What theory held that the earth was the center of the universe and that the sun, moon, and planets revolved around it?

The conductor is initially electrically neutral, and then a charge q is placed at the center of the hollow space. Suppose the co

Anvisha [2.4K]

The total charge on the interior of the conductor is zero.

The total charge on the exterior of the conductor is 8q.

<h3>Total charge on the interior</h3>

Due to large number of electrons available for conduction in a conductor, most of the electrons moves to surface leaving zero net charge inside the conductor.

Thus, the total charge on the interior of the conductor is zero.

<h3>Total charge on the exterior</h3>

The total charge on the exterior of the conductor is calculated as follows;

Q = q + 7q = 8q

Thus, the total charge on the exterior of the conductor is 8q.

Learn more about net charge on interior and exterior of conductors here: brainly.com/question/14653264

7 0

2 years ago

A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret

Nataly [62]

Answer:

a)32.34 N/m

b)10cm

c)1.6 Hz

Explanation:

Let 'k' represent spring constant

'm' mass of the object= 330g =>0.33kg

a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.

ΣF=kx-mg=0

k=mg / x

k= (0.33 x 9.8)/ 0.1

k= 32.34 N/m

b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.

Therefore, amplitude of the oscillation is 10cm

c)frequency of the oscillation can be determined by,

f= 1/2π $\sqrt{\frac{k}{m} }$

f= 1/2π $\sqrt{\frac{32.34}{0.33} }$

f= 1.57

f≈ 1.6 Hz

Therefore, the frequency of the oscillation is 1.6 Hz

5 0

3 years ago

How can i stop loveing you if yo keep saying the things i want to hear

xxTIMURxx [149]

Answer:

....

Explanation:

5 0

2 years ago

Read 2 more answers

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

xz_007 [3.2K]

Answer:

a) $\Delta{t} = 5.39s$

b) the motorcycle travels 155 m

Explanation:

Let $t_2-t_1 = \Delta{t}$ , then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

$v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}$

where:

$v_{m2}$ is the speed of the motorcycle at time 2

$v_{c}$ is the velocity of the car (constant)

$v_{0}$ is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

$\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]$

$v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s$

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

$x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933$

3 0

3 years ago

When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting

Mrrafil [7]

Answer:

12N

Explanation:

when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied

3 0

2 years ago