M)³ / 6 = 4.2e9 m³
<span>so its mass is </span>
<span>M = 3300kg/m³ * 4.2e9m³ = 1.4e13 kg </span>
<span>and so its KE at 16 km/s = 16000 m/s is </span>
<span>KE = ½ * 1.4e13kg * (16000m/s)² = 1.8e21 J
</span># of bombs N = 1.8e21J / 4.0e16J/bomb = 44 234 bombs
<span>give or take.
</span>
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Answer:
i do belive its C
Explanation:
i remeber this question from somewhere also it makes the most sense
42.9°
Explanation:
Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:
Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at 
Solving for the angle, we get
or
![\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]](https://tex.z-dn.net/?f=%5C%3B%5C%3B%5C%3B%3D%20%20%5Csin%5E%7B-1%7D%5Cleft%5B%5Cdfrac%7B34%5C%3A%5Ctext%7BN%7D%7D%7B%285.1%5C%3A%5Ctext%7Bkg%7D%29%289.8%5C%3A%5Ctext%7Bm%2Fs%7D%5E2%29%7D%5Cright%5D)
The average act on her during the deceleration is 4.47 meters per second.
<u>Explanation</u>:
<u>Given</u>:
youngster mass m = 50.0 kg
She steps off a 1.00 m high platform that is s = 1 meter
She comes to rest in the 10-meter second
<u>To Find</u>:
The average force and momentum
<u>Formulas</u>:
p = m * v
F * Δ t = Δ p
vf^2= vi^2+2as
<u>Solution</u>:
a = 9.8 m/s
vi = 0
vf^2= 0+2(9.8)(1)
vf^2 = 19.6
vf = 4.47 m/s .
Therefore the average force is 4.47 m/s.
