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2 years ago

How does a plane mirror work

Physics

1 answer:

Nastasia [14]2 years ago

5 0

Answer:

Plane mirrors work because the light rays create a virtual image behind the mirror. Light rays from the object strike the mirror and reflect according to the law of reflection. ... Therefore, our eye and brain track the light rays backward to a position from which they appear to have come.

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Using a rope that will snap if the tension in it exceeds 379 N, you need to lower a bundle of old roofing material weighing 473

Tema [17]

Answer:

Acceleration = a = -1.94 m/s^2

Explanation:

Weight of the bundle = W = mg

m = W/g = 473/9.8

m = 48.26 kg

There are two forces acting on the bundle such as tension in the upward direction and weight in the downward direction. So, according to newton’s second law of motion:

Sum of forces = ma

T - W = ma

a = (T – W)/m

a = (379 - 473)/48.26

a = -1.94 m/s^2

Negative sign shows that the acceleration is in the downward direction.

8 0

3 years ago

This subject is MUSIC, but there is no MUSIC subject to pick in the "pick a subject " bar, so I just picked a random one...Sorry

max2010maxim [7]

It's the "Fa" syllable ... Choice B .

4 0

2 years ago

a uniform rod of length 1.5m is placed over a wedge at 0.5m from one end .a force of 100 N is applied at its one end near the we

andreev551 [17]

Explanation:

The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end. The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.

Sum the torques about the wedge (it may help to draw a diagram first). Take counterclockwise to be positive.

∑τ = Iα

W (0.25 m) − (100 N) (0.50 m) = 0

W = 200 N

Sum the forces in the y direction.

∑F = ma

F − 100 N − 200 N = 0

F = 300 N

8 0

3 years ago

I NEED HELP SOLVING THIS!!!!!!!!!!!!

Neko [114]

Answer:

Yes

Explanation:

The given parameters are;

The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)

The angle in which the fastball is hit, θ = 22°

The distance of the field = 96 m (315 ft)

The range of the projectile motion of the fastball is given by the following formula

$Range = \dfrac{u^2 \times sin(2\cdot \theta)}{g}$

Where;

g = The acceleration due to gravity = 9.81 m/s², we have;

$Range = \dfrac{49.1^2 \times sin(2\times22^{\circ})}{9.81} \approx 170.71 \ m$

Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.

7 0

2 years ago

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

jeka94

Answer:

a) t_l - t_r = 12.54 us

b) (t_l - t_r) / T = 0.0157

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

d_l = sqrt ( 2.6^2 + 0.15^2)

d_l = sqrt ( 6.7825 )

d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)

t_l - t_r = ( 0.0043 / 343 )

t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

T = 1 / f

T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

(t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

(t_l - t_r) / T = 0.0157

3 0

3 years ago