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MissTica
3 years ago
10

A ball that has a mechanical energy of 65 J has 12 J of kinetic energy. The ball has

Physics
1 answer:
bulgar [2K]3 years ago
8 0

Answer:

The ball has a potential energy of 53 J.

Explanation:

Mechanical energy, E = Kinetic energy + Potential energy

E = K.E + P.E

65 = 12 + P.E

P.E = 65 – 12

P.E = 53 J

Therefore the potential energy of the ball is 53 J

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Identify each statement as an example of melting or sublimation,
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5 0
3 years ago
A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,
Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2

now we know that

v_0 = 660 m/s

M = 5.98 \times 10^{24} kg

m = 5 \times 10^9 kg

r = 4 \times 10^9 m

R = 6.37 \times 10^6 m

now from above formula

GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2

now we have

2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2

now plug in all data

2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2

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5 0
3 years ago
"relate the fertile phase of the menstrual cycle to the process of fertilisation"​
DerKrebs [107]

Explanation:

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8 0
2 years ago
A 25,000 kg traveling east collides with a 2,000 kg truck standing still on the tracks. After the collision the train and truck
Elis [28]

Answer:

24.084 m/s

Explanation:

From the law of conservation of linear momentum

Total momentum before collision equals to the total momentum after collision

Since momentum=mv where m is mass and v is velocity

M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing}) where M_{truck} is the mass of the truck, V_{truck} is velocity of the truck, V_{common} is the common velocity of moving and standing truck after collision and M_{standing} is the mass of the standing truck

Making V_{truck} the subject we obtain

V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}

Substituting M_{truck} as 25000 Kg, V_{common} as 22.3 m/s, M_{standing} as 2000 Kg we obtain

V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s

Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive

The truck was moving at 24.084 m/s

3 0
3 years ago
A referee will toss up the ball between to opponents.what is this called
OlgaM077 [116]
This happens in basketball. It is known as "jump ball".
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3 years ago
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