Answer:
Frequency of the light will be equal to 
Explanation:
We have given wavelength of the light 
Velocity of light is equal to 
We have to find the frequency of light
We know that velocity is equal to
, here
is wavelength and f is frequency of light
So frequency of light will be equal to 
So frequency of the light will be equal to 
By Newton's second law, the net force on the object is
∑ <em>F</em> = <em>m</em> <em>a</em>
∑ <em>F</em> = (2.00 kg) (8 <em>i</em> + 6 <em>j</em> ) m/s^2 = (16.0 <em>i</em> + 12.0 <em>j</em> ) N
Let <em>f</em> be the unknown force. Then
∑ <em>F</em> = (30.0 <em>i</em> + 16 <em>j</em> ) N + (-12.0 <em>i</em> + 8.0 <em>j</em> ) N + <em>f</em>
=> <em>f</em> = (-2.0 <em>i</em> - 12.0 <em>j</em> ) N
Answer:
1. 75N
2. 67,983 J (=67.98 kJ)
Explanation:
1. Work = Force x Distance
we are given that Work = 1,500J and Distance = 20m
hence,
Work = Force x Distance
1,500 = Force x 20
Force = 1,500 ÷ 20 = 75N
2. Potential Energy, PE = mass x gravity x change in height
we are given that mass = 165 kg and change in height = 42m
assuming that gravity, g = 9.81 m/s²
Potential Energy, PE = mass x gravity x change in height
Potential Energy, PE = 165 x 9.81 x 42 = 67,983 J (=67.98 kJ)
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
Answer:
a) 35.44 mm
b) 17.67 mm
Explanation:
u = Object distance = 3.6 m
v = Image distance
f = Focal length = 35 mm
= Object height = 1.8 m
a) Lens Equation

The CCD sensor is 35.34 mm from the lens
b) Magnification


The person appears 17.67 mm tall on the sensor