The wind-chill index is modeled by the function W = 13.12 + 0.6215T − 11.37v0.16 + 0.3965Tv0.16 where T is the temperature (°C) and v is the wind speed (km/h). When T = 12°C and v = 18 km/h, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1°C? (Round your answers to two decimal places.)

Answers

Answer 1

Answer:

This question involves the concepts of derivative, apparent temperature, actual temperature,and wind speed.

The drop in apparent temperature will be "1.25°C".

The apparent temperature (W) is given in terms of actual temperature (T) and wind speed (v) is given by the following function:

$W = 13.12 + 0.6215\ T-11.37\ v^(0.16)+0.3965\ Tv^(0.16)$

Taking the derivative with respect to actual temperature, we get:

$(dW)/(dT)=0.6215+0.3965\ v^(0.16)\n\n$

where,

dW = drop in apparent temperatures = ?

dT = drop in actual temperature = - 1°C

v = wind speed = 18 km/h

Therefore,

$dW=(-1)(0.6215-0.3965(18)^(0.16))$

dW = - 1.25°C

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Answer 2

Answer:

Answer:

Δw=1.25°C

Explanation:

Given that

$w=13.12 +0.6215 T-11.37 v^(0.16)+0.3965 T v^(0.16)$

Given that T= 12°C and v=19 km/h

Now to find the drop in the apparent temperature w

$(dw)/(dT)=0.6215 +0.3965v^(0.16)$

$(\Delta w)/(\Delta T)= 0.6215 +0.3965 v^(0.16)$

Now by putting the values v=19 km/hr and ΔT=1

$(\Delta w)/(1)=0.6215 +0.3965* 18^(0.16)$

Δw=1.25°C

So we can say that when temperature is decrease by 1°C then apparent temperature will decrease by 1.25°C at given velocity.

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A force of 240.0 N causes an object to accelerate at 3.2 m/s2. What is the mass of the object?

Answers

the mass would be 75kg

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross section as J(r) = Br, where r is in meters, J is in amperes per square meter, and B = 2.35 ✕ 105 A/m3. This function applies out to the wire's radius of 2.00 mm. How much current is contained within the width of a thin ring concentric with the wire if the ring has a radial width of 11.5 μm and is at a radial distance of 1.20 mm?

Answers

The current contained within the width of a thin ring concentric is 18.1 x 10⁻⁶A

What is Current?

This is defined as electric charges moving through an electric conductor or space.

Parameters

Current density of J(r) = Br, where B = 2.35 x 10⁵ A/m³.

I = Jₓ A

where I is current, A is area and J is current density

A= 2rΔr

where 2r = circumference, Δr = width,

Substitute the values into the equation.

I=J(2rΔr)

I=2Br^2Δr

I= 2(2.35 x 10⁵)(1.2 x 10⁻³)^2(11.5 x 10⁻⁶)

= 18.1 x 10^-6 A

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Answer:

18.1 x 10^-6 A

Explanation:

A cylindrical wire carries a current density of J(r) = Br, where B = 2.35 x 10^5 A/m^3, to find the current within a certain area we multiply the current density with the are of this area:

I = J*A

for a ring with r distance from the center and width Δr, where Δr<< r, the area is:

A= 2 $\pi$ rΔr

where 2 $\pi$ r is the circumference and Δr is the width, substitute to get:

I=J(2 $\pi$ rΔr)

I=2 $\pi$ Br^2Δr

substitute with the given values to get:

I= 2 $\pi$ (2.35 x 10^5)(1.2 x 10^-3)^2(11.5 x 10^-6)

= 18.1 x 10^-6 A

A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find the initial speed of the rock.a. 3m/s
b. 30.3 m/s
c. None of the above

Answers

Answer:

so initial speed of the rock is 30.32 m/s

correct answer is b. 30.3 m/s

Explanation:

given data

h = 15.0m

v = 25m/s

weight of the rock m = 3.00N

solution

we use here work-energy theorem that is express as here

work = change in the kinetic energy ..............................1

so it can be written as

work = force × distance ...................2

and

KE is express as

K.E = 0.5 × m × v²

and it can be written as

F × d = 0.5 × m × (vf)² - (vi)² ......................3

here

m is mass and vi and vf is initial and final velocity

F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s

so put value in equation 3 we get

m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²

solve it we get

(vi)² = 919

vi = 30.32 m/s

so initial speed of the rock is 30.32 m/s

When a body falls freely under gravity, then the work done by the gravity is ___________

Answers

Answer: positive

Explanation:

Gravity can be defined as the force with which the body is attracted towards the center of the earth, or towards any other body. If the force acting on the body is in the direction of displacement then the word done by the applicable force is positive. This causes the free fall of the ball under the influence of gravity is also positive.

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Answers

Answer:

STRING held by the man

Explanation:

Wave is defined as a disturbance that travels through a MEDIUM and transfer energy from one point to another without causing any permanent displacement of the medium itself.

The medium through which wave travels varies e.g water, string, air etc

According to the diagram, wave travels through the string held by the man. This string is referred to as the MEDIUM through which the wave moves. The wave generated produces both crest (D) and trough (C) when displaced from its initial position (A)

Rope or one could consider it to be C to B or A to D

Parallel Plates Consider a very large conducting plate at potential V0 suspended a distance d above a very large grounded plane. Find the potential between the plates. The plates are large enough so that they may be considered to be infinite. This means that one can neglect fringing fields.

Answers

Answer:

$V = (V_0x)/(d)$

Explanation:

Since the field lines are parallel and the electric field is uniform between two parallel plates, a test charge would experience the same force of attraction or repulsion no matter where it is located in the field,

I attached an image that could help to understand the representation of the field. The formula used to calculate it is given by,

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