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2 years ago

A particle starts to move in a straight line with a velocity of 10 m/s

Physics

2 answers:

Naddik [55]2 years ago

8 0

Answer: Zero after 5 s and -10 m/s after 10s

Explanation:

Gre4nikov [31]2 years ago

5 0

Answer:

Explanation:

Lol man

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I NEED HELP ASAP!!!

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1 year ago

Read 2 more answers

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 $u = 2.05 *10^{4} \ m/s$

Generally the total energy possessed by the space probe when on earth is mathematically represented as

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

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

$T__{E}} = KE_p$

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

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

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2 years ago