Answer:
<h3>The heat capacity and the specific heat are related by C=cm or c=C/m. The mass m, specific heat c, change in temperature ΔT, and heat added (or subtracted) Q are related by the equation: Q=mcΔT. Values of specific heat are dependent on the properties and phase of a given substance.</h3>
Answer:
6.575
Explanation:
T1 = 30C = 30 + 273 = 303 K
T2 = - 10 C = - 10 + 273 = 263 K
The coefficient of performance of heat pump
k = T2 / (T1 - T2)
k = 263 / (303 - 263) = 6.575
Answer:
1. Distance travelled = 12 km.
2. Displacement = 8.6 km
Explanation:
From the question given above, the following data were obtained:
Distance 1 (d₁) = 7 km
Distance 2 (d₂) = 5 km
Total distance =?
Displacement =?
1. Determination of the distance travelled.
Distance 1 (d₁) = 7 km
Distance 2 (d₂) = 5 km
Total distance (dₜ) =?
dₜ = d₁ + d₂
dₜ = 7 + 5
dₜ = 12 km
2. Determination of the displacement.
In the attached photo, R is the displacement.
We can obtain the value of R by using the pythagoras theory as illustrated below:
R² = 7² + 5²
R² = 49 + 25
R² = 74
Take the square root of both side
R = √74
R = 8.6 km
Answer:
0.739
Explanation:
If we treat the four tire as single body then
W ( weight of the tyre ) = mass × acceleration due to gravity (g)
the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r
where v is speed 25.6 m/s and r is the radius of the circle
centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²
net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²
coefficient of static friction between the tires and the road = frictional force / force of normal
frictional force = m × net acceleration / m×g
where force of normal = weight of the body in opposite direction
coefficient of static friction = (7.2567 × m) / (9.81 × m)
coefficient of static friction = 0.739