<span>The distance between two objects is increased by three times the oringinal distance. Since they were already separated by one time the original distance,
the additional three times the oringinal distance now puts them four times the original distance apart.
Whether we're talking about the gravitational forces of attraction or
the electrical forces of attraction, either one is inversely proportional
to the square of the distance between the objects.
So changing the distance to four times the original distance causes
the forces to become 1/4</span>² as strong as they were originally.
The forces become 1/16 of their original magnitude.<span>
</span>
Answer:
Explanation:
potential energy of compressed spring
= 1/2 k d²
= 1/2 x 730 d²
= 365 d²
This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .
Kinetic energy after crossing the rough patch
= 1/2 x 1.2 x 2.3²
= 3.174 J
Loss of energy
= 365 d² - 3.174
This loss is due to negative work done by frictional force
work done by friction = friction force x width of patch
= μmg d , μ = coefficient of friction , m is mass of block , d is width of patch
= .44 x 1.2 x 9.8 x .05
= .2587 J
365 d² - 3.174 = .2587
365 d² = 3.4327
d² = 3.4327 / 365
= .0094
d = .097 m
= 9.7 cm
If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .
Answer:
a-
V= IR
9V = I ×( 12+6)
I = 9/ 18 A = 0.5 A
b
V=IR
240 = 6 A ×( 20 + R)
40 = 20 + R
R = 20 ohm
c
resultant resistance of the 2 parallel resistances= Ro
1/Ro = 1/ 5 + 1/ 20
1/Ro =( 20+5)/100
= 1/Ro = 1/4
Ro= 4 ohm
V=IR
V = 2A × ( 1+ 4 OHM)
V = 10V
d
equivalent resistance = Ro
1/Ro = 1/(2+8) + 1/(5+5)
1/Ro = 1/10 +1/10
2/10 = 1/ Ro
Ro= 10/2 = 5 ohm
V = IR
12V = I × 5Ohm
I=2.4 A
