1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
amm1812
3 years ago
11

A negative charged particle of known magnitude q is placed in a uniform electric field of known strength E. What additional info

rmation is needed to determine the magnitude of the force on the particle?
a. The distance between the charged plates producing the uniform field.
b. The distance from the nearest plate to the charged particle.
c. The magnitude of the charges producing the field.
d. The mass of the negatively charged particle.
d. No additional information is needed.
Physics
1 answer:
olga_2 [115]3 years ago
8 0

Answer:d

Explanation:

Electric field is a region around charged particle where charged particle experience a force .

Magnitude of Force experienced by charged particle is given by

F=q\cdot E

where F=force experienced by charged particle

q=charge of particle

E=strength of electric field

If q and E are given then no additional information is needed to determine the magnitude of force.      

You might be interested in
How will the motion of the arrow change after it leaves the bow?
Pavel [41]

The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.

pls thank me and brainliest me

4 0
3 years ago
2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
  • M is magnification.

{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
4 0
2 years ago
What is the least force you could exert on the refrigerator to move it?
Ulleksa [173]
 <span>If the refrigerator weights 1365 and you are not exerting any vertical force on it, then the normal force is also 1365N. so Fn=1365 

Fsf = Static frictional force = (coefficient of static friction) * (Normal force) 

So the least for you could exert to move it is equal to the Fsf. 
Fsf = (0.49)(1365N)</span><span>
</span>
8 0
2 years ago
A cart travels down a ramp at an average speed of 5.00 centimeters/second. What is the speed of the cart in miles/hour? (Remembe
viva [34]
Set up the problem with the conversion rates as fractions where when you multiply the units cancel out leaving the desired units behind.

( \frac{ 5 cm}{1 sec} ) *( \frac{1m }{100cm})*( \frac{1km}{1000m} )*( \frac{1mi}{1.6km} ) = 0.00003125 mi/sec \\  \\ ( \frac{0.00003125 mi}{1sec} )*( \frac{60sec}{1min} )*( \frac{60min}{1hr} )=0.1125 mi/hr
7 0
3 years ago
Read 2 more answers
I need help quick pls . Its either physical or chemical change !
Trava [24]
Physical, chemical, chemical, physical, I’m pretty sure. Natural things, like you starting to breathe heavier are physical most of the time!!!
8 0
3 years ago
Other questions:
  • What questions should be considered when evaluating the reliability of a website?
    5·2 answers
  • Coherent light with wavelength 599 nm passes through two very narrow slits with separation of 20 μm, and the interference patter
    14·1 answer
  • Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th
    9·2 answers
  • On my bike, I was able to travel 40 miles in one hour. From this information, we can determine the bike's
    10·2 answers
  • A 0.0200-Ω ammeter is placed in series with a 10.00-Ω resistor in a circuit.(a) Draw a circuit diagram of the connection.(b) Cal
    8·1 answer
  • How do we gather most of the information we get about the universe around us?
    13·1 answer
  • A 92-kg water skier floating in a lake is pulled from rest to a speed of 12 m/s in a distance of 25 m. What is the net force exe
    12·2 answers
  • "a force sets an object in motion. when the force is multiplied by the time of its application, we call the quantity impulse, an
    14·1 answer
  • In what ways are meteorites different from meteors? What is the probable origin of each?
    6·1 answer
  • A 1.6 Kg bird that is flying through the air has 220 J of kinetic energy. How fast is the bird flying?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!