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3 years ago

A 2.5-kg rock is dropped

Physics

2 answers:

Butoxors [25]3 years ago

7 0

Answer:

Okay it dropped now what..........................................

Explanation:

ipn [44]3 years ago

3 0

Is the rock okay I’m concerned

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A ___ is when too much water moves into an area

Marina86 [1]

A. Flood.
Hope it helps!

6 0

4 years ago

Read 2 more answers

collection of such objects we observe that they emit electromagnetic radiation of three different energies: 0.7 eV (infrared), 2

UkoKoshka [18]

It 45 because it has to be this and that and ur gay

4 0

3 years ago

A series of pulses, each of amplitude 0.1 m, is sent down a string that is attached to a post at one end. The pulses are reflect

AleksAgata [21]

Answer:

A_resulting = 0.2 m

Explanation:

Let's analyze the impact of the pulse with the pole, this is a fixed obstacle that does not move therefore by the law of action and reluctant, the force that the pole applies on the rope is of equal magnitude to the force of the rope on the pole (pulse), but opposite directional, so the reflected pulse reverses its direction and sense.

With this information we analyze a point on the string where the incident pulse is and each reflected with an amplitude A = 0.1 m, the resulting is

A_res = 2A

A_resultant = 2 .01

A_resulting = 0.2 m

8 0

3 years ago

A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 103 ft3/min, meaning that about 1.85 ✕ 103 cubic feet

densk [106]

Answer:

0.8726 $m^3/s$

Explanation:

We are to convert 1.85 x $10^3$ $ft^3/min$ to $m^3/s$

First, let us convert the numerator from ft3 to m3

1 ft3 = 0.0283 m3

Hence,

1.85 x $10^3$ ft3 = 1.85 x $10^3$ x 0.0283 m3

= 52.355 m3

Now, let us convert the denominator from minutes to seconds

1 min = 60 sec

Therefore;

1.85 x $10^3$ $ft^3/min$ = 52.355/60 $m^3/s$

= 0.8726 $m^3/s$

7 0

3 years ago

Light rays in a material with index of refrection 1.35 1.35 can undergo total internal reflection when they strike the interface

nekit [7.7K]

Answer:

1.30

Explanation:

To calculate the critical angle we have ti use the formula:

$sin\theta_c=\frac{n_2}{n_1}$

where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.

By taking n_2 and replacing we obtain:

$n_2=n_1sin\theta_c=(1.35)sin75.1\°=1.30$

hope this helps!!

6 0

3 years ago