East component: 3.9 m/s
South component: 1.8 m/s
Explanation:
We have to resolve the velocity vector along the east and south axis.
Taking east as positive x-direction and south as positive y-direction, the components of the velocity are given by:
where
v = 4.3 m/s is the magnitude of the velocity
is the angle between the direction of the velocity and of the x-axis
Substituting into the equations, we find:
East component:
South component:
Learn more about vector components:
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"increments of 8" means the major divisions are 0,8,16,24 ?
<span>x axis, calculate the moment arms from 0 </span>
<span>3x4, 2x12, 1x20 </span>
<span>from an arbitrary C </span>
<span>3(c-4) + 2(c-12) + (c-20) = 0 </span>
<span>3c - 12 + 2c -24 + c - 20 = 0 </span>
<span>6c = 56 </span>
<span>c = 9.33 </span>
<span>y axis </span>
<span>3x3, 1x12, 2x20 </span>
<span>3(c-4) + 1(c-12) +2 (c-20) = 0 </span>
<span>3c - 12 + c - 12 + 2c - 40 = 0 </span>
<span>6c = 64 </span>
<span>c = 10.67 </span>
<span>so center is x = 9.33, y = 10.67 </span>
Power is defined as the rate of doing work or the work per unit of time. The first step to solve this problem is by calculating the work which can be determined by the equation:
W = Fd
where:
F = force exerted = ma
d = distance traveled
m = mass of object
a = acceleration
Acceleration is equivalent to the gravitational constant (9.81 m/s^2) if the force exerted has a vertical direction such as lifting.
W = Fd = mad = 63(9.81)(7) = 4326.21 Joules
Now that we have work, we can calculate power.
P = W/t = 4325.21 J / 5 seconds = 865.242 J/s or watts
Convert watts to horsepower (1 hp = 745.7 watts)
P = 865.242 watts (1hp/745.7 watts) = 1.16 hp
Answer:
A or B depending on your precision
Explanation:
In real life, applying to solid will shrink it depending on the amount of pressure applied, but due to the strength of solid bonds, the volume change is often negligible in practical situation and would only be accounted under astronomically high pressure phenomena (eg. at Jupiter's core or near black hole).
In high school level, and for many applications, it is entirely viable to completely neglect the change in solid volume under pressure. Thus, A is a legitimate answer. However, bear in mind that <em>in theory</em> the volume still decreases by a very slight amount. So B is correct as well, theoretically. The most correct option depends on the precision you needed.
