There is one mistake in the question.The Correct question is here
A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.
Answer:
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.
Explanation:
Given data
time=1/2 sec to 1 sec
v(t)=-9.8t m/s
To find
Distance
Solution
As the acceleration as first derivative of velocity with respect to time
So
acceleration(-g)= dv/dt
Solve it
dv = a dt
dv = -g dt
v - v₀ = -gt
v= dy/dt
dy = v dt
dy = ( v₀ - gt ) dt
y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]
y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]
y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.
Answer:
Gases and particles in Earth's atmosphere scatter sunlight in all directions. Blue light is scattered more than other colors because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.
Answer:
W = 14.8 kJ
Explanation:
W = F S cos ∅
W = 4113 x 3.99 x cos 25.5
W = 16410.87 x 0.9025 = 14810.8 J or 14.8 kJ
Answer:
Answer:This organism may be identified by its color, the spines on its back, the antennae, and therefore the long, thin body. There are many other characteristics that might even be wont to identify this organism.
Explanation:
We can solve the problem by applying Newton's second law, which states that the resultant of the forces acting on an object is equal to the product between its mass and its acceleration:
We should consider two different directions: the direction perpendicular to the inclined plane and the direction parallel to it. Let's write the equations of the forces along the two directions, decomposing the weight of the object (mg):
(parallel direction) (1)
(perpendicular direction) (2)
where
is the angle of the inclined plane, N is the normal reaction of the plane,
is the frictional force, with
being the coefficient of friction.
From eq.(2), we find
and if we substitute into eq.(1), we can find the acceleration of the block:
from which