A housefly walking across a surface may develop a significant electric charge through a process similar to frictional charging.

When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the

Maru [3345]

Answer:

a

The value at a point inside is Zero

b

The electric field is $E = 2.7*10^{6} \ N/C$

Explanation:

We know from the problem that

The charge magnitude is $q = 3.0 \mu C = 3.0 *10^{-6} \ C$

The radius of the spherical ball is $r = 5.0 \ mm = 0.005 \ m$

According to Gauss’s law, the enclosed charge within a conductor is zero which indicates that the electric field within the spherical ball is zero

On the outside, the electric field around the spherical ball is mathematically expressed as

$E = \frac{kq}{ a^2}$

Here a denotes a point outside the spherical ball with its value of $a = 10 \ cm = \frac{10}{100} = 0.1 \ m$

and k represents Coulomb's constant, valued at

$k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

=> $E = \frac{ 3 *10^{-6} * 9*10^9 }{ (0.1)^2}$

=> $E = 2.7*10^{6} \ N/C$

5 0

2 months ago

In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t

ValentinkaMS [3465]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

$T\cos\theta=mg$

$T=\dfrac{mg}{\cos\theta}$

Vertical component

$T\sin\theta=m\omega^2 r$

Substituting the tension value

$mg\tan\theta=m\omega^2(d+L\sin\theta)$

$\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}$

Substituting the value into the equation

$\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}$

$\omega=1.34\ rad/s$

Thus, the angular speed of rotation computes to 1.34 rad/s.

7 0

4 months ago

A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha

serg [3582]

Answer:

$a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

Explanation:

Data provided

initial velocity v₀=20 cm/s at time t=3s

final velocity vf=0 at time t=8 s

Required

Average Acceleration for the interval from 3s to 8s

Solution

Acceleration can be defined as the first derivative of velocity concerning time

$a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\ a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\ a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

8 0

3 months ago

Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme

Ostrovityanka [3204]

Upon comparison, it is evident that the student with the highest percent error is A. Student 4, who measured 9.61 m/s². Four students recorded the acceleration of gravity, with the accepted local value being 9.78 m/s². Now let's find out which student's measurement exhibited the greatest percent error.

4 0

3 months ago

Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of t

Softa [3030]

Answer:

1/7 kg

Explanation:

Refer to the attached diagram for enhanced clarity regarding the question.

One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.

g denotes the acceleration due to gravity.

Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.

Given M = 1.0 kg and a = 3/4g.

By applying Newton's second law; $\sum fy = ma_y$

For the body with mass m;

T - mg = ma... (1)

For the body with mass M;

Mg - T = Ma... (2)

Combining equations 1 and 2 gives;

+Mg -mg = ma + Ma

Ma-Mg = -mg-ma

M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into this equation leads to;

3/4 g-g = -m(3/4 g+g)

3/4 g-g = -m(7/4 g)

-g/4 = -m(7/4 g)

1/4 = 7m/4

Multiplying gives: 28m = 4

m = 1/7 kg

Hence, the mass of the other box is 1/7 kg

3 0

3 months ago