where \mu is the coefficient of friction and N is the normal force.

When the student pulls on the block with force F_p at an angle \theta, the normal force on the block becomes

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Now, as we increase \theta, sin(\theta) increases which as a result decreases the normal force Mg- F_psin(\theta), which also means the frictional force decreases; Hence choice B stands true.

<em>P.S: Choice D is tempting but incorrect since the weight </em>W=mg<em> is independent of the external forces on the block. </em>

6 0
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