Based on Newton's principle, whenever objects A and B interact with each other, they exert forces upon each other.
When a horse pulls on a cart, t<span>he horse exerts a force only to the cart. But that force applies only to the cart, not to the horse.
The cart in turn exerts a force on the horse. But that force applies only to the horse, not the cart also.
</span>
There are two forces resulting from this interaction - a force on the horse and a force on the cart. T<span>he net force on the cart remains as it was --- a positive force in the direction of the horse's movement. Therefore, the cart begins to accelerate and move.</span><span>
</span>
W = _|....F*dx*cos(a)........With F=force, x=distance over which force acts on object,
.......0.............................and a=angle between force and direction of travel.
Since the force is constant in this case we don't need the equation to be an integral expression, and since the force in question - the force of friction - is always precisely opposite the direction of travel (which makes (a) equal to 180 deg, and cos(a) equal to -1) the equation can be rewritted like so:
W = F*x*(-1) ............ or ............. W = -F*x
The force of friction is given by the equation: Ffriction = Fnormal*(coeff of friction)
Also, note that the total work is the sum of all 45 passes by the sandpaper. So our final equation, when Ffriction is substituted, is:
W = (-45)(Fnormal)(coeff of friction)(distance)
W = (-45)...(1.8N).........(0.92).........(0.15m)
W = ................-11.178 Joules
In physical chemistry, the terms body-centered cubic (BCC) and face-centered cubic (FCC) refer to the cubic crystal system of a solid. Each solid is made up simple building blocks called lattice units. There are different layouts of a lattice unit.
It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.
Answer:
markers are 29.76 m far apart in the laboratory
Explanation:
Given the data in the question;
speed of particle = 0.624c
lifetime = 159 ns = 1.59 × 10⁻⁷ s
we know that; c is speed of light which is equal to 3 × 10⁸ m/s
we know that
distance = vt
or s = ut
so we substitute
distance = 0.624c × 1.59 × 10⁻⁷ s
distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s
distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s
distance = 29.76 m
Therefore, markers are 29.76 m far apart in the laboratory
