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Leona [35]
3 years ago
14

Which refers to the area on the thermometer marked with the letter A? scale bulb mercury strip number line

Physics
2 answers:
expeople1 [14]3 years ago
8 0

Answer:

number line

Explanation: hope this helps lol

alexandr1967 [171]3 years ago
6 0

Answer:

its A

Explanation:

Scale,

i got it right on edge

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The brakes on your automobile are capable of creating a deceleration of 5.0 m/s2. If you are going 135 km/h and suddenly see a s
poizon [28]

Answer:

3.33 seconds

Explanation:

We can use the velocity formula [ v = u + at ] to solve.

Find the value "u".

135km/h -> 135km*1000m/3600s -> 37.5m/s

Find the value "v".

75km/h -> 75km*1000m/3600s -> 20.83m/s

Keep in mind we are dealing with "deceleration" so when we input 5.0m/s into the formula, it will be a negative value.

Now, find "t" which is the value we aren't given with the values we're given in the question.

20.83 = 37.5 - 5t

-16.67 = -5t

3.33 = t

Best of luck!

8 0
3 years ago
Why does a purple flower appear purple when white light shines on it?
grandymaker [24]
The flower absorbs all light but purple, making it appear, purple!
5 0
3 years ago
A dragster runs the quarter mile in 8.96 s. What is the car's velocity (in ft/s) at the finish line?
lara [203]
The first thing you should know to answer this question is the following conversion:
 1mi = 5280feet
 We have then that the speed is:
 v = ((1/4) * (5280)) / (8.96)
 v = 147.32 feet / s
 Answer:
 the car's velocity (in ft / s) at the finish line is 147.32 feet / s
7 0
3 years ago
a. Define Valency. b. Write the symbol and valency of the following : (i) Oxygen (ii) potassium 7. a. Magnesium burns in Oxygen
sweet-ann [11.9K]

Answer:

Oxygen-Symbol-O Potassium Symbol-K

6 0
3 years ago
Read 2 more answers
You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle
nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If  θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for  90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function  

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A  

4 0
4 years ago
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