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pantera1 [17]
2 years ago
5

A 25kg ball is thrown into the air. when thrown it is going 10 m/s. calculate how high it travels?

Physics
1 answer:
Darina [25.2K]2 years ago
3 0

Answer:

5m

Explanation:

V^2 = U^2 - 2gh

V=0 , U=10m/s, g=10m/s^2

0^2 = 10^2 - 2×10×h

0 = 100 - 20h

100=20h

H = 100/20 = 5m

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It's 300! I just got it correct:)

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Substance A has a heat capacity that is much greater than that of substance B. If 10.0 g of substance A initially at 30.0 ∘C is
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Explanation:

In a heat exchange, the temperature change is inversely proportional to the specific heat capacity. Since substance A has a heat capacity that is much greater than that of substance B, the temperature change of substance A will be less than the temperature change of substance B. Therefore, the final temperature is closer to that of 30^\circ C than 80^\circ C.

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State Pascal's principle of pressure . please help due tomorrow​
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3 years ago
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
Elanso [62]

Answer:

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

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

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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

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

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

        KE_p =  \frac{1}{2}  *  m *  v^2

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}}

=>     v  =  2.3359 *10^{4} \ m/s

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Answer:

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