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zhuklara [117]
2 years ago
12

What is the magnitude of the displacement of a car with an acceleration of 3.75 m/s2 as it increases its speed from 5.20m/s to 1

3.5m/s?
Physics
1 answer:
NeX [460]2 years ago
7 0

Answer:

20.7 m

Explanation:

using v^2=u^2+2as

s=(v^2-u^2)/2a

v=13.5

u=5.2

a=3.75

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S Four point charges each having charge Q are located at the corners of a square having sides of length a. Find expressions for(
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The total electric potential at the center of the square due to the four charges is V  = √2Q/πÈa.

<h3>What do you mean by electric potential? </h3>

The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.       

V = kq/r

Where V represents electric potential, K is coulomb constant, q  is Charge and r is distance between any  two around charge to the point charge.

Electric potential at O due to four charges is given by,

V = 4KQ/ r

where, r = √2a/2 = a/√2

V = 4k × Q√2/a

V  = √2Q/πÈa

The total electric potential at the center of the square due to the four charges is V  = √2Q/πÈa.

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Find its moment of inertia about an axis perpendicular to its plane and passing through the midpoint of the line connecting its
antoniya [11.8K]

A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem$\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

\mathrm{x}=\frac{\sqrt{2}}{4} L$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$$I_x=I_8+M x^2$$

$$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$, Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

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