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2 years ago

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Two point charges are brought closer together, increasing the force between them by a factor of 20. By what factor was their sep

aration changed?

1 answer:

olga55 [171]2 years ago

8 0

The separation between them is $\frac{r}{\sqrt{20} }$

Concept :

If the force increases, distance between charges must decrease. Force is indirectly proportional to the distance squared.

Given,

Two point charges are brought closer together, increasing the force between them by a factor of 20.

Original force is

F = $\frac{kq_{1} q_{2} }{r^{2} }$ -------- ( 1 )

Here, $q_{1} , q_{2}$ are charges and r is the distance between them

New force F' = $\frac{kq_{1q_{2} } }{r'^{2} }$ ----------- (2 )

Divide ( 1 ) and ( 2 )

$\frac{F'}{F}$ = $\frac{\frac{kq_{1}q_{2} }{r'^{2} } }{\frac{kq_{1}_{2} }{r^{2} } }$

20 = $\frac{r^{2} }{r'^{2} }$

r' = $\frac{r}{\sqrt{20} }$

Given that force between them are increasing and therefore distance between them decrease by $\frac{r}{\sqrt{20} }$

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You observe a plane approaching overhead and assume that its speed is 600 miles per hour. The angle of elevation of the plane is

scZoUnD [109]

Answer:4.34 miles

Explanation:

first Elevation = $19^{\circ}$

After 1 minute Elevation changes to $59^{\circ}$

Ditsance travelled in 1 minute = $600\times \frac{1}{60}$ =10 mile

Now

tan59= $\frac{H}{x}$

H=xtan59

tan19= $\frac{H}{10+x}$

H= $\left ( 10+x\right )tan19$

Equating H

we get

1.319x=10tan19

x=2.61 miles

H= $2.61\times tan59$ =4.34 miles

4 0

3 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

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3 years ago

Is it accurate to describe the physical universe as composed of only matter and energy?

Dennis_Churaev [7]

Yes, it is <span>accurate to describe the physical universe as composed of only matter and energy. Some people might argue about the dark matter, but it is not yet defined properly. Different universes can be made up of different compositions but it is a fact that our universe is made of matter and energy. </span>

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3 years ago

Read 2 more answers

Balboa Park in San Diego has an outdoor organ. When the air temperature increases, the fundamental frequency of one of the organ

musickatia [10]

The fundamental frequency of one of the organ pipes will go up or increase.

When pressured air is forced into an organ pipe, it echoes at a particular pitch, generating the sound of the pipe organ. Each pipe has been adjusted to a particular pitch on the musical scale.

A musical instrument called an outdoor pipe organ is used to perform music. It produces some calming tones and has a really serene sound. The organ pipe produces the sound of the outdoor organ. The wavelength of the sound is also dependent on the length of the pipe. The fundamental frequency of one of the organ pipes will grow as the speed of the sound increases as the ambient air temperature rises.

The correct option is (c).

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1 year ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

irinina [24]

288.51 N is the magnitude of the force that the beam exerts on the hi.nge.

Given

Mass 0f beam = 40 Kg

The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N

Angle between the beam and cable is = 90°

Angle between beam and the horizontal component = 31°

As the system of the beam, hi_nge and cable are in equilibrium.

The magnitude of the force that the beam exerts on the hi_nge can be calculated by -

F =The horizontal component of force + the vertical component of force

F = 86.62 N + 40 × 9.8 × sin 31°

F =86.62 N + 201.89 N

F = 288.51 N

Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51 N.

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2 years ago