Answer:
The value is 
Explanation:
From the question we are told that
The initial speed is 
Generally the total energy possessed by the space probe when on earth is mathematically represented as

Here
is the kinetic energy of the space probe due to its initial speed which is mathematically represented as
=>
=> 
And
is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

Here
is the escape velocity from earth which has a value 
=> 
=> 
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

Generally from the law energy conservation we have that
So

=> 
=> 
=> 
44 x 12. I got the 12 from the total of 12 months in a year.
44 > 40
x
12 > 10
----------
The way my teacher taught me how to estimate is look at the neighbor to 44 and 12. The only time 44 can become 50, is when the neighbor is 5 or up. Same thing for 12. Now, multiply 40 and 10.
40 x 10 = 400.
Therefore, your estimate is 400.
The real answer is 520 breaths.
Answer:
<em>B. 68.6m</em>
Explanation:
<u>Free Fall Motion
</u>
When a body is left to move in the air with no friction, the motion is ruled only by the force of gravity. The vertical distance a body travels in the air after a time t is
.

We know the egg takes 3.74 seconds to reach the ground. The height it was launched from is


The closest correct option is
B. 68.6m
A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.
<h3>Why does the first ball comes to rest after collision ?</h3>
Let m be the mass of the two identical balls.
u1 = velocity before the collision of ball 1
u2 = 0 = velocity of second ball that is at rest
v1 and v2 are the velocities of the balls after the collision.
From the conservation of momentum,
∴ mu1 + mu2 = mv1 + mv2
∴ mu1 = mv1 + mv2
∴ u1 = v1 + v2
In an elastic collision, the kinetic energy of the system before and after collision remains same.

∴ 
∴ 
∴
₁
₂ = 0
- It is impossible for the mass to be zero.
- Because the second ball moves, velocity v2 cannot be zero.
- As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>
An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.
Can learn more about elastic collision from brainly.com/question/12644900
#SPJ4
Q = mc<span>∆t, where:
q = energy flow
m = mass, 120 000 g
c = specific heat capacity, 4.81 J/gC
</span><span>∆t = change in temperature, ~75 (100 - 25, which is room temperature)
Substituting in the values, we get:
q = 120000 x 4.81 x 75 = 43290000 Joules = 43.29 MJ
Hope I helped!! xx
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