Answer: 900
Explanation: Force equals mass x acceleration F=M×A
Answer:
You will fly forward in the bus until you hit something.
Explanation:
While standing there on the bus, you are traveling at the same speed as the bus. If the bus suddenly stops, you will still be traveling at the same speed you started with. That is until you hit something hard enough or big enough to stop you.
The correct answer is 1.07m.
The area surrounding an electric charge where its impact may be felt is known as the electric field. When another charge enters the field, the presence of an electric field may be felt. The electric field will either attract or repel the charge depending on its makeup. Any electric charge has a property known as the electric field. The charge and electrical force working in the field determine the strength or intensity of the electric field.
Here, is the charge per unit length, r is the distance from the wire, and
is the free space permittivity ε_0. Electric field due to the long straight wire is,
E= λ/2πε_0r
Rearrange the equation for r.
r=λ/2πε_0E
Substitute 2.41 N/C for E,
E=1.44×10^-10C/m
λ=8.85×10^-12C^2/Nm^2
r=(1.44×10^-10C/m)/(2(3.14)(8.85×10^-12C^2/Nm^2)(2.41N/C))
r=1.07m
At a distance of 1.07 m the magnitude of electric field is 2.41 N/C.
To learn more about electric field refer the link:
brainly.com/question/12821750
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<span><span>C. People who work hard are superior to those who are lazy.</span><span> </span></span>
Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
