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a_sh-v [17]
3 years ago
5

The acceleration due to gravity on the surface of Jupiter is about 2.5 times the acceleration due to gravity on Earth’s surface.

What would be the weight of a space probe on the surface of Jupiter?
A.2.5 times lighter than on Earth
B.6.25 times heavier than on Earth
C.2.5 times heavier than on Earth
D.6.25 times lighter than on Earth
Physics
1 answer:
Ann [662]3 years ago
6 0

Answer: The correct answer is option C.

Explanation:

Weight = Mass × Acceleration

Let the mass of the space probe be m

Acceleration due to gravity on the earth = g

Weight of the space probe on earth = W

W=m\times g

Acceleration due to gravity on the Jupiter = g' = 2.5g

Weight of the space probe on earth = W'

W'=mg'=m\times 2.5g

\frac{W'}{W}=\frac{m\times 2.5g}{m\times g}

W'=2.5\times W

The weight of the space probe on the Jupiter will be 2.5 times the weight of the space probe on earth.

Hence, the correct answer is option C.

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A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo
Zinaida [17]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2

E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

3 0
3 years ago
A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e
IgorLugansk [536]

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm  - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

6 0
3 years ago
John walks 1.00 km north, then turns right and walks 1.00 km east. His speed is 1.50 m/s during the entire stroll.a) What is the
avanturin [10]

Answer:

(a) 1.414 km

(b) 1.06 m/s

Explanation:

(a) For John:

Distance = 1 km north and then 1 km east

Speed = 1.5 m/s

total distance traveled = 1 + 1 = 2 km = 2000 m

Time taken to travel = Distance / speed

t = 2000 / 1.5 = 1333.3 seconds

Displacement = \sqrt{1^{2}+1^{2}}=1.414 km

(b) For jane :

Time is same as john = 1333.33 second

Distance = 1.414 km = 1414 m

Speed = distance / time = 1414 / 1333.33 = 1.06 m/s

3 0
3 years ago
1. Una pelota rueda hacia la derecha siguiendo una trayectoria en línea recta de modo que recorre una distancia de 10m en 5 s ,
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Answer:

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3 years ago
Bádminton is played to a score of
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