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Art [367]
2 years ago
5

What is the mass of an asteroid with a speed of 200 m/s and a momentum of 2,000 kg m/s?

Physics
2 answers:
Bezzdna [24]2 years ago
4 0

Answer:10 kg

Explanation: momentum p = mv and m = p/v

Mass m = 2000 kgm/ s / 200 m/s

r-ruslan [8.4K]2 years ago
3 0

Answer:

jhnnmhunn vgu fhbiy f

Explanation:

hjbgyvfvubhjgvg

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supose you have two wires of equal length made from same material. how is it possible for the wires to have different resistance
Ivenika [448]
I'm not sure but I had this question on a benchmark I think its the density of the wire  you need to  find the density or the mass I'm not sure but i do remember this question 
6 0
3 years ago
5. You head downstream on a river in an outboard.
Elena-2011 [213]

Answer:

hope this helps you're welcome

5 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

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag
allochka39001 [22]
In order to get the propoerty of work you need to use the following formula
 <span>work = force times distance
</span>replacing data you will get:
W = (1.500) (3)
W =  4.500 NM
The answer should be in NM. So it will be 4500 NM againts the force of gravity
4 0
3 years ago
How did Aristotle's inability to detect parallax lead him to propose a geocentric model of the Solar System.
sasho [114]
He reasoned that since parallax could not be observed for celestial objects near the sun, then the earth was stationary. This erroneous assumption was because at the time he had no way of knowing that celestial objects were so far away that their parallax angles were too small to detect.
6 0
3 years ago
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