Answer:
The gravitational potential energy equals the work needed to lift the object.
Explanation:
here we know that

work done is given as

Potential energy is given as

force due to gravity is given as

now here if we plug in the value of distance and force in the formula of work done then we will have

so here we got

so we can concluded that
The gravitational potential energy equals the work needed to lift the object.
The first law of thermodynamics says that the variation of internal energy of a system is given by:

where Q is the heat delivered by the system, while W is the work done on the system.
We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system
So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J
So, the variation of internal energy of the system is
Answer:
The current needed to transmit Power of 4 W is 28.47 A
Solution:
As per the question:
Length of the antenna, 
Frequency, 
Power transmitted, 
Now,
For a monopole antenna:

where
= wavelength transmitted by the antenna
c = speed of light in vacuum

Now,
Since, the value of
>>
thus the monopole is a Hertian monopole.
The resistance is calculated as:




Now, the current I is given by:

<h3>pressure = force / area</h3>
<h3>force = 84 N</h3><h3>pressure = 6 × 10 - 5 = 55 m2</h3>
<h3>pressure = 84 / 55</h3>
<h3>pressure = 1.53 pascals</h3>
hope that helps and please tell me if i am wrong :)
Answer:
r= 2.17 m
Explanation:
Conceptual Analysis:
The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:
E= 2k*(λ/r) Formula (1)
Where:
E: electric field .( N/C)
k: Coulomb electric constant. (N*m²/C²)
λ: linear charge density. (C/m)
r : distance from the charge line to the surface where E calculates (m)
Known data
E= 2.9 N/C
λ = 3.5*10⁻¹⁰ C/m
k= 8.99 *10⁹ N*m²/C²
Problem development
We replace data in the formula (1):
E= 2*k*(λ/r)
2.9= 2*8.99 *10⁹*(3.5*10⁻¹⁰/r)
r =( 2*8.99 *10⁹*3.5*10⁻¹⁰) / (2.9)
r= 2.17 m