A 300-kg bear grasping a vertical tree slides down at constant velocity. The friction force between the
tree and the bear is
The sphere’s Electric potential energy is 1.6*
J
Given,
q=6. 5 µc, V=240 v,
We know that sphere’s Electric potential energy(E) = qV=6.5*
=1.6*
J
<h3>Electric potential energy</h3>
The configuration of a certain set of point charges within a given system is connected with the potential energy (measured in joules) known as electric potential energy, which is a product of conservative Coulomb forces. Two crucial factors—its inherent electric charge and its position in relation to other electrically charged objects—can determine whether an object has electric potential energy.
In systems with time-varying electric fields, the potential energy is referred to as "electric potential energy," but in systems with time-invariant electric fields, the potential energy is referred to as "electrostatic potential energy."
A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is the sphere’s potential energy?
Learn more about Electric potential energy here:
brainly.com/question/24284560
#SPJ4
A string with linear density 0.500 g/m.
Tension 20.0 N.
The maximum speed 
The energy contained in a section of string 3.00 m long as a function of
.
We are given following data for string with linear density held under tension :
μ = 0.5 
= 0.5 x 10⁻³ 
T = 20 N
If string is L = 3m long, total energy as a function of
is given by:
E = 1/2 x μ x L x ω² x A²
= 1/2 x μ x L x 
= 7.5 x 10⁻⁴ 
So, The total energy as a function of
= 7.5 x 10⁻⁴ 
Learn more about linear density problem here:
brainly.com/question/17190616
#SPJ4
Answer:
A) 
B) 
Explanation:
Given:
- temperature of air,

- temperature of lungs,

- specific Heat exchanged from the lungs ,

- specific heat of air,

- mass of 1 L air,

- breath rate,

A)
Now,
amount of heat needed to warm the air of lungs to the body temperature:



B)
Amount of heat lost per hour:
<u>No. of breaths per hour:</u>



<u>Now the total loss of energy in 1 hr.:</u>


