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almond37 [142]
2 years ago
11

object P1 and P2 are in a straight line with the normal to a plane mirror.If P1 and P2 are 18m and 21m away from the mirror. Cal

culate 1)distance between P1 and it's image I1​
Physics
1 answer:
VARVARA [1.3K]2 years ago
8 0

The distance between object P1 and its image formed is determined as 36 m.

<h3>Distance of the image</h3>

The distance of the image formed by object P1 is calculated as follows;

In a plane mirror; object distance = image distance

image distance of P1 = 18 m

distance between object and image = 18m + 18 m = 36 m

Thus, the distance between object P1 and its image formed is determined as 36 m.

Learn more about plane mirrors here: brainly.com/question/1126858

#SPJ1

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What is the mechanical advantage of the screw shown below? O A. 14.1 O B. 2 O C. 12.6 O D. 8.2.​
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Assume (unrealistically) that a TV station acts as a point source broadcasting isotropically at 3.2 MW. What is the intensity of
Dahasolnce [82]

Answer:

I=1.5\times10^{-28}W/m^2

Explanation:

The intensity is related to the power and surface area by I=\frac{P}{A}=\frac{P}{4\pi r^2}. We need to calculate the surface area of a sphere of radius r=4.3ly.

Since 4.3ly is the distance light travels in 4.3 years at 299792458m/s, we can obtain it in meters by doing:

r=vt=(299792458m/s)(4.3\times365\times24\times60\times60s)=4.1\times10^{16}m

So we have:

I=\frac{P}{4\pi r^2}=\frac{3.2\times10^6W}{4\pi (4.1\times10^{16}m)^2}=1.5\times10^{-28}W/m^2

4 0
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Why can’t a real machine ever have 100% efficiency
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3 years ago
A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do t
Marrrta [24]

Answer:

a) U = 652.545\,kJ, b) v \approx 659.568\,\frac{m}{s}

Explanation:

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Q = U

The internal energy for a monoatomic ideal gas is:

U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T

Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:

T = \frac{P\cdot V}{n\cdot R_{u}}

T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}

T = 52.325\,K

The thermal energy contained by the gas is:

U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)

U = 652.545\,kJ

b) The physical model for the cat is constructed from Work-Energy Theorem:

U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:

v = \sqrt{\frac{2 \cdot U}{m_{cat}}}

v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }

v \approx 659.568\,\frac{m}{s}

3 0
3 years ago
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