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Physics

1 answer:

FrozenT [24]3 years ago

7 0

Answer:

11250Joules

Explanation:

1 half times 100kg times (15m/s) ^2

1. Squared the 15 =225m/s

2.times the kg to m/s- 100kg x 225m/s=22500

3. Divided by 2 = 11250Joules

KE= 11250J

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A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit

elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

= 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

$T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }$ second

Squaring the terms and solving it for 'g'

g = 4 π² $\frac{(R+h)^{3} }{R^{2}T^{2} }$ m/s²

Substituting the values in the above equation

g = 4 π² $\frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}$

g = 4.84 m/s²

Therefore, the weight

w = m x g newton

= 5832 Kg x 4.84 m/s²

= 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N

8 0

3 years ago

Look at the diagram. Emily needs to complete the circuit. What two points should be connected to complete the circuit and make t

pogonyaev

Answer:

The points 2 and 4 should be connected.

Explanation:

To complete the circuit, we need to connect the two points which when connected, encompass the battery and the bulb in the circuit. The points 2 and 4 do the job, since they connect the terminal of the battery and the terminal of the bulb, and thus complete the circuit.

Therefore, the choice C is correct.