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:
3 m/s
Explanation:
We'll begin by calculating the change in displacement of the jogger. This can be obtained as follow:
Initial displacement (d₁) = 4 m
Final displacement (d₂) = 16 m
Change in displacement (Δd) =?
Δd = d₂ – d₁
Δd = 16 – 4
Δd = 12 m
Finally, we shall determine the determine the average velocity. This can be obtained as follow:
Change in displacement (Δd) = 12 m
Time (t) = 4 s
Velocity (v) =?
v = Δd / t
v = 12 / 4
v = 3 m/s
Thus, the average velocity of the jogger is 3 m/s
I believe the answer is "When a neutral atom looses an electron to another neutral atom, two charged atoms are created."
Frequency of the wave is 2 per second
Explanation:
- Frequency is the number of times waves pass at a particular point of time. Here, time period = 0.5 s
- Frequency is given by the formula
f = 1/T, where f is the frequency and T is the time period
⇒ f = 1/0.5 = 2 per second
Answer/Explanation:
They have a relatively small mass compared to Protons and Neutrons. Protons are electrochemically positive in charge and the Neutrons are electrochemically neutral in charge.
