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dlinn [17]
2 years ago
11

If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g

ravitational field? (Note: The radius of the moon is roughly 1080 miles)
Physics
1 answer:
yulyashka [42]2 years ago
5 0

Answer:

178200 g mile pounds

Explanation:

Work= Force * Distance= Fh

F=ma=mg where m is mass and g is acceleration due to gravity

Work= 165 pounds *g* 1080 m=  178200 g mile pounds

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Calculate the flux of the vector field F⃗ =−6i⃗ +5x2j⃗ −5k⃗ , through the square of side 8 in the plane y=1, centered on the y-a
Tasya [4]

Answer:

The flux is 682.6 Wb.

Explanation:

Given that,

Vector field F=-6i+5x^2j-5k

We need to calculate the flux

Using formula of flux

\phi=\int_{-4}^{4}\int_{-4}^{4}(F\cdot j\ dxdz)

Put the value into the formula

\phi=\int_{-4}^{4}\int_{-4}^{4}(-6i+5x^2j-5k)1\ dxdz

\phi=\int_{-4}^{4}\int_{-4}^{4}(5x^2)dxdz

\phi=2(\dfrac{x^3}{3})_{-4}^{4}\times(z)_{-4}^{4}

\phi=682.6\ Wb

Hence, The flux is 682.6 Wb.

7 0
2 years ago
Estimate the change in the equilibrium melting point of copper caused by a change in pressure of 10 kbar. The molar volume of co
mr Goodwill [35]

Answer:

The change in the equilibrium melting point is 4.162 K.

Explanation:

Given that,

Pressure = 10 kbar

Molar volume of copperV=8.0\times10^{-6}\ m^3

Volume of liquid V=7.6\times10^{-6}\ m^3

Latent heat of fusion L= 13.05 kJ

Melting point =1085°C

We need to calculate the change temperature

Using Clapeyron equation

\dfrac{\Delta P}{\Delta T}=\dfrac{\Delta H}{T\Delta V}

Put the value into the formula

\dfrac{1000\times10^{5}}{\Delta T}=\dfrac{13050}{(1085+273)\times(8.0-7.6)\times10^{-6}}

\Delta T=\dfrac{1000\times10^{-5}\times(1085+273)\times(8.0-7.6)\times10^{-6}}{13050}

\Delta T=4.162\ K

Hence, The change in the equilibrium melting point is 4.162 K.

5 0
2 years ago
A test car and its driver, with a combined mass of 600 kg, are moving along a straight,horizontal track when a malfunction cause
ANEK [815]

Answer:

The two of the following measurements, when taken together, would allow engineers to find the total mechanical energy dissipated during the skid

B. The contact area of each tire with the track.

C. The co-efficent of static friction between the tires and the track.

D. The co-efficent of static friction between the tires and the track.

Explanation:

4 0
2 years ago
A snowball is melting at a rate of 324π mm3/s. At what rate is the radius decreasing when the volume of the snowball is 972π mm3
Oduvanchick [21]

Answer:

The radius is decreasing at 4 mm/s

Explanation:

The volume of a sphere is:

V = 4/3*\pi *r^3   So, when the volume is 972π mm^3 the radius r is:

r = 9mm

Now, the change rate is given by the derivative:

dV/dt = 4/3*\pi *3*r^2*dr/dt  

Where: dV/dt = -324π mm^2/s

            r = 9mm

Solving for dr/dt:

dr/dt = -4mm/s

5 0
2 years ago
A small block of wood of inertia mb is released from rest a distance h above the ground, directly above your head. you decide to
Zinaida [17]

Since I'm assuming that its perfectly elastic, considering there's not enough information given, so I think that no energy is dissipated in the collision


hmax = h - d + { [ mpvp - mb√(2gd) ] / (mp+mb) }² / (2g)

7 0
2 years ago
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