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3 years ago

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Suppose you have a dipole that's free to move in any way (including rotate - imagine it floating in space). And there's an objec

t with charge Q a distance r away. That distance r would be much larger than d, the distance between the charges of the dipole, so we draw the dipole small.

1 answer:

Setler [38]3 years ago

5 0

Complete Question

The complete question shown on the first uploaded image

Answer:

a)

The force on Q due to dipole is Attractive

b)

The charge Q exerts attractive force on the dipole

c)

Yes from the above parts, force depends on the sign of charge

d)

$F = kQq[\frac{d^{2}+2rd}{r^{2}(d+r)^{2}} ]$

e)

The magnitude o force decrease by a factor of 8.0 times

Explanation:

The explanation is shown on the second uploaded image

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Sam, whose mass is 75 kg , takes off across level snow on his jet-powered skis. The skis have a thrust of 160 N and a coefficien

Volgvan

Answer:

top speed = 17.25

Total height = 281.19 m

Explanation:

given data

mass = 75 kg

thrust = 160 N

coefficient of kinetic friction = 0.1

solution

we get here frictional force acting that is

frictional force = $\mu *m*g$ .............1

frictional force = 0.1 × 75 × 9.8

frictional force = 73.5 N

and

Net force acting will be F = 160 - 73.5 N

F = 86.5 N

so

Acceleration in the First 15 second will be

F = ma .........2

86.5 = 75 × a

a = 1.15 m/s²

and

now After 15 second the velocity will be as

v = u + at ..........3

here u is 0

so v will be

V = 1.15 × 15

v = 17.25

and

now we get travels distance S in 15 s

s = u × t + 0.5 × a × t²

here u is 0

so distance s will be

s = 0.5 × a × t²

s = 0.5 × 1.15 × 15²

s = 129.37 m

and

now acceleration acting is

F = $\mu *m*g$

m a = $\mu *m*g$

a = $\mu* g$

a = - 0.98

here it is negative it mean downward nature of acceleration

and

now we get distance s by this formula

V² - u² = 2 a s

here v velocity is 0 and

u initial velocity is 17.25 m/s

put here value

0 - 17.25² = 2 × (-0.98) × s

solve it we get

s = 151.82 m

so

Total height is

Total height = 129.37 m + 151.82 m

Total height = 281.19 m

7 0

3 years ago

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

3 years ago

For two traveling waves, if the crest of one wave coincides with a trough of another, what type of interference occurs?

kogti [31]

When crest of one wave interferes with the trough of other wave, the amplitude of the resultant wave formed is less. Hence the type of interference is destructive interference.

3 0

3 years ago

The electric field strength in the space between two closely spaced parallel disks is 1.0 105 N/C. This field is the result of t

balu736 [363]

Answer:

$D=2.996\times 10^{-2} m$

Explanation:

*Assume the parallel disks have equal diameters.

Given the electric strength as $1.0\times 10^5 N/C.$ transferring $3.9\times 10^9$ electrons, the disk's Area can be calculated using the formula:

$E=\frac{\eta}{\epsilon_o}=\frac{Q}{A\epsilon_o}\\\\A=\frac{Q}{E\epsilon_o}\\\\=\frac{(3.9\times 10^9)\times (1.6\times10^{-19})}{(1.0\times 10^5 )\times (8.85\times10^{-12})}\\\\A=7.0508\times 10^{-4} \ m^2$

#We now calculate the disks diameter:

$A=\pi(D/2)^2\\\\2\sqrt{\frac{A}{\pi}}=D\\\\=2\sqrt{7.0508\times 10^{-4}/\pi}\\\\D=2.996\times 10^{-2} \ m$

Hence, the diameter of the disks is $D=2.996\times 10^{-2} m$

8 0

3 years ago

What are the units for mass and weight?

Luda [366]

The kilogram is the SI unit of mass and it is the almost universally used standard mass unit. The associated SI unit of force and weight is the Newton, with 1 kilogram weighing 9.8 Newtons under standard conditions on the Earth's surface.

6 0

3 years ago

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