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Dafna1 [17]
4 years ago
15

An electric field of magnitude e is measured at a distance r from a point charge q. if the charge is doubled to 2q, and the elec

tric field is now measured at a distance of 2r from the charge, what is the value of the new electric field?
Physics
1 answer:
pav-90 [236]4 years ago
5 0
Electric field is proportional to the charge and inversely proportional to the square of distance:
E\propto\frac{Q}{r^{2}}
With charge 2q and distance 2r, the electric field is proportional to:
E_{q}\propto\frac{2q}{(2r)^{2}}=\frac{2q}{4r^{2}}=\frac{1}{2}\frac{q}{r^{2}}
The new electric field is half of that of the original measured field.
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A toy gun uses a spring to project a 4.5-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu
marishachu [46]

Answer:

1.93 m/s

Explanation:

Parameters given:

Mass = 4.5g = 0.0045kg

Spring constant = 8.0 N/m

Length of barrel = 13 cm = 0.013m

Frictional force = 0.035N

Compression = 5.8 cm = 0.058m

First, we find the P. E. stored in the spring:

P. E. = ½*k*x²

P. E. = ½ * 8 * 0.058² = 0.013J

Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:

Work = Force * distance

The distance here is the length of the barrel.

Work = 0.035 * 0.13 = 0.0046 J

The kinetic energy of the sphere can now be found:

K. E. = P. E. - Work done

K. E. = 0.013 - 0.0046 = 0.0084J

We can now find the speed using the formula for K. E.:

K. E. = ½*m*v²

0.0084 = ½ * 0.0045 * v²

v² = 0.0084/0.00255 = 3.733

=> v = 1.93 m/s

4 0
3 years ago
Read 2 more answers
A light-year is the distance light travels in one year (at speed = 2.998 × 108 m/s). (a) how many meters are there in 11.0 light
larisa [96]
<span>The answers are as follows:

(a) how many meters are there in 11.0 light-years?

11.0 light years ( 365 days / 1 year ) ( 24 h / 1 day ) ( 60 min / 1 h ) ( 60 s / 1 min ) ( 2.998x10^8 m/s ) = 1.04x10^17 m

(b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years?

1.04x10^17 m ( 1 au / </span>1.50 × 10^8 km <span>) ( 1 km / 1000 m) = 693329.472 au

(c) what is the speed of light in au/h? au/h

</span>2.998 × 10^8 m/s ( 1 au / 1.50 × 10^8 km ) ( 1 km / 1000 m) ( 3600 s / 1 h ) = 7.1952 au/h

8 0
3 years ago
A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative
Vesnalui [34]
If the current takes him downstream we must find the resultant vector of the velocities: V res= \sqrt{1^{2}+0.91^{2}  } = \sqrt{1.8281}= 1.3520747 Then if the river is 3000 m-wide the swimmer will have to pass:
 1.3520747 · 300 = 4056.14 m                t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.  
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.
5 0
3 years ago
When does carbon dioxide absorb the most heat energy? uring freezing during deposition during sublimation during condensation?
ololo11 [35]
The answer for the given question above would be the third option. Carbon dioxide absorbs the most heat energy during SUBLIMATION. By definition, sublimation is <span>the transition of a substance from the solid to the gas phase without passing through the intermediate liquid phase. Hope this answers your question.</span>
3 0
3 years ago
Read 2 more answers
A man standing 1.54 m in front of a shaving mirror produces a real, inverted image 15.2 cm from it. What is the focal length of
zavuch27 [327]

Answer:

The focal length is 16.86 cm and the distance of the man  if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

Explanation:

Given that,

Object distance u=1.54 m =154 cm

Image distance v = 15.2 cm

Magnification = 2

We need to calculate the focal length

Using formula of mirror

\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{f}=\dfrac{1}{15.2}+\dfrac{1}{-154}

\dfrac{1}{f}=\dfrac{347}{5852}

f=16.86\ cm

We need to calculate the focal length

Using formula of magnification

m= \dfrac{-v}{u}

Put the value into the formula

2=\dfrac{v}{u}

v = -2u

Using formula of for focal length

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{16.86}=\dfrac{1}{u}-\dfrac{1}{2u}

\dfrac{1}{16.86}=\dfrac{1}{2u}

2u=16.86

u=\dfrac{16.86}{2}

u=8.43\ cm

Hence, The focal length is 16.86 cm and the distance of the man  if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

3 0
3 years ago
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