The work done on the car is -20 J.
Work done on the car is negative, meaning that the car actually does work on the external system.
<h3>Energy and law of conservation of energy</h3>
- Energy is the ability to do work
- the law of conservation of energy states that the total energy in a system is conserved
From the law of conservation of energy, the initial energy of the car before it moves down the road remains constant or unchanged.
- Initial energy = 100 J
- Initial energy = Final energy - work done on car
- Final Energy = Work done on car + initial energy
80J = Work done on car + 100 J
Work done on car = 80 - 100J
Work done on car = -20 J
Hence, the work done on the car is -20 J
Work done on car is negative.
Since work done on the car is negative, it means that the car actually does work on the external system. Hence, the decrease in the energy of the car.
Learn more about energy and work at: brainly.com/question/13387946
The properti that you didnt share
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
Change minutes to hrs, divide by 60:
30 min = .50 hrs
45 min = .75 hrs
12 min = .20 hrs
----------------
total + 1.45 hrs, total travel time
:
let a = average speed for the trip
:
Write a dist equation, dist = speed * time
:
80(.5) + 100(.20) + 40(.75) = 1.45a
40 + 20 + 30 = 1.45a
90 = 1.45a
a =
a = 62.069 km/h, for the entire trip
and
90 km is the total distance
Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V
2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V
R= 12 ohm
2) circuit Q
R = 2.3 ohm
