a. 46 m/s east
The jet here is moving with a uniform accelerated motion, so we can use the following suvat equation to find its velocity:
where
v is the velocity calculated at time t
u is the initial velocity
a is the acceleration
The jet in the problem has, taking east as positive direction:
u = +16 m/s is the initial velocity
is the acceleration
Substituting t = 10 s, we find the final velocity of the jet:
And since the result is positive, the direction is east.
b. 310 m
The displacement of the jet can be found using another suvat equation
where
s is the displacement
u is the initial velocity
a is the acceleration
t is the time
For the jet in this problem,
u = +16 m/s is the initial velocity
is the acceleration
t = 10 s is the time
Substituting into the equation,
Answer:
A
Explanation:
hope it helped a lot
pls mark brainliest with due respect .
The 'strength' of the electric field is the force on 1C of charge at that point.
At this 'certain location', the field is 40/5 = 8 newtons per coulomb = <u>8 volts</u>
Answer:
8400m
Explanation:
The engine that falls off would have the same constant horizontal velocity as the airplane's when if falls off if we ignore air resistance. So it would have a horizontal velocity of 280m/s for 30seconds before it hits the ground.
Therefor the horizontal distance the engine travels during its fall is
280 * 30 = 8400m
Answer: Speed = 4 m/s
Explanation:
The parameters given are
Mass M = 60 kg
Height h = 0.8 m
Acceleration due to gravity g= 10 m/s2
Before the man jumps, he will be experiencing potential energy at the top of the table.
P.E = mgh
Substitute all the parameters into the formula
P.E = 60 × 9.8 × 0.8
P.E = 470.4 J
As he jumped from the table and hit the ground, the whole P.E will be converted to kinetic energy according to conservative of energy.
When hitting the ground,
K.E = P.E
Where K.E = 1/2mv^2
Substitute m and 470.4 into the formula
470.4 = 1/2 × 60 × V^2
V^2 = 470.4/30
V^2 = 15.68
V = square root (15.68)
V = 3.959 m/s
Therefore, the speed of the man when hitting the ground is approximately 4 m/s
