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Law Incorporation [45]
4 years ago
5

Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th

e right hand and has magnetic field of 4.00 N/C .If an ion of charge +2e is released at rest at the left hand plate .What its K.E when reaches the right hand plate?tell the correct answer with explanation..(a)4ev (B) 32ev (C)64ev(D)16ev
Physics
1 answer:
yuradex [85]4 years ago
3 0

Answer:

C

Explanation:

Formula E=F/C also E=V/d

In this case use the second formula; E=V/d

Data given; E=4N/C d=8m

So v=E X d

     V=4x8=32V

k.e=eV= 2X32=64eV

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3. Discuss the main purpose of Experiment" Explain the greatest advantage of experiment in psychology?
Scrat [10]

Answer:

Social psychologists utilize experimental techniques to study how people are influenced by groups. Health psychologists rely on experimentation and research to better understand the factors that contribute to wellness and disease.

Explanation:

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3 years ago
An airplane dropped a flare from a height of 2860 feet above a lake. How many seconds did it take for the flare to reach the wat
KATRIN_1 [288]

Answer: 13.2 seconds.

Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0

S= distance travelled

a = acceleration due gravity

t= time.

1 foot = 0.305m so,

S= 2860 feet =872.3m

S= ut+1/2 at²

872.3 = 0×t + 1/2×10 × t²

872.3 =0 + 5t²

T²= 872.3/5

T²= 174.46

Take the square root of T we then have;

t = 13.2 seconds to one decimal place.

8 0
3 years ago
Read 2 more answers
The electric potential in a region of space is \[V=350/\sqrt{x ^{2}+y ^{2}}\] where x and y are in meters. what is the strength
VARVARA [1.3K]
So the given value or the formula in getting the electric potential region of space is V=350/sqrt of x^2+y^2. So the given data is x and y is equals to 2.6 and 2.8. So in my calculation i came up with an answer of 91.6
8 0
3 years ago
What is the greatest distance an image can be located behind a convex spherical mirror?
Afina-wow [57]

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}

\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}

so here maximum value of image distance is equal to focal length of the mirror

6 0
3 years ago
Question 4 of 10
Alexus [3.1K]

Answer:

22.3 kg•m/s

Explanation:

Apex;)

7 0
3 years ago
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