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zubka84 [21]
3 years ago
5

What color would the sky be if the atmosphere was 100% large molecules and particles like dust and water?

Physics
1 answer:
DedPeter [7]3 years ago
3 0

Answer:

Gray-ish looking like as if the atmosphere were full of only pollution.

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A guitar string is 90 cm long and has a mass of 3.7 g . The distance from the bridge to the support post is L=62cm, and the stri
zvonat [6]

To solve this problem we will apply the concept of frequency in a string from the nodes, the tension, the linear density and the length of the string, that is,

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

Here

n = Number of node

T = Tension

\mu = Linear density

L = Length

Replacing the values in the frequency and value of n is one for fundamental overtone

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

f = \frac{1}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})

\mathbf{f = 281.2Hz}

Similarly plug in 2 for n for first overtone and determine the value of frequency

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

f = \frac{2}{2(62*10^{-2})}(\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}})

\mathbf{f = 562.4Hz}

Similarly plug in 3 for n for first overtone and determine the value of frequency

f = \frac{n}{2L}(\sqrt{\frac{T}{\mu}})

f = \frac{3}{2(62*10^{-2})}\bigg (\sqrt{\frac{500}{(\frac{3.7*10^{-3}}{90*10^{-2}})}} \bigg)

\mathbf{f= 843.7Hz}

4 0
3 years ago
Mbbnmzmzkdjhxhxndmfmfmckcjcjcncncngmgkgjcjcj
Amiraneli [1.4K]
Gmsisient djaektnfidiaus wjrnrksiauenrnsiaur msiwlwkrjsuakeurjrndkaiakejrjd laismrnff
8 0
3 years ago
Read 2 more answers
Please help it’s easy will give brainlist !! Thank youu worth a lot be serious pls
Genrish500 [490]

Answer:

Part 1

Stationary

Part 2

20 Newtons

Part 3

Force

Part 4

4.0 m/s²

Part 5

Normal

Part 6

Cart m

Part 7

The gravitational force is less than magnetic force

Explanation:

Part 1

The position time graph of the object is an horizontal straight line passing across the top of the position 3 boxes vertically up from the origin

As the time increases by the units of number boxes to the left, the position of the object does not change and remains at the 3 boxes up above the origin, therefore, the object is stationary

Part 2

By Newton's third law of motion, the action action obtained from a force is equal to the reaction given to the force, therefore, we have;

The force exerted by the student on the scale = The force exerted by the scale on the student = 20 N

Part 3

A force is a the directional push on an object or pull from the object as a form of interaction with another object which tends to alter or maintain the motion of the object

Part 4

The given parameters are'

The mass of block A = 1.0 kg

The mass of block B = 2.0 kg

Both blocks, "A" and "B" are initially at rest

The applied horizontal force, F = 12-N

The nature of the surface over which the blocks move = Smooth surface

Force, F = Mass, m × Acceleration, a

F = m × a

The blocks two blocks experience a common acceleration, a

The combined mass of the two blocks, m = 1.0 kg + 2.0 kg = 3.0 kg

m = 3.0 kg

Therefore, a = F/m = 12-N/(3.0 kg) = 4 m/s²

Part 5

A normal force is a force acting perpendicularly to a surface that supports the weight of an object

Part 6

The given parameters are;

M = 2 kg

The mass of the left cart attached to the spring = M = 2 kg

The mass of the right cart attached to the spring = 2·M = 2 × 2 kg = 4 kg

Therefore, given that the force exerted by one cart on the other after the spring is removed, we have;

Force, F = M × a₁ = 2·M × a₂

Where;

a₁ = The average acceleration of the cart with mass, M

a₂ = The average acceleration of the cart with mass, 2·M

M × a₁ = 2·M × a₂

∴ a₁ = 2·a₂

The acceleration, a₁, of car M = 2 × The acceleration, a₂, of car 2·M

The acceleration of cart M is two times the acceleration of cart 2·M

Cart M will experience a greater average acceleration

Part 7

For the top magnet of two magnets placed two magnets arranged so that one is on top the other, to be held in the air with a space between the two magnets, the gravitational force pulling the top magnet down, is less than the repelling magnetic force between the two magnets.

7 0
3 years ago
You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi
Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0
3 years ago
What part of the scientific method tests the hypothesis?
blagie [28]

Answer:

A hypothesis is a conjecture, based on knowledge obtained while seeking answers to the question. The hypothesis might be very specific, or it might be broad. Scientists then test hypotheses by conducting experiments or studies.

Explanation:

mark me as the brainliest

have a great day

6 0
3 years ago
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