In picture 1, heat is flowing from the ____ to the _____ In picture 2, heat is flowing from the _______ to the ____​

Answer:

0.248

Explanation:

Initial speed u = 13.6

Final speed v = 0

Distance s = 38.1

We have umg = ma

We make u subject of the formula

u = a/g

V² = u² + 2as

a = v²-u²/2s

We substitute the values into the above

a = 0-(13.6)²/2*38.1

a = 184.96/76.2

a = 2.427m/sec

Remember that

u = a/g

u = 2.427/9.8

= 0.2476

This is approximately 0.248

This is the minimum coefficient of friction required to keep the crate from sliding.

Answer:

predict annual flooding

Answer:

I'm pretty sure it is b make their calander

Answer:

      I =  [tex]\frac{1}{4\pi \ r^2}[/tex]

we see the intensity decreases with the inverse of the distance squared

Explanation:

Intensity is defined as power per unit area,

           I = P / A

in this case we have that the sound is emitted in a spherical form therefore the area is

           A = 4 pi r2

therefore the intensity is

          I =  [tex]\frac{1}{4\pi \ r^2}[/tex]

as we see the intensity decreases with the inverse of the distance squared

Answer:

Explanation:

Current  has speed vW with respect to the shore and boat has speed vB with respect to water or current so speed of boat  with respect to shore

vW + vB .

Distance travelled with respect to shore by boat = D

time ( tout ) = distance / speed with respect to shore

tOut = D / ( vW + vB )

When the boat travels upstream , its velocity with respect to shore

= ( vB - vW ) , vB must be higher .

tin = D /  ( vB - vW )

3 ) tin = D /  ( vB - vW )

170 = 120 / (vB - 0.3 )

(vB - 0.3 ) = 12 / 17 = .706

vB = 1.006 m / s

4 )

tOut = D / ( vW + vB )

= 120 / ( .3 + 1.006 )

= 92.26 s

Time taken by a body is ratio of the distance traveled by it to the speed.

          [tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]

           [tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]

The net speed of the boat is upstream speed. The difference of the speed of the boat is downstream speed.

Given information-

The speed of the boat with respect to shore is [tex]v_w[/tex].

The speed of the boat in downstream with respect to water is [tex]v_B[/tex].

The distance traveled by the boat is [tex]D[/tex] in time [tex]t_{out}[/tex].

Time taken by a body is ratio of the distance traveled by it to the speed.

        [tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]

        [tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]

Now the distance is 120 m, the value of [tex]t_{in}[/tex] is 170 s and [tex]v_W[/tex] 0.3 m/s. Thus,

         [tex]170=\dfrac{120}{v_B-0.3}\\v_B-0.3=\dfrac{120}{160} \\v_B=0.706+0.3\\v_B=1.006[/tex]

Hence the speed of the boat with respect to the water is 1.006 m/s.

          [tex]t_{out}=\dfrac{120}{1.006+0.3}\\t{out}=\dfrac{120}{1.306} \\t{out}=91.9[/tex]

Hence the time it takes the boat to move a distance D downstream is 91.9 seconds.

Thus,

          [tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]

           [tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]

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Answer:

The number of electrons that surround the nucleus will determine whether or not it is electrically charged or electrically neutral

Explanation:

Answer:

letra A segundo o couculo a divisão e completa

Answer:

a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c)  F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N

Explanation:

a) q1 is negative and the charge of the bar is positive therefore the force is attractive

b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x

           dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]

where k is a constant, Q₁ the charge at the origin, x the distance

c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L

         ∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]

as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density

          λ = dQ₂ / dx

          DQ₂ = λ dx

we substitute

         F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]

         F = k Q1 λ ([tex]-\frac{1}{x}[/tex])  

we evaluate the integral

        F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]

        F = k Q₁ λ  [tex]( \frac{L}{d \ (d+L)})[/tex]

we change the linear density by its value

      λ = Q2 / L

       F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]

d) we calculate the magnitude of F

       F =9 10⁹ (-4.2 10⁻⁶)   [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]

       F = -1.09 N

the sign indicates that the force is attractive

Answer:

a)Toward the rod

b)|dF| = k|Q1|Q2(dx/L)/x^2

c)|F| = k|Q1|Q2/(d(d+L))

d)Plug in for answer c and solve

Explanation:

A)

Q1 is negative and Q2 is positive so it is an attractive force to  where the rod is located.

B)

The formula for Force due to electric charges is F=kQ1Q2/r^2

In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.

The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.

The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.

The final formula is |dF| = k|Q1|Q2(dx/L)/x^2

C)

Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:

F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2

factor out constants

F = kQ1Q2/L * integral d to d+L(1/x^2)dx

F = kQ1Q2/L * (-1/x)| from d to d+L

F = kQ1Q2/L * (-1/d+L - -1/d)

F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))

F = kQ1Q2/L * (L)/(d(d+L))

F = kQ1Q2/(d(d+L))

D)

Plug in the given values into c and you have your answer.

