Missing part in the text of the problem:
"<span>Water is exposed to infrared radiation of wavelength 3.0×10^−6 m"</span>
First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
where
m=1.8 g is the mass of the water
is the specific heat capacity of the water
is the increase in temperature.
Substituting the data, we find
We know that each photon carries an energy of
where h is the Planck constant and f the frequency of the photon. Using the wavelength, we can find the photon frequency:
So, the energy of a single photon of this frequency is
and the number of photons needed is the total energy needed divided by the energy of a single photon:
Answer:
Second drop: 1.04 m
First drop: 1.66 m
Explanation:
Assuming the droplets are not affected by aerodynamic drag.
They are in free fall, affected only by gravity.
I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.
We can use the equation for position under constant acceleration.
X(t) = x0 + v0 * t + 1/2 * a *t^2
x0 = 0
a = 9.81 m/s^2
v0 = 0
Then:
X(t) = 4.9 * t^2
The drop will hit the floor when X(t) = 1.9
1.9 = 4.9 * t^2
t^2 = 1.9 / 4.9
That is the moment when the 4th drop begins falling.
Assuming they fall at constant interval,
Δt = 0.62 / 3 = 0.2 s (approximately)
The second drop will be at:
X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m
And the third at:
X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m
The positions are:
1.9 - 0.86 = 1.04 m
1.9 - 0.24 = 1.66 m
above the floor
Explanation:
Given parameters:
Distance hopped = 84m
Displacement = 84m due east
Time = 7s
Unknown:
Speed of kangaroo = ?
Velocity of kangaroo = ?
Solution:
To solve this problem,
Speed = 
= 
= 12m/s
Velocity = 
= = 12m/s due east
Answer:
m1/m2 = 0.51
Explanation:
First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:
V = √F/u
This is the equation that describes the relation between speed of a pulse and a force exerted on it.
the value of "u" is:
u = m/L
Where m is the mass of the rod, and L the length.
Now, for the rod 1:
V1 = √F/u1 (1)
rod 2:
V2 = √F/u2 (2)
Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:
1.4V2 = √F/u1 (3)
Replacing (2) in (3):
1.4(√F/u2) = √F/u1 (4)
Now, let's solve the equation 4:
[1.4(√F/u2)]² = F/u1
1.96(F/u2) =F/u1
1.96F = F*u2/u1
1.96 = u2/u1 (5)
Now, replacing the expression of u into (5) we have the following:
1.96 = m2/L / m1/L
1.96 = m2/m1 (6)
But we need m1/m2 so:
1.96m1 = m2
m1/m2 = 1/1.96
m1/m2 = 0.51
Answer:
<h3>2,321.62Joules</h3>
Explanation:
The formula for calculating workdone is expressed as;
Workdone = Force * Distance
Get the force
F = nR
n is the coefficient of friction = 0.5
R is the reaction = mg
R = 46 ( 9.8)
R = 450.8N
F = 0.5 * 450.8
F = 225.4N
Distance = 10.3m
Get the workdone
Workdone = 225.4 * 10.3
Workdone = 2,321.62Joules
<em>Hence the amount of work done is 2,321.62Joules</em>
