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ratelena [41]
2 years ago
11

A 15.2 kg mass has a gravitational potential energy of -342 J. How high from the ground is it? Group of answer choices GPE canno

t be negative 2.3 m 22.5 m 530 m
Physics
1 answer:
vredina [299]2 years ago
4 0

A. The height above the ground reached by the mass is 2.3 m.

<h3>Height traveled by the mass</h3>

P.E = mgh

where;

  • m is mass
  • h is height reached by the object
  • g is acceleration due to gravity

h = P.E/mg

h = (342)/(15.2 x 9.8)

h = 2.3 m

Thus, the height above the ground reached by the mass is 2.3 m.

Learn more about gravitational potential energy here: brainly.com/question/1242059

#SPJ1

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A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
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Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

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             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

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Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

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Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

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3 years ago
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