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
Answer:
Explanation:
a ) work done by gravitational force
= mg sinθ ( d + .21)
Potential energy stored in compressed spring
= 1/2 k x²
= .5 x 431 x ( .21 )²
= 9.5
According to conservation of energy
mg sinθ ( d + .21) = 9.5
3.2 x 9.8 x sin 30( d + .21 ) = 9.5
d = 40 cm
b )
As long as mg sin30 is greater than kx ( restoring force ) , there will be acceleration in the block.
mg sin30 = kx
3.2 x 9.8 x .5 = 431 x
x = 3.63 cm
When there is compression of 3.63 cm in the spring , block will have maximum velocity. there after its speed will start decreasing.
Answer : The height is 188 meters
Explanation : When the cart reached at the end from top of hill then the cart have potential energy .
Given that,
Potential energy = 88435 J
Mass of cart = 48 kg
We know that,
The potential energy is
So, the height of the top is 188 meters.
Answer:
Lens at a distance = 7.5 cm
Lens at a distance = 6.86 cm (Approx)
Explanation:
Given:
Object distance u = 12 cm
a) Focal length = 20 cm
b) Focal length = 16 cm
Computation:
a. 1/v = 1/u + 1/f
1/v = 1/20 + 1/12
v = 7.5 cm
Lens at a distance = 7.5 cm
b. 1/v = 1/u + 1/f
1/v = 1/16 + 1/12
v = 6.86 cm (Approx)
Lens at a distance = 6.86 cm (Approx)
