Answer:The velocity of the object will be 5
m/s or 13.23m/s
Explanation:
force exerted by the object= 30N
distance displayed by the object by the action of force=6.0m
mass of object=10kg
velocity gained by the object=?
![\frac{1}{2}mv^{2}= forcexdisplacement\\\frac{1}{2}10v^{2} = 30x6\\ 5v^{2}=180\\ v^{2}= 180-5\\ v^{2} =175\\v=\sqrt{175} \\v=5\sqrt{7} or 13.23](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E%7B2%7D%3D%20forcexdisplacement%5C%5C%5Cfrac%7B1%7D%7B2%7D10v%5E%7B2%7D%20%3D%2030x6%5C%5C%205v%5E%7B2%7D%3D180%5C%5C%20v%5E%7B2%7D%3D%20180-5%5C%5C%20v%5E%7B2%7D%20%3D175%5C%5Cv%3D%5Csqrt%7B175%7D%20%5C%5Cv%3D5%5Csqrt%7B7%7D%20or%2013.23)
Answer:
Time taken, t = 30.15 minutes
Explanation:
It is given that,
Distance from the center of planet, r = 40 miles
Speed of the shuttle, v = 500 miles per hour
Let t is the time taken by it to complete one full orbit. The displacement of the shuttle is,
![d=2\pi r](https://tex.z-dn.net/?f=d%3D2%5Cpi%20r)
![d=2\pi \times 40](https://tex.z-dn.net/?f=d%3D2%5Cpi%20%5Ctimes%2040)
d = 251.32 miles
Now time can be calculated using the expression for the velocity as :
![v=\dfrac{d}{t}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%7D%7Bt%7D)
![t=\dfrac{d}{v}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bd%7D%7Bv%7D)
![t=\dfrac{251.32\ miles}{500\ miles\ per\ hour}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B251.32%5C%20miles%7D%7B500%5C%20miles%5C%20per%5C%20hour%7D)
t = 0.50264 hours
or
t = 30.15 minutes
So, the shuttle will take 30.15 minutes to complete one full orbit. Hence, this is the required solution.
Answer:
76969.29 W
Explanation:
Applying,
P = F×v............. Equation 1
Where P = Power, F = force, v = velocity
But,
F = ma.......... Equation 2
Where m = mass, a = acceleration
Also,
a = (v-u)/t......... Equation 3
Given: u = 0 m/s ( from rest), v = 12.87 m/s, t = 3.47 s
Substitute these values into equation 3
a = (12.87-0)/3.47
a = 3.71 m/s²
Also Given: m = 1612 kg
Substitute into equation 2
F = 1612(3.71)
F = 5980.52 N.
Finally,
Substitute into equation 1
P = 5980.52×12.87
P = 76969.29 W