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k0ka [10]
4 years ago
11

Light waves will bend when passing from one transparent substance to another unless:

Physics
2 answers:
-BARSIC- [3]4 years ago
5 0
Unless the light ray strikes the medium at an angle of 90°

The phenomenon of bending of light when going from one medium to another is referred to as refraction. However, refraction is only possible when the light ray is not perpendicular to the medium. If the light is perpendicular, it passes through without refracting.
mojhsa [17]4 years ago
5 0

Light waves will bend when passing from one transparent substance to another unless the light ray strikes the medium at an angle of 90°.

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1. A 61-kg woman doing pull-ups lifts her body a distance of 0.32 meters in 1.8 seconds. What is the power provided by her bicep
Allushta [10]

Answer:

P = mgh/t = 61(9.8)(0.32)/1.8 = 106.275555... ≈ 110 W

Explanation:

Power is the rate of doing work.

The work changes her potential energy.

4 0
3 years ago
A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and
cricket20 [7]

Answer:

The frequency f = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = \frac{d}{2} =\frac{44}{2} = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = \frac{d}{2} =\frac{3.0}{2} = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = \frac{d}{2} = \frac{2.0}{2} = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that  the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2

pgy_1=\frac{1}{2}pv^2_2 +pgy_2

v_2=\sqrt{2g(y_1-y_2)}

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = \sqrt{2*9.8*(0.35-0)}

v₂ = \sqrt {6.86}

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

v_3 = \frac{v_2r_2^2}{v_3^2}

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = (2.62  m/s)* (\frac{1.5mm^2}{1.0mm^2} )

v₃ = 0.0589 m/s

∴ velocity  of the fluid in the cylinder =  0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

f=\frac{v_s}{4(h-v_3t)}

where;

v_s = velocity of sound

h = height of the fluid

v₃ = velocity  of the fluid in the cylinder

f=\frac{343}{4(0.40-(0.0589)(0.4)}

f= \frac{343}{0.6576}

f = 521.59 Hz

∴ The frequency f = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})

\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)

\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}

where;

v_s (velocity of sound) = 343 m/s

v₃ (velocity  of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s  when the cylinder has been filling for 4.0 s.

8 0
3 years ago
The graphs show the motion of an object in both the x and y directions. Classify the motion of this object.
Natali5045456 [20]

B It is moving in one dimension

6 0
3 years ago
Read 2 more answers
Suppose a shrimp has been put on the ground that has just been taken out of water.Now touch the shrimp from a distance by a stic
Alex777 [14]

Answer:

yes

Explanation:

I would say yes because it's exerting some type of force as it reacts to the stick to make it jump straight in an upwards direction

4 0
3 years ago
Read 2 more answers
If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add
mr Goodwill [35]

Each Al^+^3 ion contains three extra protons. Hence, the extra charge on each  Al^+^3 = 3 \times 1.6 \times 10^-^1^9 C

Total charge = 0.035 pC

Total charge (Q) = 0.035 \times 10^-^1^2 C

Let the number of Al^+^3 ions be n.

According to question:

n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2

n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}

n = 7.29167 \times 10^4

n = 72917

Hence, the total number of ions needed to be transferred is 72917

3 0
3 years ago
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