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Tom [10]
3 years ago
10

Can a plane mirror produce a real image? Explain.

Physics
1 answer:
Komok [63]3 years ago
4 0
I don't think so, because in order to produce an image, you need a surface behind the mirror. The light will hit the mirror, then it will bounce it back in your eyes and you see the image.
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When operated on a household 110.0 V line, typical hair dryers draw about 1650 W of power. The current can be modeled as a long,
Andre45 [30]

Explanation:

Given that,

Voltage of household line, V = 110 V

Power of the hairdryer, P = 1650 W

During use, the current is about 1.95 cm from the user's hand.

(a) Power is given by :

P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1650\ W}{110\ V}\\\\I=15\ A

(b) Again the power is given by :

P=\dfrac{V^2}{R}

R is resistance of the dryer

R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{1650}\\\\R=7.34\ \Omega

(c) The magnetic field produced by the dryer at the user's hand is given by :

B=\dfrac{\mu_o I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 15}{2\pi \times 1.95\times 10^{-2}}\\\\B=1.53\times 10^{-4}\ T

Hence, this is the required solution.

4 0
2 years ago
What is the independent variable of the graph above?
Makovka662 [10]

Answer:

Height of 15 kg Object is the correct answer.

Explanation:

Because the IV (Independent Variable) is always on the x-axis.

6 0
3 years ago
A guitar string is 0.620m long, and oscillates at 234Hz. What is the velocity of the waves in the string? m/s
9966 [12]

Answer:

v = 72.54 m/s

Explanation:

We have,

Length of a guitar string is 0.62 m

Frequency of a guitar string is 234 Hz

For guitar string,

L=2\lambda\\\\\lambda=\dfrac{L}{2}\\\\\lambda=\dfrac{0.62}{2}\\\\\lambda=0.31\ m

The velocity of the wave in the string is given by :

v=f\lambda\\\\v=234\times 0.31\\\\v=72.54\ m/s

So, the velocity of the waves in the string is 72.54 m/s.

3 0
3 years ago
If the object represented by the FBD below has a mass of 2.5 kg, what is the acceleration of the object?
Debora [2.8K]

Answer:

4 m/s² down

Explanation:

We'll begin by calculating the net force acting on the object.

The net force acting on the object from the left and right side is zero because the same force is applied on both sides.

Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:

Force up (Fᵤ) = 15 N

Force down (Fₔ) = 25 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fᵤ

Fₙ = 25 – 15

Fₙ = 10 N down

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass (ml= 2.5 Kg

Net force (Fₙ) = 10 N down

Acceleration (a) =?

Fₙ = ma

10 = 2.5 × a

Divide both side by 2.5

a = 10 / 2.5

a = 4 m/s² down

Therefore, the acceleration of the object is 4 m/s² down

6 0
3 years ago
A sound source producing 1.00 kHz waves moves toward a stationary listener at one-half the speed of sound.
SIZIF [17.4K]

Answer:

2000Hz and 1500Hz

Explanation:

Using

a) f = f0((c+vr)/(c+vs))

=>>> f0((c)/(c-0.5c))

=>>>1000/0.5 = 2000Hz

b) f = f0((c+vr)/(c+vs))

=>>>f0((c+0.5c)/(c))

=>>>>1000 x 1.5 = 1500Hz

5 0
3 years ago
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