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3 years ago

When can an electric field be measured at any point from the force on a stationary test charge at that point?

Physics

1 answer:

bogdanovich [222]3 years ago

8 0

Answer:

Electric field be measured at any point from the force on a stationary test charge at that point, when the force on the test charge is generated only by the coulomb field (E) of static charges.

Explanation:

Total force on a test charge is given as;

F = qE + q(V x B)

F = q(E + V x B)

Where;

F is the total force on the test charge

q is the test charge

E is the electric field

B is the magnetic field

V is the velocity of the test charge

F = q(E + V x B)

for a stationary test charge, V = 0

The equation above reduces to;

F = qE

Thus, electric field be measured at any point from the force on a stationary test charge at that point, when the force on the test charge is generated only by the coulomb field (E) of static charges.

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A ball is bouncing to a height that is 1/3 of the previous height after every bounce. What kind of sequence is this? How could y

tamaranim1 [39]

Answer and Explanation:

The ball is bouncing to a height of 1/3 of its previous height this is a type of geometric sequence the total distance can be found by the sum of geometric sequence

For example let the initial height is 243 fit

After one bounce it will reach 243/3 =81 feet

After second bounce 81/3=27 feet

After third bounce 27/3 =9 feet

After fourth bounce 9/3 =3 feet

So a sequence is formed that is 243,81,27,9,3..........

Here $r=\frac{3}{9}=\frac{1}{3}$

Sum of infinite GP = $\frac{a}{1-r}$

From this formula we can find the total distance traveled by the ball

3 0

3 years ago

Recall from Chapter 1 that a watt is a unit of en- ergy per unit time, and one watt (W) is equal to one joule per second ( J·s–1

harkovskaia [24]

Answer:

Explanation:

The energy of a photon is given by the equation $E_p=h f$ , where h is the Planck constant and f the frequency of the photon. Thus, N photons of frequency f will give an energy of $E_N=N h f$ .

We also know that frequency and wavelength are related by $f=\frac{c}{\lambda}$ , so we have $E_N=\frac{N h c}{\lambda}$ , where c is the speed of light.

We will want the number of photons, so we can write

$N=\frac{\lambda E_N}{h c}$

We need to know then how much energy do we have to calculate N. The equation of power is $P=E/t$ , so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be $E_N$ .

Putting this paragraph in equations:

$E_N=(\frac{4}{100})E=0.04Pt=(0.04)(100W)(1s)=4J$ .

And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

$N=\frac{\lambda E_N}{h c}=\frac{(520 \times10^{-9}m) (4J)}{(6.626\times10^{-34}Js) (299792458m/s)}=1.047 \times10^{19}$

3 0

3 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

2 years ago

a spherical mirror is cut in half horizontally will an image be formed by the bottom half of the mirror how

monitta

Answer:

Explanation:

the spherical mirror can form an image even if it is cut in half horizontally , but the image formed may be blurred.

pls mark as brainliest if you find it helpful

7 0

3 years ago

In still water, a boat averages 18 18 miles per hour. it takes the same amount of time to travel 16 miles 16 miles downstream,

Vladimir79 [104]

The current is 6 miles per hour.
Let's create a few equations:
Traveling with the current:
(18 + c)*t = 16

Traveling against the current:
(18 - c)*t = 8

Let's multiply the 2nd equation by 2
(18 - c)*t*2 = 16

Now subtract the 1st equation from the equation we just doubled.
(18 - c)*t*2 = 16
(18 + c)*t = 16

(18 - c)*t*2 - (18 + c)*t = 0
Divide both sides by t
(18 - c)*2 - (18 + c) = 0

Now solve for c
(18 - c)*2 - (18 + c) = 0
36 - 2c - 18 - c = 0
36 - 2c - 18 - c = 0
18 - 3c = 0
18 = 3c
6 = c

So the current is 6 mph.
Let's verify that.
(18 + 6)*t = 16
24*t = 16
t = 16/24 = 2/3

(18 - 6)*t = 8
12*t = 8
t = 8/12 = 2/3

And it's verified.

4 0

3 years ago