All categories

3 years ago

What type of mirror can produce both converging and diverging rays?

Physics

2 answers:

Ivahew [28]3 years ago

4 0

Answer:

A convex mirror is a diverging mirror (f is negative) and forms only one type of image. It is a case 3 image—one that is upright and smaller than the object, just as for diverging lenses.

Explanation:

hope this helps have a good night :)

belka [17]3 years ago

4 0

Convex mirror
i hope this helps :) i learned about this last year

You might be interested in

Calcula qué tan lejos está Saturno del Sol, sabiendo que la luz solar necesita una hora y veinte segundos para llegar a Saturno.

Ksju [112]

Answer:

mi gusta

Explanation:

6 0

3 years ago

What distance will be traveled if you are going 120km/hr for 30 min?

nikklg [1K]

Answer:

60 km

Explanation:

For an object (or a person, such as in this case) moving at constant speed, the speed is equal to the ratio between the distance travelled and the time taken:

$v=\frac{d}{t}$

where

v is the speed

d is the distance

t is the time taken

In this case, we have:

v = 120 km/h is the speed

t = 30 min = 0.5 h is the time taken

Therefore, we can rearrange the equation to find the total distance travelled:

$d=vt=(120)(0.5)=60 km$

3 0

3 years ago

Plzzzzzzzzzz!!!!!!! Hurryyyyy

Scilla [17]

Answer:

student A or B

Explanation:

A common demonstration is to put a ringing alarm clock or bell in the bell jar, and when the vacuum is created, you can no longer hear the sound of the clock/bell.

The bell is connected to a lab pack or batteries and rung to show pupils it can be heard under normal circumstances. The bell jar is then connected to a vacuum pump using a vacuum plate (see Fig 2) and the air is removed from inside creating a near vacuum. The bell is then again rung. This time however, it cannot be heard.

Small low voltage buzzers can be used as a bell replacement for the bell and work in exactly the same way though teachers generally prefer bells as students may be able to see the hammer moving, proving that it is actually ringing even though they cannot hear it.

Some vacuum pumps are better than others at keeping a strong vacuum though if you cannot completely lose the sound, you will at least notice the volume decreasing.

Sound is simply a series of longitudinal waves travelling from the source, through the air to our ears. Without air present, these waves cannot form and therefore sound cannot be conveyed.

In a longitudinal wave the particles oscillate back and forth in the direction of the wave movement unlike transverse waves which like waves on the sea, single particles travel up and down and not in the direction of the wave.

Because you will not be able to create a perfect vacuum, you may still be able to hear the bell ring slightly. Vibrations from the ringing bell can also travel up to the bung in the bell jar which in turn may resonate the jar slightly. This means you may hear the bell ring, however strong the vacuum. To compensate for this, try to insulate the bell as much as possible from the bell jar. Hanging the bell using elastic cord means some of the vibrations will be absorbed by the cord and not be transferred to the bell jar.

3 0

2 years ago

Two bulbs are connected in parallel across a source of EMF = 8.0V with a negligible internal resistance. One bulb has a resistan

Sonja [21]

<h2>Answer:</h2>

(a) 3.18Ω

(b) 3.18Ω

<h2>Explanation:</h2>

Let the two bulbs be A and B

Given;

$R_{A}$ = Resistance in bulb A = 3.0Ω

$R_{B}$ = Resistance in bulb B = 2.5Ω

Since the two bulbs are connected in parallel;

i. their effective resistance ( $R_{X}$ ) is given by

$\frac{1}{R_{X}}$ = $\frac{1}{R_{A} }$ + $\frac{1}{R_{B} }$ ---------------(i)

Substitute the values of $R_{A}$ and $R_{B}$ into equation (i)

=> $\frac{1}{R_{X}}$ = $\frac{1}{3.0}$ + $\frac{1}{2.5}$

Solve for $R_{X}$

$R_{X}$ = 1.36Ω

ii. voltage (potential difference), V, across them is the same;

Therefore we can get the total current (I) that will flow through them if the voltage to be supplied is 2.4V.

Use the Ohm's law;

V = I x R -----------------(ii)

Where;

V = voltage across them = 2.4V

I = total current flowing through them

R = their effective resistance = $R_{X}$ = 1.36Ω

Substitute these values into equation (ii) as follows;

2.4 = I x 1.36

I = 2.4 / 1.36

I = 1.76A

(a) Now get the value of R

Since the voltage across the two bulbs is 2.4V out of the 8.0V supplied by the source, then the remaining (8.0 - 2.4 = 5.6)V will pass across the resistor R.

Also, since the two bulbs make a series connection with the resistor R, the same total current (I = 1.76A) that flows through these bulbs will flow through the resistor R.

Therefore, to get the value of R, we use the relation

V = I x R ------------------------------(iii)

Where;

V = voltage across the resistor = 5.6V

I = current through the resistor = 1.76A

<em>Substitute these values into equation (iii)</em>

=> 5.6 = 1.76 x R

=> R = 5.6 / 1.76

=> R = 3.18Ω

Therefore, the value of R to be chosen in order to supply each bulb with a voltage of 2.4V is 3.18Ω

(b) The potential difference and voltage across refer to the same thing. Therefore, the value of R that would make the potential difference across each of the bulbs be 2.4V is the same as the one calculated in (a) which is 3.18Ω

3 0

3 years ago

Four different objects are placed on a number line at 0. The chart describes the motion of each object.

Ket [755]

Answer:

Explanation:

8 0

3 years ago