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kkurt
11 days ago
5

jenny's model train is set up on a circular track. There are six telephone poles evenly spaced around the track. It takes the en

gine of her train 10 seconds to go from the first pole to the third pole. How long would it take for the engine to go the entire distance around the track?
Physics
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As a rough approximation, the human body may be considered to be a cylinder of length L=2.0m and circumference C=0.8m. (To simpl
inna [3103]
Thermal Power is approximately 460W. According to the Stephan-Boltzmann Law Formula: P = єσT⁴A, where: P = radiation energy, σ = Stefan-Boltzmann Constant, T = absolute temperature in Kelvin, є = emissivity of the material, and A = the surface area. Given that σ = 5.67 x 10^(-8), ε = 0.6, and T = 30°C which converts to Kelvin as 303K, with the human body dimensions of 2m length and 0.8m circumference leading to an area of 1.6m², so thermal power equals 0.6 x 5.67 x 10^(-8) x 303⁴ x 1.6 = 458.8W. Rounding gives about 460W.
4 0
2 months ago
Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.
Yuliya22 [3333]

The time required for the water balloon to reach the ground is given as

h = \frac{1}{2} gt^2

Here we understand that

h = 18 m

g = 9.8 m/s^2

Now applying the earlier mentioned formula

18 = \frac{1}{2}*9.8* t^2

18 = 4.9 t^2

t = 1.92 s

Now in the same time frame, we can conclude the distance covered will be

d = v_x * t

d = 8.2 * 1.92 = 15.7 m

Thus, it will land at a distance of 15.7 m from where it started

5 0
2 months ago
A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle
Keith_Richards [3271]

To determine the average net force, we can calculate acceleration using:

x = 0.5*a*t^2

v = a*t

where x=3.6m and v=185 m/s.

Thus,

t=v/a and therefore x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a

which gives us a= (0.5*v^2)/x

Since we have the known values of v and x, we can compute a by substituting these numbers.

The average net force is then given as:

F = m*a,

with m=7.5kg.


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2 months ago
What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?
Keith_Richards [3271]
Based on my findings, within a period of 2 hours, there are certain atoms remaining. N = N0 * 2^(-t/6.020) = N = N0 * 2^-0.33223 = 0.7943 N0 Thus, the quantity of atoms that undergo disintegration is N0 - N = N0 * (1 - 0.79430) = 0.2057 N0 This must equate to 15 mCi = 15 * 3.7 * 10^7 = 5.55 * 10^8 atoms N0 = 5.55 * 10^8 / 0.2057 = 2.698 * 10^9 atoms Consequently, 2.698 * 10^9 atoms represents the value of N0.
4 0
2 months ago
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