I believe that the answer is C<span />
Answer:
A) d = 11.8m
B) d = 4.293 m
Explanation:
A) We are told that the angle of incidence;θ_i = 70°.
Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;
tan 70° = d/4.3m
Where d is the distance from point B at which the laser beam would strike the lakebottom.
So,d = 4.3*tan70
d = 11.8m
B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)
So,
n1*sinθ_i = n2*sinθ_r
Thus; sinθ_r = (n1*sinθ_i)/n2
sinθ_r = (1 * sin70)/1.33
sinθ_r = 0.7065
θ_r = sin^(-1)0.7065
θ_r = 44.95°
Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;
d = 4.3 tan44.95
d = 4.293 m
<h2>Hello</h2>
The answer is:
<h2>Why?</h2>
Momentum is the quantity of movement of an object, and it's calculated using the mass and the velocity of the object. Momentum is expressed by the following formula:
Where:
So, calculating we have:
Remember,
Have a nice day!
The magnitude of the action–reaction pair between the two boxes (A and B) will be "18.2 N".
According to the question,
Mass of box A,
Mass of box B,
Horizontal force,
From the Newton's law,
→ 
or,
→ 
Bu substituting the values, we get
→ 
→ 
→ 
We can see that between the two boxes, the action-reaction pair exist.
then,
→ 
→ 
→ 
(magnitude)
Thus the above solution is appropriate.
Learn more about the magnitude here:
brainly.com/question/13545862
Answer:
a) When R is very small R << r, therefore the term R+ r will equal r and the current becomes
b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes
Explanation:
<u>Solution :</u>
(a) We want to get the consumed power P when R is very small. The resistor in the circuit consumed the power from this battery. In this case, the current I is leaving the source at the higher-potential terminal and the energy is being delivered to the external circuit where the rate (power) of this transfer is given by equation in the next form
P=∈*I-I^2*r (1)
Where the term ∈*I is the rate at which work is done by the battery and the term I^2*r is the rate at which electrical energy is dissipated in the internal resistance of the battery. The current in the circuit depends on the internal resistance r and we can apply equation to get the current by
I=∈/R+r (2)
When R is very small R << r, therefore the term R+ r will equal r and the current becomes
I= ∈/r
Now let us plug this expression of I into equation (1) to get the consumed power
P=∈*I-I^2*r
=I(∈-I*r)
=0
The consumed power when R is very small is zero
(b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes
I=∈/R
The dissipated power due toll could be calculated by using equation.
P=I^2*r (3)
Now let us plug the expression of I into equation (3) to get P
P=I^2*R=(∈/R)^2*R
=∈^2/R
