Answer:
Kinetic energy of the projectile at the vertex of the trajectory: 
.
Work done when firing this projectile: 
.
Explanation:
Since the drag on this projectile is negligible, the horizontal velocity 
of this projectile would stay the same (at 
) throughout the flight.
The vertical velocity 
of this projectile would be 
at the vertex (highest point) of its trajectory. (Otherwise, if 
, this projectile would continue moving up and reach an even higher point. If 
, the projectile would be moving downwards, meaning that its previous location was higher than the current one.)
Overall, the velocity of this projectile would be 
when it is at the top of the trajectory. The kinetic energy 
of this projectile (mass 
) at the vertex of its trajectory would be:
.
Apply the Pythagorean Theorem to find the initial speed of this projectile:
.
Hence, the initial kinetic energy of this projectile would be:
.
All that energy was from the work done in launching this projectile. Hence, the (useful) work done in launching this projectile would be .
E = mc²
E = 0.235 kg · (3×10⁸ m/s)² = 0.235 · 9×10¹⁶ kg·m/s²
E = 2.115×10¹⁶ J
The answer is d) 2.12×10¹⁶ J
Answer:
The required steady force of each rocket is 28.79 N
Explanation:
mass of the satellite, M=3900 kg
radius, r=4.3 m
mass of rocket, m=210 kg
time, t=5.4 min
Moment of Inertia:
I = 1/2 (Mr^2) + 4mr^2
I = 1/2 ( 3900* (4.3)^2) + 4 (210)*(4.3)^2
I = 51587.1 kg m^2
the angular acceleration is:
a= w/t
here w= 2*π*30
so,
a= 2*π*30 / 5.4* 3600
a=0.0096 rad/ s^2
the Torque becomes:
T=I*a = 4r*F
( 51587.1 )*(0.0096) = 4*4.3* F
F= 28.79 N
the required steady force of each rocket is 28.79 N
learn more about steady force here:
<u>brainly.com/question/13841147</u>
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To calculate the center of gravity, divide total weight distance moment by total mass of the system. Thus, the center of gravity is 13 meter from left-hand side.
Were you able to fix that problem?
