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3 years ago

How to calculate the slope of position time graph ?

Physics

1 answer:

meriva3 years ago

8 0

Answer:

Slope= $\frac{x2-x1}{t2-t1}$

Explanation:

We are supposed to find the slope in a position time graph,

For that we need atleast two points $(t1,x1),(t2,x2)$ .( $(x,y )$ represents there coordinates in respective axis directions.)

In the given case the y axis would be the position and the x axis as time .

Slope is defined as the ratio of change in y coordinate to change in x coordinate :- $\frac{y2-y1}{x2-x1}$

So subsituting the coordinates of position and time we get slope as:-

$\frac{x2-x1}{t2-t1}$

[When we have more than two points which are not collinear then to calculate the slope we take the average.]

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A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and

sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

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2 years ago

If the kinetic energy of an electron is 4.1e-18 j, what is the speed of the electron? (you can use the approximate (nonrelativis

arlik [135]

The kinetic energy of the electron is
$K= \frac{1}{2}mv^2$
where $m=9.1 \cdot 10^{-31} kg$ is the mass of the electron and v its speed. Since we know the value of the kinetic energy, $K=4.1 \cdot 10^{-18} J$ , we can find the value of the speed v:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2\cdot 4.1 \cdot 10^{-18}J}{9.1 \cdot 10^{-31}kg} } = 3\cdot 10^6 m/s$

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3 years ago

3. If I run 150m in 30 seconds, what speed will I have been running at?

Radda [10]

Answer:

speed = distance/time

Explanation:

speed = 150/30

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you were running fast .....5m/s is a good speed

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2 years ago