For the answer to the question above,
<span>Q = amount of heat (kJ) </span>
<span>cp = specific heat capacity (kJ/kg.K) = 4.187 kJ/kgK </span>
<span>m = mass (kg) </span>
<span>dT = temperature difference between hot and cold side (K). Note: dt in °C = dt in Kelvin </span>
<span>Q = 100kg * (4.187 kJ/kgK) * 15 K </span>
<span>Q = 6,280.5 KJ = 6,280,500 J = 1,501,075.5 cal</span>
Answer:
at the top
Explanation:
Potential energy is the stored energy, mechanical energy,
or energy possessed by by virtue of the position of an object.an example of potential energy is the energy that a ball possesses by virtue of its sitting at the top of the stairs it being about to roll down the stairs.
Answer:
a). va=17.23
or 38.54 mph
b). v=38.54 mph and limit is 35 mph
c). Completely inelastic
d). Eka=192.967 kJ
Ekt=76.071 kJ
Explanation:

The motion is an inelastic collision so

The force of the motion is contrarest by the force of friction so

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

So the velocity final can be find using this time

a).
Replacing in the first equation the final velocity can find the initial velocity



b).

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s
so is exceeding the speed limit in about 1.58 m/s
or in miles per hour
3.5 mph
c).
The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic
d).

Answer:
Explanation:
We define the linear density of charge as:

Where L is the rod's length, in this case the semicircle's length L = πr
The potential created at the center by an differential element of charge is:

where k is the coulomb's constant
r is the distance from dq to center of the circle
Thus.

Potential at the center of the semicircle
Answer:
I = E/R e^{-t/RC}
Explanation:
In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit,
E -q / c-IR = 0
we replace the current by its expression and divide by the resistance
I = dq / dt
dq / dt = E / R -q / RC
dq / dt = (CE -q) / RC
we solve the equation
dq / (Ce-q) = -dt / RC
we integrate and evaluate for the charge between 0 and q and for the time 0 and t
ln (q-CE / -CE) = -1 /RC (t -0)
eliminate the logarithm
q - CE = CE
q = CE (1 + 1/RC e^{-t/RC} )
In general the teams measure the current therefore we take the derivative to find the current
i = CE (e^{-t/RC} / RC)
I = E/R e^{-t/RC}
This expression is the one that describes the charge of a condensate in a DC circuit