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: sheet of charge
Explanation:
a )
Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.
b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.
C ) Electric field which is almost uniform near the sheet of charge is equal t the following
E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²
E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²
= .3389 x 10³
= 338.9 V / m
spacing between 1 V
= 1 / 338.9 m
= 2.95 X 10⁻3 m
= 2.95 mm.
Answer:
Plzzzzzzzzzzzzzzzz brainliest
Explanation:
In static friction, the frictional force resists force that is applied to an object, and the object remains at rest until the force of static friction is overcome. In kinetic friction, the frictional force resists the motion of an object. ... The frictional force itself is directed oppositely to the motion of the object.
From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)
Distance traveled is
d = 100 km = 10⁵ m
Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s
Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s
The difference in time is
302.12 - 19.49 = 282.63 s
Answer: 283 s (nearest second)
Answer:
=24.25 ^−1
Explanation:
Let and be initial and final velocity of the body respectively,
be acceleration due to gravity ( 9.8^−2 ), ℎ be the height of the body.
=0 ^ −1
ℎ=30
we know that, ^2−^ 2=2ℎ
^2=2∗9.8∗30
^2=588
=24.25 ^−1
