What is the moment produced by 15n force ?

A python can detect thermal radiation from objects that differ in temperature from their environment as long as the received int

yanalaym [24]

Answer:

10.52 m

Explanation:

The power radiated by a body is given by

P = σεAT⁴ where ε = emissivity = 0.97, T = temperature = 30 C + 273 = 303 K, A = surface area of human body = 1.8 m², σ = 5.67 × 10⁻⁴ W/m²K⁴

P = σεAT⁴ = 5.67 × 10⁻⁸ W/m²K⁴ × 0.97 × 1.8 m² × (303)⁴ = 834.45 W

This is the power radiated by the human body.

The intensity I = P/A where A = 4πr² where r = distance from human body.

I = P/4πr²

r = (√P/πI)/2

If the python is able to detect an intensity of 0.60 W/m², with a power of 834.45 W emitted by the human body, the maximum distance r, is thus

r = (√P/πI)/2 = (√834.45/0.60π)/2 = 21.04/2 = 10.52 m

So, the maximum distance at which a python could detect your presence is 10.52 m.

3 0

3 years ago

When a guitar string is plucked, in what direction does the wave travel? In what directions does the string vibrate?

Alex

Answer: the waves travel in an horizontal direction while the strings vibrate in a vertical direction.

4 0

2 years ago

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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field

Katyanochek1 [597]

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength at the midpoint between the two rings is given as;

$E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0$