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2 years ago

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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?

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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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

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Put the value into the formula

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

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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

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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!

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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

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Answer:

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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.

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

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3 years ago

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Answer:

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

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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> $y\approx\frac{m\lambda D}{d} =\frac{m2\lambda D}{2d}$

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