A car is traveling at a velocity of 22 m/s when the driver puts on the brakes
1 answer:
The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4
- Distance, S = 110 meters
To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;
Substituting the values into the formula, we have;
<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Answer:
0.02585 BTU
Explanation:
Given: Spring constant, k = 270 lbf/in
Initial force, f =100 lbf
Compression, x = 1 in
Work done can be calculated as follows:
The H field is in units of amps/meter. It is sometimes called the auxiliary field. It describes the strength (or intensity) of a magnetic field. The B field is the magnetic flux density. It tells us how dense the field is. If you think about a magnetic field as a collection of magnetic field lines, the B field tells us how closely they are spaced together. These lines (flux linkages) are measured in a unit called a Weber (Wb). This is the analog to the electric charge, the Coulomb. Just like electric flux density (the D field, given by D=εE) is Coulombs/m², The B field is given by Wb/m², or Tesla. The B field is defined to be μH, in a similar way the D field is defined. Thus B is material dependent. If you expose a piece of iron (large μ) to an H field, the magnetic moments (atoms) inside will align in the field and amplify it. This is why we use iron cores in electromagnets and transformers.
So if you need to measure how much flux goes through a loop, you need the flux density times the area of the loop Φ=BA. The units work out like
Φ=[Wb/m²][m²]=[Wb], which is really just the amount of flux. The H field alone can't tell you this because without μ, we don't know the "number of field" lines that were caused in the material (even in vacuum) by that H field. And the flux cares about the number of lines, not the field intensity.
I'm way into magnetic fields, my PhD research is in this area so I could go on forever. I have included a picture that also shows M, the magnetization of a material along with H and B. M is like the polarization vector, P, of dielectric materials. If you need more info let me know but I'll leave you alone for now!
So if you need to measure how much flux goes through a loop, you need the flux density times the area of the loop Φ=BA. The units work out like
Φ=[Wb/m²][m²]=[Wb], which is really just the amount of flux. The H field alone can't tell you this because without μ, we don't know the "number of field" lines that were caused in the material (even in vacuum) by that H field. And the flux cares about the number of lines, not the field intensity.
I'm way into magnetic fields, my PhD research is in this area so I could go on forever. I have included a picture that also shows M, the magnetization of a material along with H and B. M is like the polarization vector, P, of dielectric materials. If you need more info let me know but I'll leave you alone for now!
Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C