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Firlakuza [10]
3 years ago
13

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

ative plate with a speed of 45000 m/s.
What will be the final speed of an electron released from rest at the negative plate?
Express your answer to two significant figures and include the appropriate units
V=_________
Physics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:

2.1x10^6m/s

Explanation:

One electron has a charge of –1.602e-19 C

mass of electron is 9.1e-31 kg

mass of proton is 1.6726e−27 kg

mass ratio is 1.6726e−27 / 9.1e-31 = 1838

The force is constant, F

distance is constant, d

a = F/m

a increases by a factor 1838, as m decreases by that factor

a = a₀1838

v₀² = 2a₀d

v² = 2a₀d1838

v²/v₀² = 2a₀d1838 / 2a₀d = 1838

v² = 1838v₀² = 1838(45000)²

v = 45000√1838 = 2.1e6 m/s

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A force of 350N is applied to a body. If the work done is 40kJ, what is the distance through which the body moved?
Studentka2010 [4]

The distance covered by the body is 114.3 m

Explanation:

The work done by a force exerted on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the object in this problem, we have

F = 350 N is the force applied

W=40 kJ = 40,000 J is the work done

\theta=0^{\circ} if we assume that the force is applied parallel to the motion of the object

Solving for d, we find the distance covered by the object:

d=\frac{W}{F cos \theta}=\frac{40,000}{(350)(cos 0)}=114.3 m

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

7 0
3 years ago
A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in
miv72 [106K]

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is \mu_s =  0.60

The value for static friction is \mu_k =  0.70

Explanation:

From the question we are told that

The mass of the clock is m  =  95 \  kg

The first horizontal force is F _s  =  560 \  N

    The second horizontal force is    F _k  =  650  \  N

Generally the static frictional force is equal to the first  horizontal force

So

     F _s  =  m  *  g  *  \mu_s

=>   560  =  95  *  9.8  *  \mu_s

=>    \mu_s =  0.60

Generally the kinetic frictional force is equal to the second horizontal force

So

      F _k  =  m  *  g  *  \mu_k

      650 =  95  *  9.8  *  \mu_k

     \mu_k =  0.70

3 0
3 years ago
A 42.6kg lamp is hanging from wires as shown in figure.The ring has negligible mass. Find tensionsT1, T2,T3 if the object is in
IRISSAK [1]

Answer:

T1 = 417.48N

T2 = 361.54N

T3 = 208.74N

Explanation:

Using the sin rule to fine the tension in the strings;

Given

amass = 42.6kg

Weight = 42.6 * 9.8 = 417.48N

The third angle will be 180-(60+30)= 90 degrees

Using the sine rule

W/Sin 90 = T3/sin 30 = T2/sin 60

Get T3;

W/Sin 90 = T3/sin 30

417.48/1 = T3/sin30

T3 = 417.48sin30

T3 = 417.48(0.5)

T3 = 208.74N

Also;

W/sin90 = T2/sin 60

417.48/1 = T2/sin60

T2 = 417.48sin60

T2 = 417.48(0.8660)

T2 = 361.54N

The Tension T1 = Weight of the object = 417.48N

8 0
3 years ago
What do all elements in a column in the periodic table have in common?
JulsSmile [24]

Answer:

1, their atoms have the same number of valence electron. because valence electron determine the group of elements.

6 0
3 years ago
What happens to parallel light rays that strike a concave lens?
zhenek [66]
Due to the shape of the lens , parallel rays will be deviated
5 0
3 years ago
Read 2 more answers
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