Example of scalar: speed. Example of vector: velocity
Explanation:
In physics, there are two types of quantities:
- Scalar: a scalar quantity is a quantity having only magnitude, so it is just a number followed by a unit. Examples of scalar quantities in physics are:
Mass
Time
Speed
- Vector: a vector quantity is a quantity having both a magnitude and a direction. Examples of vector quantities in physics are:
Force
Acceleration
Velocity
The two types of quantities can be used in the same event, but in a different way. One of the most common example is the difference between speed and velocity.
In fact, let's consider an object moving in a uniform circular motion: it means that it is moving in a circle at a constant speed. The speed of the object measures only how fast the object is moving, but without telling anything about its direction of motion. The velocity, viceversa, also takes into account the direction of motion, and exactly for this reason, the velocity in a uniform circular motion is not constant, because the direction (it is a vector) is constantly changing. So, in a uniform circular motion, the speed is constant but the velocity is not.
Learn more about vectors:
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Image<span> formed by a </span>plane mirror is<span> always </span>virtual<span> (meaning that the light rays </span>do<span> not actually come from the </span>image<span>), upright, and of the same shape and size as the object it </span>is<span> reflecting.
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<h2>
Spring constant is 6.73 N/m</h2>
Explanation:
For a spring mass system we have
Mass x Acceleration due to gravity = Spring constant x Elongation
mg = kl
Mass, m= 0.454 kg
Acceleration due to gravity, g = 9.81 m/s²
Spring constant,k = ?
Elongation, l = 0.662 m
Substituting
0.454 x 9.81 = k x 0.662
k = 6.73 N/m
Spring constant is 6.73 N/m
Answer:
E = k λ₀ / x₀, the field is in thenegative direction of the x axis (-x)
Explanation:
In this problem the electric field of a line of charge is requested, the expression for the electric field is
E = k ∫ dq / r²
where k is the Coulomb constant that you are worth 9 10⁹ N m²/C², that the charge and r the distance to the point of interest, in this case it is the origin (x = 0)
let's use the definite linear density
λ₀ = dq / dx
dq = λ₀ dx
we replace and integrate
E = k λ₀ ∫ dx / x²
E = k λ₀ ( -1 / x)
we evaluate the integral from the lower limit of load x = x₀ to the upper limit x = ∞
E = - k λ₀ (1 /∞ - 1 / x₀)
E = k λ₀ / x₀
as the field is positive the direction is away from the charges, so it is in the negative direction of the x axis (-x)
