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:
5. dispersion
6. 49.8°
Explanation:
5. Dispersion is the name given to the phenomenon of light of different wavelengths being bent differently. A rainbow is the result of light from a point source (the sun) being spread out by wavelength (color), a nice example of dispersion.
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6. n = 1.31 is the ratio of the sine of the angle of refraction to the sine of the angle of incidence (for light passing to a medium of n = 1). When the angle of refraction is 90°, the angle of incidence is the "critical angle." So, ...
sin(90°)/sin(critical) = 1.31
critical angle = arcsin(1/1.31) ≈ 49.8°
Answer:
x=4.06m
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem
Vf=7.6m/s
t=1.07
Vo=0
we can use the ecuation number one to find the acceleration
a=(Vf-Vo)/t
a=(7.6-0)/1.07=7.1m/s^2
then we can use the ecuation number 2 to find the distance
{Vf^{2}-Vo^2}/{2.a} =X
(7.6^2-0^2)/(2x7.1)=4.06m
Answer:
0.76 rad/s^2
Explanation:
First, we convert the original and final velocity from rev/s to rad/s:
Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:
Now, we just use the formula:
