If an object has 180 J of PE and a mass of .5kg, what is its height?

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

3 years ago

A spy satellite uses a telescope with a 1.7-m-diameter mirror. It orbits the earth at a height of 180 km.

WINSTONCH [101]

Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

Explanation:

Given that;

diameter of the mirror d = 1.7 m

height h = 180 km = 180 × 10³ m

wavelength λ = 500 nm = 5 × 10⁻⁹ m

Now Angular separation from the peak of the central maximum is expressed as;

sin∅= 1.22 λ / d

sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7

sin∅ = 3.588 × 10⁻⁷

we know that;

sin∅ = object separation / distance from telescope

object separation = sin∅ × distance from telescope

object separation = 3.588 × 10⁻⁷ × 180 × 10³

object separation =6.45 × 10⁻² m

then we convert to centimeter

object separation = 6.45 cm

Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm

5 0

3 years ago

The gnaphosid spider Drassodes cupreus has evolved a pair of lensless eyes for detecting polarized light. Each eye is sensitive

BartSMP [9]

Answer:

Part A

The intensity is $I = 618 W/m^2$

Part B

The intensity is $I_1= 81.884 W/m^2$

Explanation:

From the question we are told that

The intensity of the light detected by first eye is $I = 700 W/m^2$

Now at initial state according the question the light ray is perpendicular to the eye so it means that it is at 90° the eye

Now the first question is to obtain the intensity the first eye (the first in this case is the one focused on the light )would detect when the head is rotated by 20° its previous orientation

This is mathematically evaluated as

$I = I_i cos^2 ( 20^o)$

$I = 700\ cos^2 (20)$

$I = 618 W/m^2$

Now the second question is to obtain the intensity the first eye (the first eye in this case is the one that is not focused on the light )would detect when the head is rotated by 20° its previous orientation

Now in this case the angle between the eye and the light is 90-20 = 70°

So

$I_1 = 700 \ cos^2 (70)$