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3 years ago

Does the left sink appear to be deeper, shallower, or the same depth as the right sink?

Physics

1 answer:

Debora [2.8K]3 years ago

3 0

Answer:

shallower.

Explanation:

To understand and to be able to solve this question, we have to understand the concept of Apparent depth and real depth.

If we have a bowl and water is filled to its brim, when you look inside the bowl full of water you will notice that the distance from the top of the bowl of water to the buttom seems a little bit closer this is known as as the APPARENT DEPTH and it is caused by REFRACTION.

The Snell's law can be used to show the relationship between the real depth and the apparent depth as given below;

n = real depth/apparent depth.

Refraction which is a property of wave causes the left sink which is filled with water to be able to bend. As light goes into water, it bends which will makes the left sink which is filled with water to appear SHALLOWER.

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Show that the optimal launch angle for a projectile subject to gravity is 45o by carrying out the following steps: 6. Write down

Lorico [155]

Answer:

sin 2θ = 1 θ=45

Explanation:

They ask us to prove that the optimal launch angle is 45º, for this by reviewing the parabolic launch equations we have the scope equation

R = Vo² sin 2θ / g

Where R is the horizontal range, Vo is the initial velocity, g the acceleration of gravity and θ the launch angle. From this equation we see that the sine function is maximum 2θ = 90 since sin 90 = 1 which implies that θ = 45º; This proves that this is the optimum angle to have the maximum range.

We calculate the distance traveled for different angle

R = vo² Sin (2 15) /9.8

R = Vo² 0.051 m

In the table are all values in two ways

Angle (θ) distance R (x)

0 0 0

15 0.051 Vo² 0.5 Vo²/g

30 0.088 vo² 0.866 Vo²/g

45 0.102 Vo² 1 Vo²/g

60 0.088 Vo² 0.866 Vo²/g

75 0.051 vo² 0.5 Vo²/g

90 0 0

See graphic ( R Vs θ) in the attached ¡, it can be done with any program, for example EXCEL

6 0

3 years ago

A battery charger is connected to a dead battery and delivers a current of 8.9 A for 4.7 hours, keeping the voltage across the b

adell [148]

Answer:

1807.56 kJ

Explanation:

Parameters given:

Current, I = 8.9A

Time, t = 4.7hrs = 4.7 * 3600 = 16920 secs

Voltage, V = 12V

Electrical energy is given as:

E = I*V*t

Where I = Current

V = Voltage/Potential differenxe

t = time in seconds.

E = 8.9 * 12 * 16920

E = 1807056 J = 1807.056 kJ

3 0

4 years ago

Read 2 more answers

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$