Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.
Explanation:
- Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
- Let
be the speed of the river's current given as 1.00 m/s.
- Note that this speed is the magnitude of the velocity which is a vector quantity.
- The direction of the swimmer is upstream.
Hence the resultant velocity is given as, 
= S — S 0
= 1.25 — 1
= 0.25 m/s.
Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.
Energy slowly leaks outward through the radiative diffusion of photons that repeatedly bounce off ions and electrons.
<h3>What is radiative diffusion?</h3>
A radiation zone is a layer of a star's core where energy is mostly carried toward the outside by radiative diffusion and thermal conduction rather than convection.
As photons, energy passes through the radiation zone as electromagnetic radiation.
The radiative diffusion of photons that repeatedly bounce off ions and electrons progressively drains energy outward.
Hence,radiative diffusion is correct answer.
To learn more about radiative diffusion refer:
brainly.com/question/3598352
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Answer:
The possible thickness of the soap bubble = 
Explanation:
<u>Given:</u>
- Refractive index of the soap bubble,
- Wavelength of the light taken,
Let the thickness of the soap bubble be 
.
It is given that the soap bubble appears very bright, it means, there is a constructive interference takes place.
For the constructive interference of light through a thin film ( soap bubble), the condition of constructive interference is given as:
where 
is the order of constructive interference.
Since the soap bubble is appearing very bright, the order should be 0, as 
order interference has maximum intensity.
Thus,
It is the possible thickness of the soap bubble.
Explanation:
The potential energy (Ep) = 4500 J
Ep = m. g. h
=> m = Ep/(g.h)
m = 4500/(10×10)
= 4500/100
= 45
so, the mass = 45 kgs
<span>1) The differential equation that models the RC circuit is :
(d/dt)V_capacitor </span>+ (V_capacitor/RC) = (V_source/<span>RC)</span>
<span>Where the time constant of the circuit is defined by the product of R*C
Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s
2) C<span>harge of the capacitor 1.57 time constants
1.57*(2.742) = 4.3048 s
The solution of the differential equation is
</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T) + </span>V_capac(∞)
Since the capacitor is initially uncharged V_capac(0) = 0
And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V
This means,
V_capac (t) = (-9V)e ^(-t /T) + 9V
The charge in a capacitor is defined as Q = C*V
Where C is the capacitance and V is the Voltage across
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V
V_capac (4.3048 s) = 7.1275 V
Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C
3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V
Q (∞) = 89.9mF*(9V) = 0.8091 C
