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Anna007 [38]
2 years ago
5

An astronaut landed on a far away planet that has a sea of water. To determine the gravitational acceleration on the planet's su

rface, the astronaut lowered a pressure gauge into the sea to a depth of 28.6 m. If the gauge pressure is measured to be 2.4 atm, what is the gravitational acceleration on the planet's surface?
Physics
1 answer:
TiliK225 [7]2 years ago
3 0

The concept required to solve this problem is hydrostatic pressure. From the theory and assuming that the density of water on that planet is equal to that of the earth (1000kg / m ^ 3)we can mathematically define the pressure as

P = \rho g h

Where,

\rho = Density

h = Height

g = Gravitational acceleration

Rearranging the equation based on gravity

g = \frac{P_h}{\rho h}

The mathematical problem gives us values such as:

P = 2.4 atm (\frac{101325Pa}{1atm}) = 243180Pa

\rho = 1000kg/m^3

h = 28.6m

Replacing we have,

g = \frac{243180}{(1000)(28.6)}

g = 8.5m/s^2

Therefore the gravitational acceleration on the planet's surface is 8.5m/s^2

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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
3 years ago
A scooter is accelerated from rest at the rate of 8m/s
algol13

Explanation:

time=Distance/speed

t=32/8

t=4 seconds

8 0
3 years ago
A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te
Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

\Delta L=\alpha L_{0}\Delta T

Where:

  • L(0) is the initial length
  • ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
  • ΔL is the final length minus the initial length (L(f)-L(0))
  • α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)  

So, we have:

L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})

L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})

L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)

L_{f}=0.117\: m

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0
3 years ago
A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark
Digiron [165]

Answer:

16 cm

Explanation:

Given that,

The object begins from 0 and moves 3cm towards left side followed by 7 cm towards the right and then, 6 cm towards the left side.

Let the x-axis to be the +ve and on the right side and -ve on the left

Thus, displacement would be:

= 0 -3 + 7 -6

= -2 cm

This implies that the object displaces 2cm towards the left.

While the total distance covered by the object equal to,

= 0cm + 3cm + 7cm + 6cm

= 16 cm

Thus, <u>16 cm</u> is the total distance.

3 0
3 years ago
Blue light of wavelength λ passes through a single slit of width d and forms a diffraction pattern on a screen. If we replace th
ololo11 [35]

Answer:

We can retain the original diffraction pattern if we change the slit width to d) 2d.

Explanation:

The diffraction pattern of a single slit has a bright central maximum and dimmer maxima on either side. We will retain the original diffraction pattern on a screen if the relative spacing of the minimum or maximum of intensity remains the same when changing the wavelength and the slit width simultaneously.

Using the following parameters: <em>y</em> for the distance from the center of the bright maximum to a place of minimum intensity, <em>m</em> for the order of the minimum, <em>λ </em>for the wavelength, <em>D </em>for the distance from the slit to the screen where we see the pattern and <em>d </em>for the slit width. The distance from the center to a minimum of intensity can be calculated with:

                                                    y\approx\frac{m\lambda D}{d}

From the above expression we see that if we replace the blue light of wavelength λ by red light of wavelength 2λ in order to retain the original diffraction pattern we need to change the slit width to 2d:

<em>                                                 </em>y\approx\frac{m\lambda D}{d} =\frac{m2\lambda D}{2d}

7 0
3 years ago
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