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algol13
3 years ago
14

A 2.0 m tall person is standing in front of a flat vertical mirror. What minimum height must the mirror have if the person is to

see her entire body, from the top of her head to her feet?
Physics
1 answer:
dangina [55]3 years ago
7 0

Answer:

100cm

Explanation:

Since the eyes are 6 cm below the top of her head, the point of incidence of the ray must be

200cm-3cm=197cm

Since the eyes are 194 cm from her feet, the point of incidence of this ray must be

194cm/2=97cm

So the lower edge of the mirror must be 97 cm from the ground and the vertical dimension of the mirror must be 197 cm - 97 cm = 100 cm, which is half the height of the person.

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2 years ago
What is the wavelength of a 22.75×109Hz radar signal in free space? The speed of light is 2.9979×108m/s. Express your answer to
Romashka-Z-Leto [24]

Answer:

1.318 * 10^(-2) m

Explanation:

Parameters given:

Frequency, f = 22.75 * 10^9 Hz

Velocity, v = 2.9979 * 10^8 m/s

Wavelength is given as:

Wavelength = v/f

Wavelength = (2.9979 * 10^8) / (22.75 * 10^9)

Wavelength = 0.01318 m = 1.318 * 10^(-2) m

8 0
3 years ago
Read 2 more answers
The resistivity of gold is 2.44 × 10-8 Ω · m at room temperature. A gold wire that is 1.8 mm in diameter and 11 cm long carries
tester [92]
Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:

Resistance  =  2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 <span>Ω

P = I^2R = .170^2 (</span>1.055x10^-3 ) = 3.048x10^-5 W
6 0
3 years ago
(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the ca
gulaghasi [49]

Answer:

(a) E_{ c} = 112.5 \mu J

(b) E'_{ c} = 18 \mu J

Solution:

According to the question:

Capacitance, C = 1.00\mu F = 1.00\times 10^{- 6} F

Voltage of the battery, V_{b} = 15.0 V

(a)The Energy stored in the Capacitor is given by:

E_{c} = \frac{1}{2}CV_{b}^{2}

E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}

E_{ c} = 112.5 \mu J

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

E'_{c} = \frac{1}{2}CV'_{b}^{2}

E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}

E'_{ c} = 18 \mu J

6 0
3 years ago
A ray of light is projected into a glass tube that is surrounded by air. The glass has an index of refraction of 1.50 and air ha
Tamiku [17]

Answer:

θ = 41.8º

Explanation:

This is an internal total reflection exercise, the equation that describes this process is

         sin θ = n₂ / n₁

where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂

        θ = sin⁻¹ n₂ / n₁

let's calculate

       θ = sin⁻¹ (1.00 / 1.50)

       θ = 41.8º

4 0
3 years ago
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