Answer:
height is 69.68 m
Explanation:
given data
before it hits the ground = 46 % of entire distance
to find out
the height
solution
we know here acceleration and displacement that is
d = (0.5)gt² ..............1
here d is distance and g is acceleration and t is time
so when object falling it will be
h = 4.9 t² ....................2
and in 1st part of question
we have (100% - 46% ) = 54 %
so falling objects will be there
0.54 h = 4.9 (t-1)² ...................3
so
now we have 2 equation with unknown
we equate both equation
1st equation already solve for h
substitute h in the second equation and find t
0.54 × 4.9 t² = 4.9 (t-1)²
t = 0.576 s and 3.771 s
we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls
so take t = 3.771 s
then h from equation 2
h = 4.9 t²
h = 4.9 (3.771)²
h = 69.68 m
so height is 69.68 m
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:

Answer:
i hope this will help you :)
Explanation:
mass=19kg
density=800kg/m³
volume=?
as we know that
density=mass/volume
density×volume=mass
volume=mass/density
putting the values
volume=19kg/800kg/m³
so volume=0.02375≈0.02m³
Answer:
7350 J
Explanation:
The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

where
m is the mass of the rock
g is the gravitational acceleration
h is the height of the cliff
In this problem, we have
m = 50 kg
g = 9.8 m/s^2
h = 15 m
Substituting numbers into the formula, we find:

When boat is sunk into the liquid the net buoyancy on the boat is counterbalanced by weight of the boat
So here weight of the boat = Buoyancy force
let say boat is sunk by distance "h"
now we can say


now by above force balance equation we can write




so boat will sunk by total 5 mm distance