All categories

3 years ago

Question 4 (1 point)

Physics

2 answers:

Bezzdna [24]3 years ago

5 0

Answer:

D and B

Explanation:

LekaFEV [45]3 years ago

5 0

Answer:

D and C

Explanation:

You might be interested in

A piece of rope is pulled by two people in a tug-of-war. each exerts a 400-n force. what is the tension in the rope?

Leokris [45]

Newtons 3.law: Action = Reaction

If a body exerts a force on a rope of 400 N the rope exerts a force on the body of 400N also. So the tension in the rope is 400N. See pictures below.

3 0

4 years ago

Read 2 more answers

Mice21 [21]

Answer:

The displacement in t = 0,

y (0) = - 0.18 m

Explanation:

Given f = 40 Hz , A = 0.25m , μ = 0.02 kg / m, T = 20.48 N

v = √ T / μ

v = √20.48 N / 0.02 kg /m = 32 m/s

λ = v / f

λ = 32 m/s / 40 Hz = 0.8

K = 2 π / λ

K = 2π / 0.8 = 7.854

φ = X * 360 / λ

φ = 0.5 * 360 / 0.8 = 225 °

Using the model of y' displacement

y (t) = A* sin ( w * t - φ )

When t = 0

y (0) = 0.25 m *sin ( w*(0) - 225 )

y (0) = 0.25 * -0.707

y (0) = - 0.18 m

5 0

3 years ago

I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite

e-lub [12.9K]

Answer:

The range of powers is $- 5 \ D \le P \le - 2.667\ D$

Explanation:

From the question we are told that

The far point of the left eye is $n_f = 20 cm$

The near point of the left eye is $n = 15cm$

The near point with the glasses on is $n_g =25 \ cm$

From these parameter we can see that with the glass on that for near point the

Object distance would be $u = -25 \ cm$

Image distance would be $v = -15 \ cm$

To obtain the focal length we would apply the lens formula which is mathematically represented as

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

substituting values

$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

$f = - \frac{75}{2} cm$

converting to meters

$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f}$

Substituting values

$P = - \frac{200}{75} m$

$P = - 2.667 \ D$

From these parameter we can see that with the glass on that for far point the

Object distance would be $u_f = - \infty \ cm$

Image distance would be $v_f = -20 \ cm$

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

$\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

$\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

$\frac{1}{f} = \frac{1}{-20} - 0$

$f_f = \frac{20}{1} \ cm$

converting to meters

$f_f = - \frac{20}{1} * \frac{1}{100}$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f_f}$

Substituting values

$P = - \frac{100}{20} m$

$P = - 5 \ D$

This implies that the range of powers of the lens in his glass is

$- 5 \ D \le P \le - 2.667\ D$

3 0

4 years ago

Red shift data shows that galaxies are O expanding O shrinking O moving away O moving closer

Anon25 [30]

Answer:

Should be moving away

Explanation:

Red is a longer wavelength therefore further away. Wavelength is stretched out more and on the red end. I hope this is right. I decided to research and answer since you didn’t have other answers. Are you taking this on edg? I hope I helped!

7 0

3 years ago

How long did the trip from camp wood to the pacific ocean and back again take?

maria [59]

The trip from Camp Wood to the Pacific Ocean and back again took 1.5 years to complete.

The Lewis and Clark Expedition from May 1804 to September 1806, also known as the Corps of Discovery Expedition, was the first American expedition to cross what is now the western portion of the United States.

6 0

4 years ago