Answer:
because the book's is covered by papers
Answer:
ii. Same
Explanation:
Electrostatic potential V=kQq/r
When the test charge is brought to a point A at a distance r and a net charge of +Q
i.e At point A
q₁ = q
r₁= r
V=kQq/r
At the point B, When q₂ = 2q and
r₂= 2r
V=kQ * 2q/2r
The two in the numerator and the denominator cancel to give V=kQq/r which is equal to the potential at A
To protect a material from the influence of an external magnetic field, the material should be kept in soft iron ring.
So the correct answer is A.
Hope this helps,
Davinia.
Answer:
<em>45.375m</em>
Explanation:
To get the distance travelled by the body we will use the equation of motion
S = ut + 1/2at²
u is the initial speed = 0m/s
a is the acceleration
t is the time = 5 seconds
We need to get the acceleration first:
According to Newton's second law,
Fm is the moving force = Wsin theta
Ff is the frictional force = 1500N
m is the mass = W/g
m = 4000/9.8
m = 408.16kg
Substitute the given values in the formula and get the acceleration:
Get the required distance:
Recall that S = ut+1/2at²
S = 0(5) + 1/2*3.63*5²
S = 0+0.5*3.63*25
S = 45.375m
<em>Hence the distance travelled by the body in 5 seconds if the force of friction is 1500N is 45.375m</em>
The answer for the following problem is mentioned below.
The option for the question is "A" approximately.
- <u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>
Explanation:
Given:
Spring constant (k) = 240 N/m
amount of the compression (x) = 0.40 m
To calculate:
Elastic potential energy (E)
We know;
<em>According to the formula;</em>
E = × k × x × x
<u>E = </u><u> × k ×(x)²</u>
where;
E represents the elastic potential energy
K represents the spring constant
x represents amount of the compression in the string
So therefore,
Substituting the values in the above formula;
E = × 240 × (0.40)²
E = × 240 × 0.16
E = × 38.4
E = 19.2 J or approximately 20 J
<u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>