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Sphinxa [80]
4 years ago
13

The engineering quadrangle is approximately 235 meters above sea level. If the temperature is 15 degrees Celsius, is that above,

below, or equal to the standard temperature for this altitude?
Physics
1 answer:
RUDIKE [14]4 years ago
7 0

Answer:

The 15 ⁰C measured at this altitude is above the standard temperature for the altitude.

Explanation:

The standard temperature at sea level is 15 degrees Celsius. It decreases about 2 degrees C (or 3.5 degrees F) per 1,000 feet of altitude above sea level.

235 meters is equal to 771 feet.

Using the formula below, we can estimate temperature loss due to this change in altitude, that is 771 feet above sea level.

temperature loss  =  (3.5 x Change in altitude)/1000ft

temperature loss  =  (3.5 x 771ft)/1000ft = 2.7⁰F, (32 -2.7 = 29.3 ⁰F)

this is equivalent to 1.5⁰C temperature loss.

Thus, the standard temperature of the engineering quadrangle at 235 meters above sea level is 13.5 ⁰C.

Therefore, the 15 ⁰C measured at this altitude is above the standard temperature for the altitude.

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svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength  is given by

           \lambda = \dfrac{2L}{n}

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

            f = n\dfrac{v}{2L}

now, wavelength calculation

      n = 1

           \lambda_1 = \dfrac{2\times 2}{1}

                    λ₁ = 4 m

      n = 2

           \lambda_2 = \dfrac{2\times 2}{2}

                   λ₂ = 2 m

      n =3

           \lambda_3 = \dfrac{2\times 2}{3}

                    λ₃ = 1.333 m

now, frequency calculation

      n = 1

            f = n\dfrac{v}{2L}

            f_1 =1\times \dfrac{50}{2\times 2}

                    f₁ = 12.5 Hz

      n = 2

            f = n\dfrac{v}{2L}

            f_2 =2\times \dfrac{50}{2\times 2}

                    f₂= 25 Hz

      n = 3

            f = n\dfrac{v}{2L}

            f_3 =3\times \dfrac{50}{2\times 2}

                    f₃ = 37.5 Hz

8 0
3 years ago
Raising 100 grams of water from 40 to 60 °C (the specific heat capacity of water is 1
faust18 [17]

Heat in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)
Heat = 100(1)(60-20)
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<span><span>
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<span><span>Hope this answers the question. Have a nice day.</span></span>

4 0
3 years ago
A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the t
aliina [53]

The question seems a bit incomplete. The question should be as follow:

A bucket of water with total mass 23 kg is attached to a rope, which in turn, is wound around a 0.050-m radius cylinder at the top of a well. A crank with a turning radius of 0.25 m is attached to the end of the cylinder. What minimum force directed perpendicular to the crank handle is required to just raise the bucket? (Assume the rope's mass is negligible, that cylinder turns on frictionless bearings, and that g= 9.8 m/s2.)

Answer:

45.08N

Explanation:

The question involves moment topic. Note that the equation of moment, M is

Moment, M = Force, F x Perpendicular distance from the turning point, d

M = F x d

Moment involves turning movement along the pivot. in this case, we have 2 pivots. The first one the attached near the rope, while the other is the crank.

To raise the bucket, minimum force required must be able to provide at least the same moment as the moment due to the bucket of water. i.e.

Moment due to bucket = moment due to force on the crank

First find the moment due to bucket,

Mb = F (weight) x d (radius of cylinder on top of well)

   =  (23 x 9.8) x 0.05

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Next, Find the moment due to force on the crank. Note that the force is the minimum required force for this equation.

Mc = F (min Force) x d (radius of crank)

     = F x  0.25  = 0.25F

Now we get a simple equation of

11.27 = 0.25F

Using Algebra, we'll get the minimum Force, F:

F = 11.27 / 0.25

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4 0
4 years ago
Read 2 more answers
A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.90 kg box that is sitting on the horizontal,
Lady_Fox [76]

Answer:

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Explanation:

The force is variable which is given by

F(x) = 18 - 0.53 x

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According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

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\int_{0}^{15} \left ( 18-0.53 x \right )dx=\frac{1}{2}\times m \left ( v^{2}-u^{2} \right )

\left ( 18x-0.265x^{2} \right )_{0}^{15}=\frac{1}{2}\times 8.9\times  \left ( v^{2}-0^{2} \right )

18 x 15 - 0.265 x 15 x 15 = 4.45 x v²

270 - 59.625 = 4.45 v²

v² = 47.275

v = 6.875 m/s

Thus, the final velocity of the box is 6.875 m/s.

4 0
3 years ago
When an object speeds up, it has ___________ acceleration.
Fittoniya [83]

Explanation:

when an object speeds up,it has positive acceleration.

hope it helps....

5 0
3 years ago
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