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2 years ago

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A satellite with a mass of 5.6 E 5 kg is orbiting the Earth in a circular path. Determine the satellite's velocity if it is orbi

ting at a distance of 6.8 E 5 m above the Earth's surface. Earth's mass = 5.98 E 24 kg; Earth's radius = 6.357 E 6 m.
6,800 m/s
7,200 m/s
7,500 m/s
7,900 m/s

1 answer:

solniwko [45]2 years ago

7 0

Answer:

7500 m/s

Explanation:

Centripetal acceleration = gravity

v² / r = GM / r²

v = √(GM / r)

Given:

G = 6.67×10⁻¹¹ m³/kg/s²

M = 5.98×10²⁴ kg

r = 6.8×10⁵ + 6.357×10⁶ = 7.037×10⁶ m

v = √(6.67×10⁻¹¹ (5.98×10²⁴) / (7.037×10⁶))

v = 7500

The orbital velocity is 7500 m/s.

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Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

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Time t = 0.16 s

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F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

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Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

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2 years ago

Two objects (45.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h

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a) The acceleration of the objects is $3.56 m/s^2$

b) The tension in the string is 280.8 N

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a)

We start by writing the equations of motion for the two masses attached to the pulley.

For the heavier mass, we have:

$m_1 g - T = m_1 a$ (1)

where

$m_1 = 45.0 kg$ is the mass

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For the lighter mass, we have

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where

T is the tension in the string

$m_2 = 21.0 kg$ is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

a is the acceleration of the system (here we assumed that the lighter mass accelerates upward)

From (1) we get

$T=m_1g - m_1 a$

And substituting into (2),

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b)

From the previous part of the problem we got an expression for the tension in the string:

$T=m_1g - m_1 a$

Where we have

$m_1 = 45.0 kg$

$g=9.8 m/s^2$

$a=3.56 m/s^2$ is the acceleration, found in part a)

Susbtituting, we find

$T=(45.0)(9.8)-(45.0)(3.56)=280.8 N$

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2 years ago

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q = heat required by water = ?

m = mass of water = 250 g

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$T_1$ = initial temperature of water = 293.0 K

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Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

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Now put all the given values in the above formula, we get:

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