Sometimes, a crackling sound is heard while taking off a sweater during winters. Explain

Answer:

I would say A

Explanation:

the closer you are to a planets center of gravity i feel you would weigh more.

Answer:

D. blinking ofeyea

Explanation:

not really sure

v=u+at

v=0+8.0×30

v=240

Answer:

(a) The potential difference between the spheres is 750 kVA

(b) The charge on the smaller sphere is 6.[tex]\overline 6[/tex] μC

(c) The charge on the smaller sphere, Q₁ = 13.[tex]\overline 3[/tex] μC

Explanation:

(a) The given parameters are;

The charge on the inner sphere, q = 5.00 μC

The radius of the inner sphere, r = 3.00 cm = 0.03 m

The charge on the larger sphere, Q = 15.0 μμC

The radius of the larger sphere, R = 6.00 cm = 0.06 m

The potential difference between two concentric spheres is given according to the following equation;

[tex]V_r - V_R = k \times q \times \left ( \dfrac{1}{r} - \dfrac{1}{R} \right)[/tex]

Where;

R = The radius of the larger sphere = 0.06 m

r = The radius of the inner sphere = 0.03 m

q = The charge of the inner sphere = 5.00 × 10⁻⁶ C

Q = The charge of the outer sphere = 15.00 × 10⁻⁶ C

k = 9 × 10⁹ N·m²/C²

Therefore, by plugging in the value of the variables, we have;

[tex]V_r - V_R = 9 \times 10^9 \times 5.00 \times 10^{-6} \times \left ( \dfrac{1}{0.03} - \dfrac{1}{0.06} \right) = 750,000[/tex]

The potential difference between the spheres, [tex]V_r - V_R[/tex] = 750,000 N·m/C = 750 kVA

(b) When the spheres are connected with a wire, the charge, 'q', on the smaller sphere will be added to the charge, 'Q', on the larger sphere which as follows;

[tex]Q_f[/tex] = Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C

[tex]Q_f[/tex] = 20 × 10⁻⁶ C

From which we have;

Q₁/Q₂ = R/r

Where;

Q₁ = The new charge on the on the larger sphere

Q₂ = The new charge on the on the smaller sphere

[tex]Q_f[/tex] = 20 × 10⁻⁶ C = Q₁ + Q₂

∴ Q₁ = 20 × 10⁻⁶ C - Q₂ = 20 μC - Q₂

∴ (20 μC - Q₂)/Q₂ = 0.06/(0.03) = 2

20 μC - Q₂ = 2·Q₂

20 μC = 3·Q₂

Q₂ = 20 μC/3

The charge on the smaller sphere, Q₂ = 20 μC/3 = 6.[tex]\overline 6[/tex] μC

(c) Q₁ = 20 μC - Q₂ = 20 μC - 20 μC/3 = 40 μC/3

The charge on the smaller sphere, Q₁ = 40 μC/3 = 13.[tex]\overline 3[/tex] μC.

The potential difference is 7.5*10^5V and the charge on the smaller sphere is 6.67uC while the charge on the larger sphere is 13.34uC

Data;

for the inner sphere;

[tex]v_i = \frac{kq}{r} = \frac{(9*10^9)(5*10^-^6)}{0.03} \\v_i = 1.5*10^6V[/tex]

for the outer sphere;

[tex]v_o = \frac{KQ}{R} = \frac{(9*10^9)(15*10^-^6)}{0.06} = 2.25 * 10^6V[/tex]

The difference in potential is

[tex]\delta V = v_o - v_i = 2.25*10^6 - 1.5*10^6 = 7.5*10^5V[/tex]

[tex]q + Q = 5 + 15 = 20 \mu C = 20*10^-^6C ..eq(i)[/tex]

The sharing of charge continues till they attain a point of equal potential

[tex]v_i = v_o \\\frac{kq}{r} = \frac{kQ}{R} \\\frac{q}{0.03} = \frac{Q}{0.06} \\Q= 2q ...eq(ii)\\[/tex]

let's solve for equation (i) and equation (ii)

[tex]q+2q = 20\mu C\\q = 6.67 \mu C[/tex]

The charge on the smaller sphere is 6.67uC

The charge on the larger sphere is

[tex]Q = 2q\\q = 6.67 \mu C\\Q = 2 * 6.67 = 13.34 \mu C[/tex]

The charge on the larger sphere is 13.34uC

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Answer:

2.83km/hr

Explanation:

Let stream flowing at the rat x km/hr

In still water

Speed of boat=4km/hr

Upstream speed=4-x

Downstream speed=4+x

Total time=2 hr

Time=Distance/speed

According to question

[tex]\frac{2}{4+x}+\frac{2}{4-x}=2[/tex]

[tex]\frac{8-2x+8+2x}{(4+x)(4-x)}=2[/tex]

[tex]\frac{16}{4^2-x^2}=2[/tex]

Using the formula

[tex](a+b)(a-b)=a^2-b^2[/tex]

[tex]4^2-x^2=\frac{16}{2}[/tex]

[tex]16-x^2=8[/tex]

[tex]16-8=x^2[/tex]

[tex]8=x^2[/tex]

[tex]x=\sqrt{8}=2.83km/hr[/tex]

Hence, the stream  is flowing at 2.83km/hr

impossible because fractional charges do not exust

Answer:

This is because the acceleration of objects due to gravity is independent of the mass of the object and is constant for all objects, therefore, all objects fall with the same speed.

Explanation:

The weight of an object or force of gravity acting on an object on the surface of earth is a product of its mass and acceleration due to gravity.

Mathematically, w = mg

where, m=mass of the object; g = acceleration due to gravity

Also, from newton's law of gravitation, gravitational force on the object ,F = GMm/r²

where G is the gravitational constant; M is mass of Earth; m is mass of object; r is the distance of separation between the object and the center of mass of the earth which is approximately the radius of earth.

Since the weight of an object is equal to the force of gravitation acting on it

W = F

mg = GMm/r²

g = GM/r²

The expression above is that of the relationship between the force of gravity acting on a body on the earth's surface, the weight of that body and the acceleration due to gravity, g.

It can be seen that the acceleration due to gravity g is independent of the mass of the object. Therefore, the acceleration of objects due to gravity is constant for all objects and all objects fall with the same speed.

Answer:

The time of fall of an object is dependent only on the height of fall for a given acceleration due to gravity and it is independent of the object's mass

Explanation:

The gravitational force acting on an object on Earth can be observed as the weight, 'W', of the object which is a function of the mass, 'm', of the object and the relationship between the gravitational force and the mass of the object is given as follows;

W = m × g

Where;

g = The acceleration due to gravitational force = Constant

The relationship between the speed, 'u', and therefore, the time, 't', with which an object falls, the distance or height of fall, 'h', and the acceleration of the motion, 'g' is given as follows;

h = u·t + 1/2·g·t²

Given that the initial velocity of the object, 'u', of the object allowed to fall is u = 0 m/s, we have;

h = u·t + 1/2·g·t² = 0 × t + 1/2·g·t²

h = 0 × t + 1/2·g·t² = 1/2·g·t²

h = 1/2·g·t²

t² = 2·h/g

t = √(2·h/g)

Therefore, the time, 't', it takes every object to fall from a height, 'h', is dependent only on the height, 'h', for a given acceleration due to gravity, 'g' and it is independent of the mass, 'm', of the object.

Answer: [tex]F_4[/tex]

Explanation:

The friction opposes motion. We can assume the motion to be downwards because the only force is the weight of the box mgsin(theta) aka F3.

Arrow 4 correctly indicates the direction of friction acting on the box.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and accelertaion in the different components and balancing the equation gets.

Friction is a force that resists motion. Because the only force acting on the box is its own weight, we may conclude it is moving downhill. As the force approach the downhill the friction force is acting opposite to it.

Hence, arrow 4 correctly indicates the direction of friction acting on the box.

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Work needed = 23,520 J

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object  

Work is the product of force with the displacement of objects.  

Can be formulated  

W = F x d  

W = Work, J, Nm  

F = Force, N  

d = distance, m  

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

Momentum = (mass) x (velocity)

Momentum m₁  =  (6 kg) x (13 m/s east)  =  78 kg-m/s east

Momentum m₂  =  (14 kg) x (7 m/s east)  =  98 kg-m/s east

Since they're both in exactly the same direction you can simply addum up to get their total momentum.  It's 176 kg-m/s east .

The  momentum of a system that consists of two objects should be 176 kg m/s on east hand side.

Since we know that

Momentum = (mass) x (velocity)

So,

Momentum m₁  =  (6 kg) x (13 m/s east)  =  78 kg-m/s east

Momentum m₂  =  (14 kg) x (7 m/s east)  =  98 kg-m/s east

Since they are on same direction so the total momentum should be

= 78 + 98

= 176 kg-m/s east

hence, The  momentum of a system that consists of two objects should be 176 kg m/s on east hand side.

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Answer:

For a fixed mass of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional. Or Boyle's law is a gas law, stating that the pressure and volume of a gas have an inverse relationship. If volume increases, then pressure decreases and vice versa, when the temperature is held constant.

Answer:

14945 J

Explanation:

Glad you asked this question!

The speed is  irrelevant in this question. It perhaps was put out to throw you off.

P.E = mgh = (0.5 * 9.8 * 3050) = 14945 Joules

Answer:

one quarter= 5.69 g

Explanation:

if 10 quarters = 0.1255 lbs

then 1 quarter = 0.01255 lbs (0.1255 divided by 10)

0.01255 lbs = 5.69 g

Answer:

Option b is correct - Rocky, nearly continuous, found on earth surface.

The reason is it is the rocky surface area composed of crust and upper mantle part. Basically, it is the outer shell of earth.

Answer:

Cornea

Explanation:

The outer layer of the cornea is covered by epithelial cells that protect the surface. A corneal abrasion may result from dragging a piece of debris across the surface and scraping off the epithelial cells.

Answer: I am sorry but I don't know what you're trying to say.

Explanation: Hope

Answer:

b. d. f.

Explanation:

Got it right on Edge

Answer:

2. Gina gathers information about a court case.

4. Hana interviews and advises a person who has been accused of a crime.

6. Dewayne creates the paperwork for a business contract.

Answer:

According to the Junction rule, in a circuit, the total of the currents in a junction is equal to the sum of currents outside the junction. Kirchhoff’s Voltage Law goes by several names as Kirchhoff’s Second Law and Kirchhoff’s Loop Rule. According to the loop rule, the sum of the voltages around the closed loop is equal to null.

Explanation:

Answer:

Fasle. , is correct answer hope it helps

Answer: hope this helps

Explanation:

If the final velocity of the ball is 2.5 m/s, the final velocity of the bowling pin is 7.8125 m/s. ​

According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

According to the principle of momentum conservation,  total momentum of an system remains conserved.

Total initial momentum = total final momentum

5 kg × 5 m/s + 1.6 kg × 0 m/s = 5 kg × 2.5 m/s + 1.6 kg × v

1.6 kg × v  = 5 kg × 5 m/s -  5 kg × 2.5 m/s

v = (5 kg /1.6 kg) (5 m/s - 2.5 m/s)

= 7.8125 m/s

Hence, If the final velocity of the ball is 2.5 m/s, the final velocity of the bowling pin is 7.8125 m/s. ​

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Answer:

In any collision (crash) the momentum

is ALWAYS constant

In any collision (crash) the momentum is always constant​, whether it is an elastic collision or an inelastic collision the momentum before and after the collision is conserved

It is the type of collision in which the total momentum, as well as the kinetic energy of the system, is conserved.

The momentum is always preserved before and after any collision (crash), regardless of whether it is an elastic collision or an inelastic collision.

In the inelastic collision, the momentum is conserved but the kinetic energy of the quantity is not conserved.

The momentum of a quantity is preserved regardless of the type of collision, whether it be an elastic collision or an inelastic collision; the momentum of the quantity will always be conserved.

Hence, The momentum is always preserved before and after any collision (crash), regardless of whether it is an elastic collision or an inelastic collision.

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Answer:

1. The correct way a cell phone call goes out is;

The phone call goes out, transmits through radio waves, then moves to the central hub. Then the signal go to the nearest tower to the receiver of the call

2. Long wavelength

3. Microwave

Explanation:

1. When making a call through a cell phone, the speech made is digitized such that it is converted to electrical signals which are then converted into number strings that are transmitted out of the antenna of the phone as radio waves which are then picked up by the closest cellphone tower mast which then transmit the signal to the applicable central hub from where the signal is transmitted to the tower closest to the intended call receiver

Therefore, the correct option is that the phone call goes out, transmits through radio waves, then moves to the central hub. Then the signal go to the nearest tower to the receiver of the call

2. The property of AM frequencies that allow them to travel over long distances is long wavelength

The frequency of AM radio waves is between 540 kHz and 1600 kHz, which can be propagated as ground wave or by the ionosphere at night

3. Cell phones makes use of the subcategory of microwave which is the super high frequency (SHF) band which with a range of 3 GHz to 30 GHz

Answer:

(10 L)(100° C) + (90 L)(10°) = (100 L)T

1,000 + 900 = 100T

1,900 = 100T

100T = 1,900

T = 1,900/100

Explanation:

Answer:

A device that is used to break an electric circuit is called electric switch

Answer:

b

Explanation:

Answer:

B

Explanation:

Answer:

.Cell phones convert sound waves into______waves.

-sound waves

.Electromagnetic waves used in cell phone communications are called?

-radio wave

.To send out a radio signal far and wide...it is called?

antenna

Explanation:

i might be wrong in ans c ok