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tangare [24]
2 years ago
8

Kamal said the distance from the top of the balloon to the ground in the Example is Ï···· 353 ft. What mistake might Kamal have

made? 1 2
Physics
1 answer:
snow_lady [41]2 years ago
7 0

Complete question is;

Kamal said the distance from the top of the balloon to the ground in the Example image attached is √353 ft. What mistake might Kamal have made?

Answer:

the mistake Kamal made is that she probably used 17 ft as the perpendicular side of the triangle with b as the hypotenuse instead of using 17ft as the hypotenuse

Explanation:

From the image attached, we can see that the distance from the top of the balloon which is blue in color to the ground is denoted by "b".

Now the triangle is a right angle triangle with hypotenuse = 15ft + 2ft = 17 ft; the adjacent side = 8 ft, while the opposite side is "b".

Thus, we can use pythagoras theorem to solve this as;

b = √(17² - 8²)

b = √(289 - 64)

b = √225

b = 15ft

However,we are told Kamal got b as √353 ft.

From inspection of the calculations we just did, if we had used addition instead of subtraction, we would have gotten b = √353 ft.

Thus, we can under that the mistake Kamal made is that she probably used 17 ft as the perpendicular side of the triangle with b as the hypotenuse instead of using 17ft as the hypotenuse.

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Explanation:

Given that,

Electrostatic force, F=3.7\times 10^{-9}\ N

Distance, r=5\ A=5\times 10^{-10}\ m

(a) F=\dfrac{kq^2}{r^2}, q is the charge on the ion              

q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}      

q=3.2\times 10^{-19}\ C

(b) Let n is the number of electrons are missing from each ion. It can be calculated as :

n=\dfrac{q}{e}

n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}

n = 2

Hence, this is the required solution.                        

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A chemist needs to order an element that will not react with any other element. Which element should he order
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A physics professor is pushed up a ramp inclined upward at 30.0° above the horizontal as she sits in her desk chair, which slide
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Answer:

V = 3.17 m/s

Explanation:

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Length travelled L = 2.50 m

Force applied F = 600 N

Initial Speed  u = 2.00 m/s

Solution

Work = Change in kinetic energy

F_{net}d = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}\\\frac{2F_{net}d }{m} = v^{2} -u^{2}\\ v^{2} =\frac{2F_{net}d }{m} +u^{2}\\ v^{2} =\frac{2(600cos30 - 85\times 9.8 \times sin30) \times 2.5 }{85} +2.00^{2}\\ v^{2} = 10.066\\v = 3..17m/s

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Which type of employee would most likely spend the majority of work time in a vehicle? O a Corrections Officer at a jail O a Dis
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Read 2 more answers
Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

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3 years ago
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