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Law Incorporation [45]
3 years ago
11

Why does a straw look broken viewed in a glass of water?

Physics
1 answer:
Nat2105 [25]3 years ago
7 0

Answer:

This is due to refraction of light

Explanation:

When a straw is placed in a glass it appears to be broken as a result of refraction of light on the air-water interface.

Since, the refractive index of water is more than that of glass.

Also, water is denser than air, which means light travels slower in water than in air.

Most importantly, When light travels from rarer to denser medium it bents towards the normal and deviates from its original path which explains the broken view of straw in a glass of water.  

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Current Flow and Ohm's Law

Ohm's law is the most important, basic law of electricity. It defines the relationship between the three fundamental electrical quantities: current, voltage, and resistance. When a voltage is applied to a circuit containing only resistive elements (i.e. no coils), current flows according to Ohm's Law, which is shown below.

I = V / R 

Where: 

I =

Electrical Current (Amperes)

V =

Voltage (Voltage)

R =

Resistance (Ohms)

    

Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change. Similarly, increasing the resistance of the circuit will lower the current flow if the voltage is not changed. The formula can be reorganized so that the relationship can easily be seen for all of the three variables.

The Java applet below allows the user to vary each of these three parameters in Ohm's Law and see the effect on the other two parameters. Values may be input into the dialog boxes, or the resistance and voltage may also be varied by moving the arrows in the applet. Current and voltage are shown as they would be displayed on an oscilloscope with the X-axis being time and the Y-axis being the amplitude of the current or voltage. Ohm's Law is valid for both direct current (DC) and alternating current (AC). Note that in AC circuits consisting of purely resistive elements, the current and voltage are always in phase with each other.

Exercise: Use the interactive applet below to investigate the relationship of the variables in Ohm's law. Vary the voltage in the circuit by clicking and dragging the head of the arrow, which is marked with the V. The resistance in the circuit can be increased by dragging the arrow head under the variable resister, which is marked R. Please note that the vertical scale of the oscilloscope screen automatically adjusts to reflect the value of the current.

See what happens to the voltage and current as the resistance in the circuit is increased. What happens if there is not enough resistance in a circuit? If the resistance is increased, what must happen in order to maintain the same level of current flow?


4 0
3 years ago
How many basic states of matter exist?<br> three<br> two<br> five<br> four
djverab [1.8K]

Answer:

There are four basic states of matter

5 0
2 years ago
Read 2 more answers
In the absence of air resistance, a ball of mass m is tossed upward to reach a height of 20 m. at the 10 m position, halfway up,
Mazyrski [523]

On the ball, there is mg of net force.

<h3>How would one describe air resistance?</h3>
  • Air exerts a force known as air resistance. The force works in the opposite direction of an object traveling through the air.
  • While a sports vehicle with a streamlined design will encounter reduced air resistance and experience less drag, the automobile will be able to move more quickly than a truck with a flat front.
<h3>What Causes Air Resistance?</h3>
  • Air resistance, also referred to as "drag," is a force brought on by air. When air specks collide with an object's front, it slows down.
  • The more air particles that impact the object and the larger its surface area, the more resistance it faces.

learn more about air resistance here

brainly.com/question/1385438

#SPJ4

5 0
1 year ago
Suppose the price elasticity of supply for gasoline in the short run is estimated to be 0.4. Due to an unexpected surge in the d
netineya [11]

Answer:

increase by 8 percent

Explanation:

<em>Price elasticity of supply of a product is the degree of responsiveness of supply of that product to a change in price.</em> Simply put:

Price Elasticity of supply = change in quantity supply/ change in price.

In this case, price elasticity of supply of gasoline = 0.4

Percentage price increase = 20 percent.

Hence,

           0.4 = change in supply/20

           Change in supply = 20 x 0.4 = 8 percent

<em>Therefore, the quantity supply of gasoline will increase by 8 percent</em>

5 0
3 years ago
A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

4 0
3 years ago
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