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leva [86]
3 years ago
5

If a drop is to be deflected a distance d = 0.350 mm by the time it reaches the end of the deflection plate, what magnitude of c

harge q must be given to the drop? Assume that the density of the ink drop is 1000 kg/m3 , and ignore the effects of gravity.
Physics
1 answer:
Sloan [31]3 years ago
6 0

Answer:

q = 4.87 X 10^ -14 C

Explanation:

As d=0.350 mm

The ink drop will be accelerated by the electric field between the plates:

a = F/m

d = a(D0 / v)^2 / 2 ...... 1

a = qE/m ............... 2

Substituting 2  into 1:

d = (qE/m)(D0 / v)^2 / 2

q = 2mdv^2 / [E(D0)^2]

q = 2(1.00e-11 kg)(3.50e-4 m)(15.0 m/s)^2 / [(7.70e4 N/C)(2.05e-2 m)^2]

q = 4.87e-14 C

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How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.
Amanda [17]

Answer:

Latent heatnof fusion = 417.5 J

Explanation:

Specific latent heat of fusion of water is 334kJ.kg-1.

The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.

The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.

Latent heat = ML

Latent heat= 1.25 kg * 334kJ.kg-1

Latent heat = 1.25*334 *(J/kg)*kg

Latent heat = 417.5 J

8 0
3 years ago
A sand mover at a quarry lifts 2,000 kg of sand per minute a vertical distance of 12 m. The sand is initially at rest and is dis
Marianna [84]

Answer:

<h2>E. 3.95kW</h2>

Explanation:

Power is defined as the rate of workdone.

Power = Workdone/time taken

Given Workdone = Force * distance

Power = Force * distance/time taken

Power = mgd/t (F = mg)

m = mass of the sand in kg

g = acceleration due to gravity in m/s²

d = vertical distance covered in metres

t = time taken in seconds

Given m = 2000kg, d = 12m, t = 1min = 60secs, g = 9.8m/s²

Power = 2000*9.8*12/60

Power = 3920Watts

Minimum rate of power that must be supplied to this machine is 3920Watts or 3.92kW

5 0
3 years ago
g As observed on earth, a certain type of bacteria is known to double in number every 24 hours. Two cultures of these bacteria a
tangare [24]

Answer:

86.4 hrs

Explanation:

The amount of bacteria is initially 1

It doubles every 24 hrs.

After first 24 hrs, the amount = 2

After next 24 hrs = 4

After next 24 hrs = 8

After next 24 hrs = 16

After next 24 hrs = 32

After next 24 hrs = 64

After next 24 hrs = 128

After next 24 hrs = 256

Total time taken to reach 256 = 24 x 8 = 192 hrs

For the bacteria culture on the rocket that travels at a speed of 0.893c relative to the earth, this time is contracted by the relationship

t = t'(1 - ¥^2)^0.5

Where t is the contracted time =?

t' is the time on earth

¥ = v/c

Where v is the speed of the rocket

c is the speed of light

since v = 0.893c

¥ = 0.893

Substituting, we have

t = 192 x (1 - 0.893^2)^0.5

t = 192 x 0.2025^0.5

t = 192 x 0.45 = 86.4 hrs

8 0
3 years ago
A 100 kg bungee jumper leaps from a bridge. The bungee cord has an un-streched equilibrium length of 10 m, and a spring constant
nika2105 [10]

Answer:

11.78meters

Explanation:

Given data

Mass m = 100kg

Length of cord= 10m

Spring constant k= 35N/m

At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy

That is

Us=Ug

Us= 1/2kx^2

Ug= mgh

1/2kx^2= mgh

0.5*35*10^2= 100*9.81*h

0.5*35*100=981h

1750=981h

h= 1750/981

h= 1.78

Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

6 0
3 years ago
Read 2 more answers
A vertical block-spring system on earth has a period of 6.0 s. What is the period of this same system on the moon where the acce
Vera_Pavlovna [14]

Answer:

D) 15s

Explanation:

let Te be the period of the block-spring system on earth and Tm be the period of the same system on the moon.let g1 be the gravitational acceleration on earth and g2 be the gravitational acceleration on the moon.

the period of a pendulum is given by:

T = 2π√(L/g)

so on earth:

Te = 2π√(L/g1)

     =  6s

on the moon;

Tm = 2π√(L/g2)

since g2 = 1/6 g1 then:

Tm = 2π√(L/(1/6×g1))

      = √(6)×2π√(L/(g1))

and 2π√(L/(g1)) = Te = 6s

Tm = (√(6))×6 = 14.7s ≈ 15s

Therefore, the period of the block-spring system on the moon is 15s.

5 0
3 years ago
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