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MrRa [10]
3 years ago
6

Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward force (thr

ust) of 3456 N, the craft descends at constant speed; if the engine provides only 2333 N, the craft accelerates downward at 0.39 m/s2. What is the weight of the landing craft in the vicinity of Callisto's surface
Physics
1 answer:
nordsb [41]3 years ago
5 0

Answer:

W=3456 N

Explanation:

Force 1 F_1=3456

Force 2 F_2=2333N

Acceleration at stage 2 a_2=0.39

Generally the weight of the Craft W is given as

W= upward force(thrust)

Therefore

W=3456 N

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The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres
12345 [234]

Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure(P_1)=210+101.325=311.325

T_1=25^{\circ}\approx 298 K

V=0.025 m^3

T_2=50^{\circ}\approx 323 K

as the volume remains constant therefore

\frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{311.325}{298}=\frac{P_2}{323}

P_2=337.44 KPa

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}

m_1=0.91 kg

Final mass m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}

m_2=0.839

Therefore m_1-m_2=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

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3 years ago
What type of subsurface material is able to store groundwater?
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Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
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From the law of conservation of momentum
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8 0
3 years ago
A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.470 rev/s abo
Charra [1.4K]

Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, \omega=0.47\ rev/s=2.95\ rad/s

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

I={m_1r^2}

I={52\times 3.9^2}

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

I_2=\dfrac{m_2r^2}{2}

I_2=\dfrac{118\times (3.9)^2}{2}

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

I=I_1+I_2

I=790.92+897.39

I = 1688.31 kg-m²

The angular momentum of the system is :

L=I\times \omega

L=1688.31 \times 2.95

L=4980.5\ kg-m^2/s

So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.

8 0
3 years ago
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