<span>A. </span>Let’s
say the horizontal component of the velocity is vx and the vertical is vy. <span>
Initially at t=0 (as the mug leaves the counter) the
components are v0x and v0y.
<span>v0y = 0 since the customer slides it horizontally so applied
force is in the x component only.
<span>The equations for horizontal and vertical projectile motion
are:
x = x0 + v0x t
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2 </span></span></span>
Setting the origin to be the end corner of the
counter so that x0=0 and y0=0, hence:
x = v0x t
y = - 1/2 g t^2
Given value are: x=1.50m and y=-1.15m (y is
negative since mug is going down)
<span>1.50m = v0x t
----> v0x= 1.50/t</span>
<span>-1.15m = -(1/2) (9.81) t^2 -----> t =0.4842 s</span>
Calculating for v0x:
v0x = 3.10 m/s
<span>B. </span>v0x
is constant since there are no other horizontal forces so, v0x=vx=3.10m/s
vy can be calculated from the formula:
<span>vy = v0y + at where a=-g
(negative since going down)</span>
vy = -gt = -9.81 (0.4842)
vy = -4.75 m/s
Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy
tan(90-Ø )= 3.1/4.75
Ø =
56.87˚<span> below the horizontal</span>
Lake effect is one condition
Sorry if this didn't help- I'm not quite sure what you're asking.
I can’t understand I’m sorry but if you need help with math I can help you
Answer:
Stop cheating in exam
Explanation:
Shame!!!!
I am sorry but I will have to refer you to the student conduct at UTA.
In general,
Power = (energy moved) / (time to move the energy) .
If it's mechanical power, then
Power = (work done) / (time to do the work) .
If it's electrical power, then it can be any one of these:
Power = (volts) x (amperes)
Power = (volts)² / (resistance, ohms)
Power = (amperes)² x (resistance, ohms) .
Whatever kind of energy you're dealing with, power always
turns out to be
(amount of energy produced, used, or moved)
divided by
(time taken to produce, use, or move the energy) .
