Answer:
example two
Explanation:
They have the greatest masses and close proximety relative to the rest, (If you have two black holes each with a solar mass only 1 mile away from one another, they will be highly atracted and probly
orbit each other once a second or so. But now lets try to put the earth and moon one half mile away from each other, they orbit each other much much slower then the two black holes, its becuase the gigantic mass of the black holes overwalms the closser distance between earth and the moon
Have a great day,
enjoy life.
Answer:
an 85 kg person run to equal the kinetic energy of an 8.0 g bullet fired at 410 m/s? The speed is, it might be argued,
v=4.0m/s
Explanation:
Typically, the mathematical formula for kinetic energy is as follows:
KE=
K.E= kinetic energy
M=mass
V= speed
so,
KE=1/2*0.008*(420)*420
K.E=672.4J
Therefore,
v*v=672.4/85
v*v=15.821
v=4.0m/s
Therefore
v=4.0m/s
What is Kinetic Energy?
The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.
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Answer:
Weight. Recall that the acceleration of a free-falling object near Earth's surface is approximately g=9.80m/s2 g = 9.80 m/s 2 . The force causing this acceleration is called the weight of the object, and from Newton's second law, it has the value mg.
Explanation:
Answer:
E = 9.4 10⁶ N / C
, The field goes from the inner cylinder to the outside
Explanation:
The best way to work this problem is with Gauss's law
Ф = E. dA = qint / ε₀
We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.
The flow on the faces is zero, since the field goes in the radial direction of the cylinders.
The area of the cylinder is the length of the circle along the length of the cable
dA = 2π dr L
A = 2π r L
They indicate that the distance at which we must calculate the field is
r = 5 R₁
r = 5 1.3
r = 6.5 mm
The radius of the outer shell is
r₂ = 10 R₁
r₂ = 10 1.3
r₂ = 13 mm
r₂ > r
When comparing these two values we see that the field must be calculated between the two housings.
Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is
λ = q / L
Qint = λ L
Let's replace
E 2π r L = λ L /ε₀
E = 1 / 2piε₀ λ / r
Let's calculate
E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3
E = 9.4 10⁶ N / C
The field goes from the inner cylinder to the outside
A substance that can conduct a current in solution is salt.
