Compare the mass of block A, the mass of block B is:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an

Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

$F=\dfrac{kQ^2}{r^2}$ ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

$F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}$ .....(2)

Dividing equation (1) and (2), we get :

$\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}$

Hence, the correct option is (d) i.e. " 4F/9"

7 0

3 years ago

The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum

NemiM [27]

Answer:

Explanation:

Given

Free fall acceleration on mars $g_{m}=3.7\ m/s^2$

Time Period of pendulum on earth $T=1\ s$

Time period of Pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

for earth

$1=2\pi\cdot \sqrt{\frac{L}{9.8}}$

$L=\frac{9.8}{(2\pi )^2}$

$L=0.498\ m$

(b)For same time period on mars length is given by

$L'=\frac{g_m}{(2\pi )^2}$

$L'=\frac{3.7}{39.48}$

$L'=0.0936\ m$

$L'=9.36\ cm$

3 0

3 years ago

needhelpp101 Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential ene

Vlad1618 [11]

If you remember the formula for potential energy,
then this question is a piece-o-cake.

<em>Potential energy = (mass) x (<u>acceleration of gravity</u>) x (height) .</em>

-- The object's mass is the same everywhere.
-- You said that the height is the same both times.
-- How about the acceleration of gravity ?

Compared to gravity on Earth, it's only 16.5 percent as much on the Moon.
So naturally, from the formula, you'd expect the Potential Energy to be less
on the Moon.

4 0

3 years ago

A transverse wave on a string is described by the wave function

svp [43]

The period of the transverse wave from what we have here is 0.5

<h3>How to find the period of the transverse wave</h3>

The period of a wave can be defined as the time that it would take for the wave to complete one complete vibrational cycle.

The formula with which to get the period is

w = 4π

where w = 4 x 22/7

2π/T = 4π

6.2857/T = 12.57

From here we would have to cross multiply

6.2857 = 12.57T

divide through by 12.57

6.2857/12.57 = T

0.500 = T

Hence we can conclude that the value of T that can determine the period based on the question is 0.500.