Answer:
Explanation:
mass of probe m = 474 Kg
initial speed u = 275 m /s
force acting on it F = 5.6 x 10⁻² N
displacement s = 2.42 x 10⁹ m
A )
initial kinetic energy = 1/2 m u² , m is mass of probe.
= .5 x 474 x 275²
= 17923125 J
B )
work done by engine
= force x displacement
= 5.6 x 10⁻² x 2.42 x 10⁹
= 13.55 x 10⁷ J
C ) Final kinetic energy
= Initial K E + work done by force on it
= 17923125 +13.55 x 10⁷
= 1.79 x 10⁷ + 13.55 x 10⁷
= 15.34 x 10⁷ J
D ) If v be its velocity
1/2 m v² = 15.34 x 10⁷
1/2 x 474 x v² = 15.34 x 10⁷
v² = 64.72 x 10⁴
v = 8.04 x 10² m /s
= 804 m /s
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There are 5,280 feet in 1 mile.
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Answer:
Explanation:
You can approach an expression for the instantaneous velocity at any point on the path by taking the limit as the time interval gets smaller and smaller. Such a limiting process is called a derivative and the instantaneous velocity can be defined as.#3
For the special case of straight line motion in the x direction, the average velocity takes the form: If the beginning and ending velocities for this motion are known, and the acceleration is constant, the average velocity can also be expressed as For this special case, these expressions give the same result. Example for non-constant acceleration#1
Our values can be defined like this,
The problem can be solved for part A, through the Work Theorem that says the following,
Where
KE = Kinetic energy,
Given things like that and replacing we have that the work is given by
W = Fd
and kinetic energy by
So,
Clearing F,
Replacing the values
B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.
Answer:
Explanation:
point a represents time 0 and position coordinate 30
point b represents time 10 s , and position coordinate 50 m .
time elapsed = 10 - 0 = 10 s .
displacement = 50 m - 30 m
= 20 m
average velocity = displacement / time elapsed
= 20 / 10
= 2 m /s .
