Answer:
Part a)

Part b)

Explanation:
As we know that capacitor plate is connected across 2.1 V and after charging it is disconnected from the battery
So here we can say that charge on the plates will remain conserved
So we will have

now dielectric is removed between the plates of capacitor
so new potential difference between the plates




Part b)
Now the capacitor plates are again isolated and unknown dielectric is inserted between the plates
So again charge is same so potential difference is given as




Answer:
If it is not an object in motion, all forces are balanced.
The speed of sound at T=25°C is Vs=346 m/s. So the sound has to reach the cliff and return back to you so the path it needs to travel is s=2*440 m = 880 m.
Since the speed of sound is constant s=Vs*t, and t= s/Vs=880/346=2.54335 s. You will hear the echo after t=2.54335 s after you shouted.
Answer:
velocity = 1527.52 ft/s
Acceleration = 80.13 ft/s²
Explanation:
We are given;
Radius of rotation; r = 32,700 ft
Radial acceleration; a_r = r¨ = 85 ft/s²
Angular velocity; ω = θ˙˙ = 0.019 rad/s
Also, angle θ reaches 66°
So, velocity of the rocket for the given position will be;
v = rθ˙˙/cos θ
so, v = 32700 × 0.019/ cos 66
v = 1527.52 ft/s
Acceleration is given by the formula ;
a = a_r/sinθ
For the given position,
a_r = r¨ - r(θ˙˙)²
Thus,
a = (r¨ - r(θ˙˙)²)/sinθ
Plugging in the relevant values, we obtain;
a = (85 - 32700(0.019)²)/sin66
a = (85 - 11.8047)/0.9135
a = 80.13 ft/s²
Answer:
Well the definition of an application is the act of putting to a special use or purpose so lam assuming that you want specific uses that scientists make of gravity in their work.
Well our first application has helped us to send satellites around the solar system with what Nasa calls gravity assist. Using a particular planets gravity to slingshot a satellite to another destination. Look it up.
The next application much simpler but here on Earth. There are many hydro-electric power stations in use all over the world. Water is stored at a high level and released falling 100s of metres to a turbine where it generates electricity.
Hope that helps.
Explanation: