The woman dries herself with a towel. She hangs the wet towel in the bathroom to dry. Figure 2 shows two places she could hang the towel. The towel will dry faster if it is hung from the unheated towel rail instead of the towel ring. Explain why.
2 points

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

v=u+at

v=0+8.0×30

v=240

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:

See the explanation below.

Explanation:

Work in physics is defined as the product of force by the distance that the body travels in the direction of the force.

It can be represented by means of the following equation.

[tex]W=F*d[/tex]

where:

W = work [J]

F = force [N]

d = distance [m]

In the given example the work is zero since the tree does not move, therefore the bear exerts a force on the tree. But there is no talk of movement of the tree, therefore the work is zero.

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:

C. Some waves

Explanation:

The answer is some waves because transverse waves are not the only types of waves we have on the earth. We also have longlitudal waves, which means only some waves can only be transverse. The answer to the question should be C. Some waves.

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:

KE = 1/2 m v^2

Explanation:

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

Learn more on potential difference on a charged particle here;

https://brainly.com/question/9383604

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.

Learn more about momentum here: https://brainly.com/question/24815698

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: 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. ​

Learn more about momentum here:

brainly.com/question/24030570

#SPJ5

Answer: c

Explanation:

Sound waves cannot travel through a medium

Answer:

A 50 kg ball traveling at 20 m/s would have 4 times more kinetic energy.

A 50 kg ball traveling at 5 m/s would have 4 times less kinetic energy.

A 50 kg person falling at 10 m/s would have the same kinetic energy.

Explanation:

hope this helps:)

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:

IVE BEEN TRYNA FIND THE ANSWER FOR THE PAST HOUR T-T

Explanation:

Answer:

3.2 x10^11 bits of storage

Explanation:

Answer:

it is 9m/s , hope it helps

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:

D. blinking ofeyea

Explanation:

not really sure

Answer:

80 N

Explanation:

mass = 20 kg

acceleration = 4 m/s^2

Force = mass * acceleration

         = 20 *4  

         = 80 N

Force of a crate : 300 N

Given

0.25 meters wide and 1.5 meters long

Pressure = 800 Pa

Required

Force

Solution

P = F/A

A = 0.25 x 1.5

A = 0.375 m²

P = 800 Pa = 800 N/m²

Input the value :

F = P x A

F = 800 N/m² x 0.375 m²

F = 300 N

Answer: the first is incorrect the second is correct the third is correct

Explanation: the fourth is correct the fifth is incorrect

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