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jonny [76]
3 years ago
10

When you hold a small amount of sodium chloride in a flame, you observe that the flame is bright yellow. What can you do to be s

ure that the color is due to sodium and not chlorine?
Physics
1 answer:
Anon25 [30]3 years ago
6 0

To be sure that the color is due to sodium, do a flame test. This is used to identify the presence of small number of metal ions in a compound. For sodium, the brilliant color of bright yellow is a result of an excited electron in a very high level of energy and fall to a lower level called the ground state. 

In addition, metals have different jump sizes in the energy level and this means that each different metal will have a different pattern of spectral lines, which results to different flame color.

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A stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. What is the spe
KIM [24]

Answer: V = 15 m/s

Explanation:

As  stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,

F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz

Using doppler effect formula

F = C/ ( C - V) × f

Where

F = observed frequency

f = source frequency

C = speed of light = 3×10^8

V = speed of the car

Substitute all the parameters into the formula

2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10

2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)

1.000000049 = 3×10^8/(3×10^8 - V)

Cross multiply

300000014.7 - 1.000000049V = 3×10^8

Collect the like terms

1.000000049V = 14.71429

Make V the subject of formula

V = 14.71429/1.000000049

V = 14.7 m/s

The speed of the car is 15 m/s approximately.

8 0
3 years ago
A 5000 g toy car starts from rest and moves a distance of 300 cm in 3 s under the action of a single constant force. Determine t
sveticcg [70]

Answer:

3.33 N

Explanation:

First, find the acceleration.

Given:

Δx = 3 m

v₀ = 0 m/s

t = 3 s

Find: a

Δx = v₀ t + ½ at²

3 m = (0 m/s) (3 s) + ½ a (3 s)²

a = ⅔ m/s²

Use Newton's second law to find the force.

F = ma

F = (5 kg) (⅔ m/s²)

F ≈ 3.33 N

4 0
3 years ago
A spatially challenged goldfish swims along the x-axis only. Its initial position is 7.8 m. After swimming back and forth a whil
Verdich [7]

Answer:

<em>The fish displacement was -3.4 m</em>

Explanation:

<u>Distance and Displacement</u>

A moving object constantly travels some distance at defined periods of time. The total moved distance is the sum of each individual distance the object traveled. It can be written as:

dtotal=d1+d2+d3+...+dn

This sum is obtained independently of the direction the object moves.

The displacement only takes into consideration the initial and final points of the path defined by the object in its moving. The displacement, unlike distance, is a vectorial magnitude and can be even zero if the object starts and ends the movement at the same point.

The displacement, when the object moves in one axis only is given by:

d = final position - initial position

We know the fist started at 7.8 m from a given reference along the x-axis.

After some undisclosed movements, it ends up at the position 4.4 m. Thus, the displacement is:

d = 4.4 m - 7.8 m = -3.4 m

This means the fish ended up to the left of the position it started from.

The fish displacement was -3.4 m

3 0
3 years ago
A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 152 om and makes an angl
aliya0001 [1]

Answer:

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8° , countercockwise from the positive x-axis: Direction of the second displacement

Explanation:

We find the x-y components for the given vectors:

i:  unit vector in x direction

j:unit vector in y direction

D₁: Displacement Vector 1

D₂: Displacement Vector 2

R= resulta displacement vector

D₁= 152*cos110°(i)+152*sin110°(j)=-51.99i+142.83j

D₂= -D₂(i)-D₂(j)

R=  131*cos38°(i)+ 131*sin38°(j) = 103.23i+80.65j

We propose the vector equation for sum of vectors:

D₁+ D₂= R

-51.99i+142.83j+D₂x(i)-D₂y(j) = 103.23i+80.65j

-51.99i+D₂x(i)=103.23i

D₂x=103.23+51.99=155.22 cm

+142.83j-D₂y(j) =+80.65j

D₂y=142.83-80.65=62.18 cm

Magnitude and direction of the second displacement

D_{2} =\sqrt{(D_{x})^{2} +(D_{y} )^{2}  }

D_{2} =\sqrt{(155.22)^{2} +(62.18 )^{2}  }

D₂= 167.21 cm

Direction of the second displacement

\beta = tan^{-1} \frac{D_{y}}{D_{x} }

\beta = tan^{-1} \frac{62.18}{155.22 }

β= 21.8°

D₂= 167,21 cm : Magnitude  of the second displacement

β= 21.8.° , countercockwise from the positive x-axis: Direction of the second displacement

6 0
3 years ago
Someone help this is a test and i need to finish this quick i will give brainliest of it lets me
tekilochka [14]

Sorry, I don't know but I think the correct answer is the first option.

7 0
2 years ago
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