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3 years ago

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If the Moon revolves around the Earth in a circular orbit with a period of 27 days, 7 hours, and 43 minutes, and astronomers hav

e clocked it's orbital speed at 1.03 km/s, what is the centripetal acceleration of the Moon due to its orbit around the Earth?

1 answer:

inysia [295]3 years ago

6 0

Answer:

T = (27 da * 24 hr/da + 7 hr + 43 min/60 min/hr) * 3600 sec/hr = 2.36 * 10E6 sec period of moon in sec

C = v t = 2 pi R circumference of orbit where v is orbital speed

ac (centripetal acceleration) = v^2 / R = v^2 / (v t / 2 pi) = 2 pi v / t

ac = 2 pi * 1.03 * 10E3 m/s / 2.36 * 10E6 sec = 2.74 * 10E-3 m /s^2

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A horizontal force of 350N is exerted on a 2.5 kg ball as it rotates uniformly in a horizontal circle of radius of 0.90m. Calcul

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F=mv^2/R
----> V^2=FR/m=(350x0.9)/2.5=126
----- V=11.22 m/s

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3 years ago

Marx argued that what happens when a worker is separated

kicyunya [14]

Answer:

well its simple the worker wouldn't actually be working

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3 years ago

You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo

AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

F is the applied force
Ff is the frictional force

ma = F - Ff

where;

a is acceleration of the cart
m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net) = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

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2 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

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3 years ago