When the Moon is in the position shown, how would the Moon look to an observer on the North Pole?

You leave your house at 5:02 PM and run 20 yards down the street. You don't realize that you forgot your Wallet back at home and

Sidana [21]

Answer:

The average velocity is 0.203 m/s

Explanation:

Given;

initial displacement, x₁ = 20 yards = 18.288 m

final displacement, x₂ = ¹/₃ x 18.288 = 6.096 m

change in time between 5:02 PM and 5:03 PM, Δt = 3 mins - 2 mins = 1 min = 60 s

The average velocity is given by;

V = change in displacement / change in time

V = (x₂ - x₁) / Δt

V = (18.288 - 6.096) / 60

V = 0.203 m/s

Therefore, the average velocity is 0.203 m/s

7 0

2 years ago

4.68 Steam enters a turbine in a vapor power plant operating at steady state at 560°C, 80 bar, and exits as a saturated vapor at

garik1379 [7]

Answer:

please mark me as a brainleast

Explanation:

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3 0

2 years ago

A body weighs 50 N and hangs from a spring with spring constant of 50 N/m. A dashpot is attached to the body. If the body is rai

lbvjy [14]

Answer:

a) 3.607 m

b) 1.5963 m

Explanation:

See that attached pictures for explanation.

3 0

2 years ago

10–25. The 45° strain rosette is mounted on the surface of a shell. The following readings are obtained for each gage: ε a = −20

vazorg [7]

Answer:

The answer is 380.32×10^-6

Refer below for the explanation.

Explanation:

Refer to the picture for brief explanation.

7 0

2 years ago

An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e

Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in $2.25 * 10^4 BTU$ . It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - $1 BTU = 778.17 lbf*ft$

$2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft$

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:

$Ep = m*g*(h_2 - h_1)\\ W = m*g$

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.

$h_2$ is the final elevation and $h_1$ is the initial elevation.

Replacing W in the Ep equation

$Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\$

Finally, the next step is to replace the variables of the problem.

$h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft$

The elevation at the high point of the road is 12186.5 in ft.

3 0

2 years ago