Answer:
Explanation:
I got everything but i. Don't know why but it's eluding me. So let's do everything but that.
a. PE = mgh so
PE = (2.5)(98)(14) and
PE = 340 J
b.
so
and
KE = 250 J
c. TE = KE + PE so
TE = 340 + 250 and
TE = 590 J
d. PE at 8.7 m:
PE = (2.5)(9.8)(8.7) and
PE = 210 J
e. The KE at the same height:
TE = KE + PE and
590 = KE + 210 so
KE = 380 J
f. The velocity at that height:
and
so
v = 17 m/s
g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:
590 = KE + PE and
PE = (2.5)(9.8)(11.6) so
PE = 280 then
590 = KE + 280 so
KE = 310 then
and
so
v = 16 m/s
h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:
and
26 = 0 + 9.8t and
26 = 9.8t so the time at 26 m/s is
t = 2.7 seconds. Now we use that in the equation for displacement:
Δx =
and filling in the time the object was at 26 m/s:
Δx = 0t +
so
Δx = 36 m
i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.
Answer:
The mechanical advantage of the system is equal to 19.62
Explanation:
The ratio of the force produced by a machine to the force applied to it is called mechanical advantage. In other words it is the ratio of output force to the input force.
In this problem mass=200kg
applied force=100N
input force=100N
output force=
mechanical advantage 
It gives an idea about the efficiency of a mechanical device. It is indeed a measure of force amplification. In block and tackle system an assembly of ropes and pulleys is used to lift loads. When the moving block is supported by a greater number of rope sections the force amplification will be more.
Answer:
9.03 times 10times 23 calculate it and there ya go
Answer:
speed of water is 0.0007138m/s
Explanation:
From the law of conservation of mass
Rate of mass accumulation inside vessel = mass flow in - mass flow out
so, dm/dt = mass flow in - mass flow out
taking p as density

where,
q(in) is the volume flow rate coming in
Q = is the volume of liquid inside tank at any time
But,
dQ = Adh
where ,
A = area of liquid surface at time t
h = height from bottom at time t
A = πr²
r is the radius of liquid surface

Hence,


so, the speed of water surface at height h

where,
is 75.7 L/min = 0.0757m³/min
h = 1.5m
so,

v = 0.04283 /60
v = 0.0007138m/s
Hence, speed of water is 0.0007138m/s
Endothermic: ice melting into water, and an instant ice pack turning cold
Exothermic: fireworks exploding, and gasoline burning