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cupoosta [38]
3 years ago
6

A nonconducting solid sphere of radius 11.6 cm has a uniform volume charge density. The magnitude of the electric field at 23.2

cm from the sphere's center is 1.89 103 N/C.
(a) What is the sphere's volume charge density? µC/m3

(b) Find the magnitude of the electric field at a distance of 5.00 cm from the sphere's center. N/C
Physics
1 answer:
3241004551 [841]3 years ago
5 0

Answer: a) 1.76 μC/m^3 ; b) 40.73 * 10^3 N/C

Explanation: In order to solve this problem we have to use the  Gaussian law, in this sense we have:

∫E.dS= Q inside/εo then we have:

E* 4*π*r= ρ* Volume/εo

then

ρ= E4*π* εo*r^2/Volume)= E*r^2/(k*(4/3)*π*R^3)  where R is the radius of the sphere

ρ= 0.232^2*1,89*10^3/(6.54* 10^-3*9*10^9)=1.76 μC/m^3

The electric field is given by:

E= k*Vol*ρ/r^2  for r= 5cm

E= 9*10^9*6.54~ 10^-3* 1.73*10^-6/(0.0025)=40.73 * 10^3 N/C

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Archy [21]

Answer:

The time taken by the rock to reach the ground is 0.569 seconds.

Explanation:

Given that,

A student throws a rock horizontally off a 5.0 m tall building, s = 5 m

The initial speed of the rock, u = 6 m/s

We need to find the time taken by the rock to reach the ground. Using second equation of motion to find it. We get :

s=ut+\dfrac{1}{2}gt^2\\\\5=6t+\dfrac{1}{2}\times 9.8t^2\\\\t=0.569\ seconds

So, the time taken by the rock to reach the ground is 0.569 seconds. Hence, this is the required solution.

5 0
4 years ago
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Are the units of the formula ma = mv2/2 dimensionally consistent? Select the single best answer.
Vesnalui [34]

To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency

ma = \text{Mass}\times \text{Acceleration}

ma = kg \cdot \frac{m}{s^2}

At the same time we have that

\frac{1}{2}mv^2 = \text{Mass}\times \text{Velocity}^2

\frac{1}{2}mv^2 = kg ( \frac{m}{s})^2

\frac{1}{2}mv^2 = kg \cdot \frac{m^2}{s^2}

Therefore there is not have same units and both are not consistent and the correct answer is B.

5 0
4 years ago
An object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height abo
ki77a [65]

Answer:

11.09 m/s

Explanation:

Given that an object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point.

The parameters given are:

Initial velocity U = ?

Final velocity V = 9.6 m/s

Acceleration due to gravity g = 9.8m/s^2

Let first assume that the object is thrown from rest with the velocity U, at maximum height V = 0

Using third equation of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8H

U^2 = 19.6H ........ (1)

Using the formula again for one fourth of its maximum height

9.6^2 = U^2 - 2 × 9.8 × H/4

92.16 = U^2 - 19.6/4H

92.16 = U^2 - 4.9H

U^2 = 92.16 + 4.9H ...... (2)

Substitute U^2 in equation (1) into equation (2)

19.6H = 92.16 + 4.9H

Collect the like terms

19.6H - 4.9H = 92.16

14.7H = 92.16

H = 92.16/14.7

H = 6.269

Substitute H into equation 2

U^2 = 92.16 + 4.9( 6.269)

U^2 = 92.16 + 30.72

U^2 = 122.88

U = 11.09 m/s

Therefore, the initial velocity of the object is 11.09 m/s

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4 years ago
Give an example of intense aerobics activity. Prompt must be accurate. ​
Lubov Fominskaja [6]

Answer:

Explanation:

An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.

7 0
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dem82 [27]

Answer:Surface wave

Explanation:

Longitudinal wave is perpendicular to the wave motion and transverse wave is the same as the wave motion

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