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2 years ago

How can I solve the following statement?

Physics

1 answer:

Firlakuza [10]2 years ago

7 0

Answer:

The net electric field at the midpoint is 6.85 x 10^7 N/C.

Explanation:

q = − 8.3 μC

q' = + 7.8 μC

d = 9.2 cm

d/2 = 4.6 cm

The electric field due to the charge q at midpoint is

$E = \frac{k q}{r^2}\\\\E = \frac{9\times 10^9\times 8.3\times 10^{-6}}{0.046^2}\\\\E = 3.53\times 10^7 N/C$ leftwards

The electric field due to the charge q' at midpoint is

$E' = \frac{k q}{r^2}\\\\E' = \frac{9\times 10^9\times 7.8\times 10^{-6}}{0.046^2}\\\\E' = 3.32\times 10^7 N/C$

The resultant electric field at mid point is

E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C

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1. Consider three objects: Object A is a hoop of mass m and radius r; Object B is a sphere

Zarrin [17]

a) Object C (the disk) has the greatest rotational inertia ( $\frac{3}{2}mr^2$ )

b) Object B (the sphere) has the smallest rotational inertia ( $\frac{4}{5}mr^2$ )

Explanation:

The moments of inertia of the three objects are the following:

1) For a hoop of negligible thickness, it is

$I=MR^2$

where M is its mass and R its radius. For the hoop in this problem,

M = m

R = r

Therefore, its moment of inertia is

$I=(m)(r)^2=mr^2$

2) For a solid sphere, the moment of inertia is

$I=\frac{2}{5}MR^2$

where M is its mass and R its radius. For the sphere in this problem,

M = 2m

R = r

Therefore, its moment of inertia is

$I=\frac{2}{5}(2m)(r)^2=\frac{4}{5}mr^2$

3) For a disk of negligible thickness, the moment of inertia is

$I=\frac{1}{2}MR^2$

where M is its mass and R its radius. For the disk in this problem,

M = 3m

R = r

Therefore, its moment of inertia is

$I=\frac{1}{2}(3m)(r)^2=\frac{3}{2}mr^2$

So now we can answer the two questions:

a) Object C (the disk) has the greatest rotational inertia ( $\frac{3}{2}mr^2$ )

b) Object B (the sphere) has the smallest rotational inertia ( $\frac{4}{5}mr^2$ )

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

7 0

3 years ago

Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude

aivan3 [116]

Answer:

$E=6.8Kv/m$

Explanation:

From the question we are told that

Distance b/w plate $d=10cm=>0.1m$

P_1 Potential at 7.35 $V=533v$

Generally the equation for electric field at a distance is mathematically given as

$E=\frac{v}{d}$

$E=\frac{533}{7.85*10^-^2}$

$E=6789.808917$

$E=6.8*10^3$

$E=6.8Kv/m$

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3 years ago

What substance is used by plants as a source of food

harkovskaia [24]

Answer:

dart

Explanation:

dart and sun and water so that the plant be okay

7 0

2 years ago

What does itmean to have an acceleration of 8 m/s2

Masteriza [31]

Answer:

The answer is option A.

You speed up 8 m/s every second

Hope this helps you

8 0

2 years ago

Read 2 more answers

A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,

Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

$V_{r}$ = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0

2 years ago