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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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Alexeev081 [22]

Answer:

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Explanation:

hope this help

4 0
3 years ago
Read 2 more answers
Select the correct answer
DiKsa [7]

Answer:

C or 3

Explanation:

A: no they are called sources of sound. A is incorrect.

B: It does. Many people attest to this. But this is not a property of physics.

C: A media is required is the correct answer.

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7 0
2 years ago
15. If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the
goldfiish [28.3K]

The final combined velocity after the collision is 20.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two car and of the truck must be conserved before and after the collision.

This means that we can write the following equation:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 800 kg is the mass of the sport car

u_1 = 13.0 m/s is the initial velocity of the car (taking its direction as positive  direction)

m_2 = 1200 kg is the mass of the truck

u_2 = 25.0 m/s is the initial velocity of the truck

v is the final combined velocity of the car and the truck, after the collision

Re-arranging the equation and substituting the values, we find the velocity after the collision:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(800)(13)+(1200)(25)}{800+1200}=20.2 m/s

And the positive sign indicates their final direction is the same as the initial direction of the two vehicles.

Learn more about momentum here:

brainly.com/question/7973509  

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#LearnwithBrainly

5 0
3 years ago
The gravitational force between two objects will be greatest in which of the following situations?
Kobotan [32]

Answer:

Explanation:

Gravitational law states that, the force of attraction or repulsion between two masses is directly proportional to the product of the two masses and inversely proportional to the square of their distance apart.

So,

Let the masses be M1 and M2,

F ∝ M1 × M2

Let the distance apart be R

F ∝ 1 / R²

Combining the two equation

F ∝ M1•M2 / R²

G is the constant of proportional and it is called gravitational constant

F = G•M1•M2 / R²

So, to increase the gravitational force, the masses to the object must be increased and the distance apart must be reduced.

So, option c is correct

C. Both objects have large masses and are close together.

8 0
2 years ago
if a ball with an original velocity of 0 is dropped from a tall structure and takes 7 Seconds to hit the ground what velocity do
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5 0
3 years ago
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