Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
Answer:
Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²
Explanation:
Given;
power of radio transmitter, P = 63.2 kW = 63200 W
distance of transmission, r = 30.5 km
Intensity of the transmitted radio wave is calculated as follows;
where;
I is the intensity of the transmitted radio wave
Substitute the given values and calculate the intensity of the transmitted radio wave;
Therefore, Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²
Answer:
10500 J/kg/*C
Explanation:
Quantity of heat required=mass of substance x specific heat capacity x change in temperature
Quantity of heat required=0.25 x 4200 x [30-20]
Quantity of heat required=0.25 x 4200 x 10
Quantity of heat required=10500 J/kg/*C
