Answer:
(a)
(b) 5220 j
(c) 1740 watt
(d) 3446.66 watt
Explanation:
We have given mass m = 290 kg
Initial velocity u = 0 m/sec
Final velocity v = 6 m/sec
Time t = 3 sec
From first equation of motion
v = u+at
So
(a) We know that force is given by
F = ma
So force will be
(b) From second equation of motion we know that
We know that work done is given by
W = F s = 580×9 =5220 j
(c) Time is given as t = 3 sec
We know that power is given as
(d) Time t = 1.5 sec
So
Answer:
6.32 m/s 18.43° northeast
Explanation:
We express the velocity of hawk as:
We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:
≈
And the angle with respect to the east should be with:
Is there specific factors I have to choose from?
Answer:
C) either of these depending on speed
Explanation:
In classical mechanics, the momentum of a body is defined as the product of its mass and its velocity at a given time. Although the bullet has a smaller mass than the container ship, its momentum may be greater or lesser than container ship's, depending on its speed.
Answer:
Explanation:
f =
T = 120 N
L = 3.00 m
(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m
(wow that's massive for a "rope")
f = )
f = /6 = 0.527 Hz
This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.
A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N