Answer:
a. Concave mirror, radius of curvature = 16 cm b. magnification = 2
Explanation:
a. Since the image is upright and larger than the object, we need a concave mirror.
We know image height, h'/object height, h = -image distance, d'/object distance, d
h'/h = -d'/d
Using the real is positive convention,
h'= + 5.4 cm, h = + 2.7 cm and d = + 12 cm.
So, + 5.4 cm/+ 2.7 cm = -d'/+ 12 cm
2 = -d'/12
d' = -2 × 12 cm
= -24 cm
Using the mirror formula 1/d + 1/d' = 2/r where r = radius of curvature of the mirror
1/+12 + 1/- 24 = 2/r
1/12(1 - 1/2) = 2/r
1/12(1/2) = 2/r
1/24 = 2/r
r/2 = 24
r = 2 × 24
r = 48 cm
b.
magnification = image height, h'/object height,h = + 5.4 cm/+ 2.7 cm = 2
Initial Velocity is the velocity at time interval t = 0 and it is represented by u. It is the velocity at which the motion starts. They are four initial velocity formulas: (1) If time, acceleration and final velocity are provided, the initial velocity is articulated as. u = v – at.
One atom on nitrogen because there isn’t a number directly after it’s the 3 signifies that there is three groups of those atoms as a total
Answer:
4 P
Explanation:
V = IR
when they are in series R = 2 r
when they are in parallel R = 1/2 r
so current will increase by a factor of 4 when they are in parallel
and power used will also increase by a factor of 4 to 4 P
Power = IV when I now becomes 4 I power becomes 4 P
Answer:
(A). The electric field strength inside the solenoid at a point on the axis is zero.
(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is 
.
Explanation:
Given that,
Magnetic field = 2.0 T
Diameter = 5.0 cm
Rate of decreasing in magnetic field = 5.00 T/s
(A). We need to calculate the electric field strength inside the solenoid at a point on the axis
Using formula of electric field inside the solenoid
Electric field on the axis of the solenoid
Here, r = 0
The electric field strength inside the solenoid at a point on the axis is zero.
(B). We need to calculate the electric field strength inside the solenoid at a point 1.50 cm from the axis
Using formula of electric field inside the solenoid
Hence, (A). The electric field strength inside the solenoid at a point on the axis is zero.
(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is .
