A frog can be many different colours. It appears green under normal 'white' light because it absorbs all the other colours in the light's spectrum apart from green. It reflects the green light back and that is picked up by your eye.
If the light is red, there is no green in the spectrum of the light, only red. So, the red light will be absorbed and there is no green to be reflected back for you to see. Therefore, the frog will not look green.
Answer:
The maximum value of the induced magnetic field is
.
Explanation:
Given that,
Radius of plate = 30 mm
Separation = 5.0 mm
Frequency = 60 Hz
Suppose the maximum potential difference is 100 V and r= 130 mm.
We need to calculate the angular frequency
Using formula of angular frequency

Put the value into the formula


When r>R, the magnetic field is inversely proportional to the r.
We need to calculate the maximum value of the induced magnetic field that occurs at r = R
Using formula of magnetic filed

Where, R = radius of plate
d = plate separation
V = voltage
Put the value into the formula


Hence, The maximum value of the induced magnetic field is
.
Distance of fall from rest,
without air resistance = (1/2) (gravity) (time)²
= (1/2) (9.8 m/s²) (95 sec)²
= (4.9 m/s²) (9,025 sec²)
= 44,222.5 meters .
The depth of the mine shaft is five times the height of Mt. Everest !
Answer:
<h2>59.5 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 70 × 0.85
We have the final answer as
<h3>59.5 N</h3>
Hope this helps you
Answer:
(a) f= 622.79 Hz
(b) f= 578.82 Hz
Explanation:
Given Data
Frequency= 600 Hz
Distance=1.0 m
n=120 rpm
Temperature =20 degree
Before solve this problem we need to find The sound generator moves on a circular with tangential velocity
So
Speed of sound is given by
c = √(γ·R·T/M)
............in an ideal gas
where γ heat capacity ratio
R universal gas constant
T absolute temperature
M molar mass
The speed of sound at 20°C is
c = √(1.40 ×8.314472J/molK ×293.15K / 0.0289645kg/mol)
c= 343.24m/s
The sound moves on a circular with tangential velocity
vt = ω·r.................where
ω=2·π·n
vt= 2·π·n·r
vt= 2·π · 120min⁻¹ · 1m
vt= 753.6 m/min
convert m/min to m/sec
vt= 12.56 m/s
Part A
For maximum frequency is observed
v = vt
f = f₀/(1 - vt/c )
f= 600Hz / (1 - (12.56m/s / 343.24m/s) )
f= 622.789 Hz
Part B
For minimum frequency is observed
v = -vt
f = f₀/(1 + vt/c )
f= 600Hz / (1 + (12.56m/s / 343.24m/s) )
f= 578.82 Hz