Answer:
We show added energy to a system as +Q or -W
Explanation:
The first law of thermodynamics states that, in an isolated system, energy can neither be created nor be destroyed;
Energy is added to the internal energy of a system as either work energy or heat energy as follows;
ΔU = Q - W
Therefore, when energy is added as heat energy to a system, we show the energy as positive Q (+Q), when energy is added to the system in the form of work, we show the energy as minus W (-W).
m= 60g = 60/1000 Kg = 0.06 Kg
v = 2cm3 = 2 * (0.01^3) m3 = 2 *10^-6 m3
Density= m/v = 6 * 10^-2 / 2 *10^-6 = 3 *10^4 Kg/m3
Let us say that Cp is the specific heat of the metal object.
Then we do a heat balance (heat lost by metal = heat gained by water):
- 19g * Cp * (22degC – 96degC) = 75g * 4.184J/g degC * (22degC
– 18degC)
<span>Cp = 0.893 J/g degC</span>
Answer:
205N
Explanation:
The net force (F) is the difference between the applied force(
) and the kinetic frictional force(
). i.e
F = - -----------------(i)
Note that;
= μmg
Where;
μ = coefficient of kinetic friction
m = mass of the body
g = acceleration due to gravity = 10m/s²
Equation (i) then becomes;
F = - μmg -------------------(ii)
<em>Given from question;</em>
m = mass of motorcycle = 150kg
μ = 0.03
= 250N
Substitute these values into equation (ii) as follows;
F = 250 - (0.03 x 150 x 10)
F = 250 - (45)
F = 205N
Therefore, the net force applied to the motorcycle is 205N
V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
GOOD LUCK AND HOPE IT HELPS U
