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DanielleElmas [232]
3 years ago
15

Physics:

Physics
2 answers:
Dafna1 [17]3 years ago
7 0
∆y is change in y (velocity) and ∆x is change in x (time)

So from the graph we see that ∆y is zero and ∆x is also zero (value is changing with same ratio)

∆y/∆x= slope = 0

Gwar [14]3 years ago
5 0

-- Pick any 2 points on the graph ... any two places that are convenient, and you'll be able to read their (x, y) coordinates easily.

--  See how much 'y' changes from the 1st point to the 2nd point.  Call it ' Δy '.  That little triangle is the Greek letter 'delta', and it's often used to mean "the change" in something.  When you see ' Δy ', you say "delta wye".  

--  See how much 'x' changes from the first point to the second point.           Call that  ' Δx '.  ("delta eks")

--  It's certainly possible that  Δx  or  Δy  might be zero.  That's OK.  Don't worry about it.  If there's a line on the graph and you picked two points on the line, then they can't BOTH be zero.

--  The slope of the line is ( Δy ) divided by ( Δx ).

Here are some things that you're going to find:

--  If the line is slanted and it climbs up from left to right, then  Δx  and  Δy  both have the same sign, and the slope is positive.  (If you walk the line any distance from left to right, you also have to climb some.)

-- If the line is slanted and it drops down from left to right, then  Δx  and  Δy  have opposite signs, and the slope is negative.  (If you walk the line any distance from left to right, you'll drop some.)

-- If the line is horizontal, then  Δx  is not zero,  Δy  is zero, and the slope is zero.  (If you walk the line from left to right, then no matter how far you go, you never climb or drop at all.)

-- If the line is vertical, then  Δx  is zero and  Δy  is not zero.  The slope is some number divided by zero, which we're not even allowed to do.  The slope of a vertical line is called "undefined".  It's not a number.  It's not zero.  It doesn't even have a name. It's undefined.  (There's no way to walk the line from left to right !)

On the graph in the middle of your picture, the line is horizontal.  Pick any two points on the line.  The line never gets any higher or lower, so no matter which two points you pick,  Δy  (the change in 'y' between them) is zero.  But the change in 'x' between them ...  Δx  ... is some number.

Slope of the line = (Δy) divided by (Δx)  =  (zero) divided by (something).

A fraction with zero on top is ..... zero !  The slope of that line is zero.  No matter how far you walk along it, you never climb or drop.  There's NO slope.  It's flat.  It DOESN't slope.  I don't know any more ways to say it.

It is NOT (zero/zero).  It's (zero/something) , and that's zero.

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65 miles per hour.

Explanation:

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3 years ago
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At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r
Oksana_A [137]

Answer:

<u>For proton:</u>

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

<u>For electron:</u>

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s    

Explanation:

<u>For a proton we have</u>:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.                

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

W = \Delta Vq    (1)                                        

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C    

V: is the potential    

Also, the work is equal to:

W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}     (2)      

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

v_{a}: is the velocity in the point a

v_{b}: is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

\Delta Vq = \frac{1}{2}mv_{a}^{2}

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}

v_{a} = 2.9 \cdot 10^{4} m/s

<u>For an electron we have</u>:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}

Since v_{a} = 0 (starts from rest) and m_{e} = 9.1x10⁻³¹ kg (electron's mass), we have:

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}    

v_{b} = 1.2 \cdot 10^{6} m/s

I hope it helps you!

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Oduvanchick [21]

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A rectangular field is of length 42 cm and breadth 25 m. Find the area of the field in SI unit. EXPLAIN STEP BY STEP​
lorasvet [3.4K]

Answer:

the area of the rectangular field is 10.5 m²

Explanation:

Given;

length of the rectangular field, L = 42 cm = 0.42 m

breadth of the rectangular field, b = 25 m

The area of the rectangular field is calculated as follows;

Area = Length x breadth

Area = 0.42 m x 25 m

Area = 10.5 m²

Therefore, the area of the rectangular field is 10.5 m²

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