Answer:

Permittivity, also called electric permittivity, is a constant of proportionality that exists between electric displacement and electric field intensity.

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance [tex]D_{A}[/tex] beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed [tex]v_{A}[/tex]. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed [tex]v_{B}[/tex], which is greater than [tex]v_{A}[/tex].

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]

              Part B: [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

[tex]x=x_{0}+vt[/tex]

where

[tex]x_{0}[/tex] is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is [tex]D_{A}[/tex].

The equation will be:

[tex]x_{A}=D_{A}+v_{A}t[/tex]

Car B started at the starting line. So, its equation is

[tex]x_{B}=v_{B}t[/tex]

Part A: When they meet, both car are at "the same position":

[tex]D_{A}+v_{A}t=v_{B}t[/tex]

[tex]v_{B}t-v_{A}t=D_{A}[/tex]

[tex]t(v_{B}-v_{A})=D_{A}[/tex]

[tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]

Car B meet with Car A after [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex] units of time.

Part B: With the meeting time, we can determine the position they will be:

[tex]x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )[/tex]

[tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]

Since Car B started at the starting line, the distance Car B will be when it passes Car A is [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex] units of distance.

The distance traveled by the car A and car B  should be equal to the as they meet at the same position.

The time car B will catch the car A after is,

[tex]\dfrac{D_A}{v_B-v_A}[/tex]

The distance is the product of the speed of the body and the time taken to travel the distance.

Given information-

Car A has a head start and is a distance DA beyond the starting line at,

[tex]t=0[/tex]

Car A travels at a constant speed [tex]v_A[/tex].

Car B travels at a constant speed [tex]v_B[/tex].

The distance is the product of the speed of the body and the time taken to travel the distance.

The position equation from the motion for car A can be given as,

[tex]x_A=v_At+D_A[/tex]

The position equation from the motion for car B can be given as,

[tex]x_B=v_Bt[/tex]

The distance traveled by the car A and car B  should be equal to the as they meet at the same position. Thus,

[tex]x_A=x_B[/tex]

Put the values,

[tex]v_At+D_A=v_Bt\\v_At-v_Bt=-D_A\\t(v_B-v_A)=D_A\\t=\dfrac{D_A}{v_B-v_A}[/tex]

Hence the time car B will catch the car A after is,

[tex]\dfrac{D_A}{v_B-v_A}[/tex]

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Answer:

Earth's atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by earth's atmosphere? Earth has moderate temperatures.

Explanation:

Answer:

616000 J.

Explanation:

Bi. Determination of the work done.

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Workdone is simply defined as the product of force and the distance moved in the direction of the force. Mathematically, is can be expressed as:

Workdone = Force × distance

Wd = F × s

With the above formula, we can obtain the Workdone as follow:

Force (F) = 220 N

Distance (s) = 2800 m

Workdone (Wd) =?

Wd = F × s

Wd = 220 × 2800

Wd = 616000 J.

Answer:

d

Explanation:

the US West Coast region probably has cooler summers than it would without the effect of a nearby ocean current.

ocean current can be defined as the horizontal movement of seawater which is produced by gravity, wind, and water density, it play an major role in the determination of climates of coastal regions.

The movement of ocean water is continuous which can be up three types such as Waves, Tides, Currents

The streams of water which flow continuously on the ocean surface in specific directions are called ocean currents, it affect the temperature of ocean water as Warm ocean currents increase the temperature whereas Cold ocean currents decrease the temperature.

The magnitude of the ocean currents is about  few centimeters per second to as much as 4 metres per second and the intensity of the ocean currents generally decreases with increasing depth.

There are two types of ocean currents such as Warm Ocean Currents

and Cold Ocean Currents

For more details ocean current,  visit

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#SPJ2

Answer:

Answer is explained in the explanation section below.

Explanation:

This question is very basic and easy. The answer to this question is:

Answer: If both batteries are connected we would get less amount of charge as compared to connected a single battery.

Reasoning:

If both batteries are connected in a manner of positive terminal to positive terminal and negative terminal to negative terminal then a capacitor is added to charge it from the batteries then, total electromotive force (emf) would decrease.

As a result, the capacitor added would get less amount of charge stored. But capacitor added will get more amount of charge stored when a single battery is connected.

Complete question is;

A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid. If the diameter of the rod is doubled, by how much would the rate of heat removal change?

Answer:

182.84 %

Explanation:

Formula for rate of heat transfer of an infinite log fin is given as;

q_f1 = (π/2) × (hk)^(½)) × D^(3/2)) × (T_b - T_∞)

Where D is diameter.

