Answer:
P = 2439.5 W = 2.439 KW
Explanation:
First, we will find the mass of the water:
Mass = (Density)(Volume)
Mass = m = (1 kg/L)(10 L)
m = 10 kg
Now, we will find the energy required to heat the water between given temperature limits:
E = mCΔT
where,
E = energy = ?
C = specific heat capacity of water = 4182 J/kg.°C
ΔT = change in temperature = 95°C - 25°C = 70°C
Therefore,
E = (10 kg)(4182 J/kg.°C)(70°C)
E = 2.927 x 10⁶ J
Now, the power required will be:

where,
t = time = (20 min)(60 s/1 min) = 1200 s
Therefore,

<u>P = 2439.5 W = 2.439 KW</u>
I’m not sure sorry I really wish I could help
The force required to slow the truck was -5020 N
Explanation:
First of all, we find the acceleration of the truck, which is given by

where
v is the final velocity
u is the initial velocity
t is the time
For the truck in this problem,
v = 11.5 m/s
u = 21.9 m/s
t = 2.88 s
So the acceleration is

where the negative sign means that this is a deceleration.
Now we can find the force exerted on the truck, which is given by Newton's second law:

where
m = 1390 kg is the mass of the truck
is the acceleration
And substituting,

So the closest answer among the option is -5020 N.
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Answer:
A) 1000 joules
Explanation:
In general work is given by the equation:
(1)
A) With
the displacement and
the force applied, because the force and the displacement are parallel (the crate is pushed horizontally)
is simply
, and because the path is a straight line and the force is constant work is:
(2),

B) The work-energy theorem says that the total work on a body is equal to the change on kinetic energy:
(3)
The total work on the crate is the work done by the push and plus the work of the friction
(4) , as (A) because forces are parallel to the displacement
(5) and
(6), the due friction always has negative sign because is opposite to the displacement, using (6), (5) and (4) on (3):
(3)
C) The energy is lost by friction, so the amount of energy turned into heat is the work the friction does:
(3)
If a battery with a potential difference of 1.5 volts is placed across the plates, the maximum capacitor will have a charge of 36 V.
<h3>What possible variations are there in a 1.5 volt battery?</h3>
1 V is, by definition, a potential energy differential between two places equal to one joule for every coulomb of charge. Your query is resolved by that. Between the sites where that potential difference is measured, 1.5V denotes a potential energy differential of 1.5 joules per coulomb.
<h3>How do you determine the difference in potential energy?</h3>
ΔV=VB−VA=ΔPEq. By dividing the potential energy of a charge q that has been transported from point A to point B by the charge, we may define the potential difference between points A and B as VBVA. The joules per coulomb, sometimes known as volts (V) in honor of Alessandro Volta, are the units of potential difference.
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