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babunello [35]
3 years ago
7

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential di

fference (voltage) between them of 3.22 x 104 V
Physics
1 answer:
raketka [301]3 years ago
8 0

Answer:

E=3.22*10^6 N/C

Explanation:

From the question we are told that:

Separation Distance d=1.0cm =0.01m

Potential difference V=3.22 * 10^4 V

Generally the equation for Electric Field strength is mathematically given by

 E=\frac{v}{d}

 E=\frac{3.22*10^4}{0.01}

 E=3.22*10^6 N/C

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Read 2 more answers
A river flows due east at 1.70 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
ad-work [718]

Answer:

a)

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)

the required distance is 40.98 m

Explanation:

Given that;

velocity of the river u = 1.70 m/s

velocity of boat v = 14.0 m/s

Now to get the velocity of the boat relative to shore;

( north of east), we say

a² + b² = c²

(1.70)² + (14.0)² = c²

2.89 + 196 = c²

198.89 = c²

c = √198.89

c = 14.1028 m/s  

tan∅ = v/u = 14 / 1.7 =  8.23529

∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east

Therefore, the velocity of the boat relative to shore is;

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)  

width of river = 340 m,

ow far downstream has the boat moved by the time it reaches the north shore in meters = ?

we say;

340sin( 90° - 83.0765°)

⇒ 340sin( 6.9235°)

= 40.98 m

Therefore, the required distance is 40.98 m

5 0
3 years ago
Determine the approximate force (N) used to pull a sled up a 400 m hill using 1900 J of work.
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The work done to pull the sled up to the hill is given by
W=Fd
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is W=1900 J and the distance through which the force is applied is d=400 m, so we can calculate the average force by re-arranging the previous equation and by using these data:
F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N
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4 years ago
Which situation is the best analogy for the doppler effect?
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The best scenario to describe the doppler effect would be listening to the siren of a passing ambulance or fire truck

then it is coming towards you, the pitch is higher, it gets higher as it approaches and peaks as it gets right in front of you. then it drop at once when it passes you and continues to drop till it fades away. this is a classic descrption of the doppler effect
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