Now, if the diameter of the rod is doubled, it means Diameter is now 2D.

Thus;

q_f2 = (π/2) × (hk)^(½)) × (2D^(3/2)) × (T_b - T_∞)

To find how much the rate of heat removal will change, we will calculate as follows;

((q_f2/q_f1) - 1) × 100

Plugging in the relevant expressions, we have;

([[(π/2) × (hk)^(½)) × ((2D)^(3/2)) × (T_b - T_∞)]/[(π/2) × (hk)^(½)) × (D^(3/2)) × (T_b - T_∞)]] - 1) × 100

Upon simplifying, we have;

(((2D)^(3/2))/(D^(3/2)) - 1) × 100

((2^(3/2)) - 1) × 100

This gives;

182.84 %

Answer:

Im 99.99999% sure its c

Explanation:

i cant see the pictures too well

Answer:

0.2 – 4.6 seconds   increasing speed in positive direction

4.6 - 7.8 seconds   decelerating speed in a positive direction

8 - 17.2 seconds  accelerating speed in a negative direction

Explanation:

**Plato** **Edmentum**n~ this question is pretty open ended, so its hard to get it wrong honestly, good luck <3 ~

Answer:

0.2 – 4.6 seconds   increasing speed in positive direction

4.6 - 7.8 seconds   decelerating speed in a positive direction

8 - 17.2 seconds  accelerating speed in a negative direction

Explanation:

Answer:

forcing in act

Explanation:

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

[tex]V_{tot} = V_{q1} + V_{q2}[/tex]

[tex]V_q = \dfrac{k \cdot q}{r}[/tex]

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

[tex]V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0[/tex]

The electric field at the point exactly midway between the plates, [tex]V_{tot}[/tex] = 0

3) The electric field, 'E', between plates is given as follows;

[tex]E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C[/tex]

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, [tex]F_e[/tex] = E × e

∴ [tex]F_e[/tex] = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, [tex]F_e[/tex] ≈ 1.807 × 10⁻⁷ N

Answer:

in oil film        λ = 303.57 10⁻⁹ m

in the water film    λ = 319.55 10⁻⁹ m

Explanation:

When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,

when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship

             v = λ f

in the void we have

             c = λ₀ f

we divide the two expression

            c / v = λ₀ / λ

the refractive index is

              n = c / v

              n = λ₀ /λ

              λ = λ₀ / n

let's calculate

in oil film

            λ = 425 10⁻⁹ / 1.40

            λ = 303.57 10⁻⁹ m

in the water film

            λ = 425 10⁻⁹ / 1.33

            λ = 319.55 10⁻⁹

those wavelengths are in the ultraviolet

An action which would help in preventing and coping with heat-related conditions is: A. Drinking water.

Heat can be defined as a form of energy that is transferred from a physical object (body) to another, as a result of a difference in temperature. Also, heat is a condition of weather that is generally characterized by a high degree of temperature.

This ultimately implies that, heat is most likely to cause dehydration and high body temperature.

In order to prevent and cope with heat-related conditions, you should ensure that you drink water at regular intervals for hydration.

Read more on heat here: brainly.com/question/12072129

Answer:

326,000

Explanation:

One acre-foot equals about 326,000 gallons, or enough water to cover an acre of land, about the size of a football field, one foot deep. An average California household uses between one-half and one acre-foot of water per year for indoor and outdoor use.

Answer:

20 Hz

15.8 times

Explanation:

A

Although the range of frequency for any human's ear is usually said to be between 20 Hz and 20 kHz. And since the question asked for the least intense frequency, that has to be 20 Hz. Essentially the frequency most people perceive the least intense sound is 20 Hz.

B

A 100-Hz sound must be 10^1.2 times or 15.8 times more intense compared to a 1000-Hz sound to be perceived as equal to 60 phons of loudness

Answer:

h over 2 dg

Explanation:

brainliest!!!!!!!

Answer:

a periscope use total internal reflection to allow us to see things

the reflection happens at 45°

Explanation:

Answer:

You need to know the accuracy to which you can read the ruler:

Suppose that you can read the read the ruler to the nearest milimeter

A = L * W     your calculated area of the rectangle

A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA

Or ΔA =  L ΔW + W ΔL

Where we have subtracted A = L * W and the term ΔL * ΔA is very small

So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2

Then you report A = 10 cm^2 +- .7 cm^2    including the - sign for completeness

Explanation:

this is the answer for your question. if you have any doubt.

you can send your doubt to:6369514784(what's app)

Answer:

D. Weight

Explanation:

Hope that helps:)

Answer:

C= 50km/hr west

Explanation:

150/3= 50

Because it asks for velocity, make sure to include the direction as well.