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Irina-Kira [14]
3 years ago
10

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m

, respectively. (Take V = 0 at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?
Physics
1 answer:
son4ous [18]3 years ago
3 0

Answer:

A. 30.7cm

B. 1.7*10^{-10}C

C. The electric field is directed away from the point of charge

Explanation:

A.

 because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm

B.

Considering Gauss's law

EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C

C. The electric field directed away from the point of charge when the charge is positive.

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A raging bull of mass 700 kg runs at 36 km/h .How much kinetic energy does it have ?
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Answer:

Use equation for kinetic energy: Ek=mV²/2

m=700 kg

V=10m/s

Ek=700kg*100m²7s²/2

Ek=35000 J=35kJ

Explanation:

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3 years ago
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How do prokaryotic cells replicate?
Alika [10]

Answer:

The usual method of prokaryote cell division is termed binary fission. The prokaryotic chromosome is a single DNA molecule that first replicates, then attaches each copy to a different part of the cell membrane. When the cell begins to pull apart, the replicate and original chromosomes are separated.

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3 years ago
A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the b
netineya [11]

Answer: The bottom of the ladder is moving at 3.464ft/sec

Explanation:

The question defines a right angle triangle. Therefore using pythagorean

h^2 + l^2 = 10^2 = 100 ...eq1

dh/dt = -2ft/sec

dl/ dt = ?

Taking derivatives of time in eq 1 on both sides

2hdh/dt + 2ldl/dt = 0 ....eq2

Putting l = 5ft in eq2

h^ + 5^2 = 100

h^2 = 25 = 100

h Sqrt(75)

h = 8.66 ft

Put h = 8.66ft in eq2

2 × 8.66 × (-2) + 2 ×5 dl/dt

dl/dt = 17.32 / 5

dl/dt = 3.464ft/sec

7 0
3 years ago
The magnetic field of a long, straight, and closely-wound solenoid, inside the solenoid at a point near the center, is 0.645 T.
Savatey [412]

Answer:

B'=1.935 T      

Explanation:

Given that

magnetic field ,B= 0.645 T

We know that magnetic filed in the solenoid is given as

B=\mu _0 n\ I

I=Current

n=Number of turn per unit length

μ0 =magnetic permeability

Now when the current increased by 3 factors

I'=3 I

Then the magnetic filed

B'=\mu _0 n\ I'

B'=\mu _0 n\ (3I)

B'=3 B

That is why

B' = 3 x 0.645 T

B'=1.935 T

Therefore the new magnetic filed will be 1.935 T.

3 0
3 years ago
A block of mass m = 2.0 kg lies on a rough ramp that is inclined at an angle θ = 20oto the horizontal. A force F of magnitude 5.
Marina86 [1]

Answer:

a) 0.64 b) 2.17m/s^2 c) 8.668joules

Explanation:

The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,

Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move

Frictional force = mgsin20o + 5N = 6.71+5N = 11.71

The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44

Coefficient of static friction = 11.71/18.44= 0.64

Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)

b) coefficient of kinetic friction = frictional force/ normal force

Fr = 0.4* mgcos 20o = 7.375N

F due to motion = ma = total force - frictional force

Ma = 11.71 - 7.375 = 4.335

a= 4.335/2(mass of the block) = 2.17m/s^2

C) work done = net force *distance = 4.335*2= 8.67Joules

8 0
3 years ago
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