The electrons are already there. They are freely moving through the conductor.
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
Answer:
22145.27733 ft
124984.76055 ft
Explanation:
The equation of pressure is

where,
=Atmospheric pressure = 800 mbar
k = Constant
h = Altitude = 35000 ft


Now


The altitude will be 22145.27733 ft


The elevation is 124984.76055 ft
Answer:
Explanation:
To solve this problem we use the Hooke's Law:
(1)
F is the Force needed to expand or compress the spring by a distance Δx.
The spring stretches 0.2cm per Newton, in other words:
1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm
The force applied is due to the weight

We replace in (1):
We solve the equation for m:
<span>Reducing the distance between them. In theory, also increasing the mass; but you can't really change the mass of an object. However, you can compare the forces if you replace an object by a different object, which has a different mass.
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i hope this will work..