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Sonja [21]
3 years ago
10

(8c5p80) Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward f

orce (thrust) of 3480 N, the craft descends at constant speed; if the engine provides only 2349 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:
Alexus [3.1K]3 years ago
3 0

Answer:

The weight of the landing craft in the vicinity of Callisto's surface is 3480 N.

Explanation:

The engine of the craft provides an upward thrust of 3480 N so that the space craft descends at a constant speed.

This implies that the net force on the space craft is zero.

The upward thrust will be equal to the downward gravitational pull by Callisto.

So the weight of the craft near the vicinity will be 3480 N.

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When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart
mixer [17]

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

as we know that

m_1 = 5 kg

m_2 = 2 kg

v_{1i} = 2.8 m/s

v_{2i} = 0 m/s

now from above equation we have

5(2.8) + 2(0) = m_1v+ m_2v

14 = (5+ 2) v

v = 2 m/s

so the speed of combined system is 2 m/s

8 0
3 years ago
The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic
Shkiper50 [21]

Answer:

i) 0.7

ii) 1.39

iii) 0.6

Next time, when compiling a Physics question, ensure you put the unit of each measurement.

Explanation:

i) T = time of flight =   \frac{2uSin(A)}{g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting the values, we have: T = \frac{2(4)Sin(60)}{10} = 0.7

ii) distance travel = Range =  R = \frac{u^{2}Sin(2A) }{g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: R = \frac{4^{2}Sin(2*60) }{10} = 1.39

iii) Maximum Height = H = \frac{u^{2}(Sin(A))^{2}  }{2g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: \frac{4^{2}(Sin60)^{2}  }{2*10} = 0.6

4 0
3 years ago
Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M
horsena [70]

Answer:

sin\theta - \mu_k cos\theta = \frac{m}{M}

sin\theta - \mu_k cos\theta = 1

Explanation:

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F_f = \mu Mgcos\theta

now if it is moving down the inclined plane at constant speed

so we will have

Mgsin\theta - T - \mu mgcos\theta = 0

on other side the mass "m" will go up at constant speed

so we have

T - mg = 0

so we have

Mgsin\theta = \mu Mgcos\theta + mg

so we have

sin\theta - \mu_k cos\theta = \frac{m}{M}

for special case when m = M

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sin\theta - \mu_k cos\theta = 1

5 0
3 years ago
A scientist claimed that fabric A is better able to resist fire than fabric B. Which option describes an experiment that will pr
UNO [17]

Answer:

B. Hold each type of fabric over a candle flame and time how long it takes for the fabric to start to burn.

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Explain resolution of Force​
Margarita [4]

Answer:

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Explanation:

7 0
3 years ago
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