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fiasKO [112]
3 years ago
8

We are going to focus on the gravitational force between Earth and our Moon. The Earth has a mass of about 6 x 10/24 kg and the

Moon has a
mass of about 7.3 x 10/22 kg. They are also currently about 4 x 10/8 meters apart.

The gravitational constant, G, is 6.67 x 10-/11 as always.

Question 1
If you were to weight yourself in the moon and the Earth, which would have the higher weight?

Physics
1 answer:
Basile [38]3 years ago
7 0

Answer

Gravity is what holds us down on the earth's (or moon's) surface. If you were to weigh yourself on a scale on Earth and then on the moon, the weight read on the moon would be 1/6 your earth weight

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Which of the following statements about flux are true?
vekshin1
All of these statements about flux are true, except for the last one - <span>When turning a surface inside of an electric field, the maximum flux is achieved if the electric field vector and the surface vector are perpendicular. 
These vectors don't have to be perpendicular, which is why this statement is incorrect, whereas the rest of them are true. </span>
8 0
3 years ago
A concrete highway is built of slabs 12 m long .How wide should be the expansion cracks between the slabs at 15 Celsius to preve
morpeh [17]

Answer:

x=4.2\ mm

Explanation:

Given:

  • length of the concrete slab, l=12\ m
  • temperature of observation, T_o=15^{\circ}C
  • we've the coefficient of linear expansion for concrete, \alpha=10^{-5}\ ^{\circ}C^{-1}
  • lower temperature limit, T_l=-30^{\circ}C
  • upper temperature limit, T_u=50^{\circ}C

<u>Change in length due to temperature can be  given as:</u>

\Delta l=l.\alpha.(T_u-T_l)

\Delta l=12\times 10^{-5}\times (50-(-30))

\Delta l=0.0096\ m=9.6\ mm

<u>Now at temperature 15°C:</u>

\Delta l'=12\times 10^{-5}\times (15-(-30))

\Delta l'=0.0054\ m=5.4\ mm

Hence the expansion crack between the slabs at this temperature must be:

x=\Delta l-\Delta l'

x=9.6-5.4

x=4.2\ mm

3 0
3 years ago
Water flows past a flat plate that is oriented parallel to the flow with an upstream velocity of 0.4 m/s. (a) Determine the appr
Trava [24]

Answer:

1.12 m

0.08291 m

Explanation:

u = Upstream velocity = 0.4 m/s

Re = Reynold's number = 5\times 10^5 (turbulent)

\nu = Viscosity of water = 1.12\times 10^{-6}\ Pas

Here the flow is turbulent so we have the relation

Re_{xcr}=\frac{ux_{cr}}{\nu}\\\Rightarrow x_{cr}=\frac{Re_{xcr}\nu}{u}\\\Rightarrow x_{cr}=\frac{5\times 10^5\times 1.12\times 10^{-6}}{0.4}\\\Rightarrow x_{cr}=1.4\ m

The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m

Boundary layer thickness relation is given by

\delta={\frac{\nu x}{u}}^{\frac{1}{5}}\\\Rightarrow \delta={\frac{1.12\times 10^{-6}\times 1.4}{0.4}}^{\frac{1}{5}}\\\Rightarrow \delta=0.08291\ m

The boundary layer thickness is 0.08291 m

4 0
4 years ago
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her
Marysya12 [62]

Answer:

 μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

          fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ  = 0

          fr sin θ  - cos θ  (W / 2 + 0,3 W_painter) = 0

          fr = cotan θ  (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

     

X axis

        F1 - fr = 0

        F1 = fr

the friction force has the expression

       fr = μ N

Y Axis

       N - W - W_painter = 0

       N = W + W_painter

       

we substitute

      fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

     μ (W + W_painter) = cotan θ  (W / 2 + 0,3 W_painter)

      μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values ​​they give

      μ = cotan θ  (12/2 + 0.3 55) / (12 + 55)

      μ = cotan θ  (22.5 / 67)

      μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

       cotan 45 = 1 / tan 45 = 1

the result is

    μ = 0.336

5 0
4 years ago
What is number 30,31 in this picture?
Maslowich
On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C

Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)

on the DT graph im going to assume that movement B is a positive velocity with A and C being negative. 
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity. 

Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question. 
i'll skip my working out as its the same as before:

C, B, A and then D

the same idea as on Q30 applies to Q31 part b, 

D,C,B then A 
6 0
3 years ago
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