If it's a small object near another gigantic object, then we usually call that the small object's "weight".
Answer:
Option 10. 169.118 J/KgºC
Explanation:
From the question given above, the following data were obtained:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1.61 KJ
Mass of metal bar = 476 g
Specific heat capacity (C) of metal bar =?
Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:
1 kJ = 1000 J
Therefore,
1.61 KJ = 1.61 KJ × 1000 J / 1 kJ
1.61 KJ = 1610 J
Next, we shall convert 476 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
476 g = 476 g × 1 Kg / 1000 g
476 g = 0.476 Kg
Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1610 J
Mass of metal bar = 0.476 Kg
Specific heat capacity (C) of metal bar =?
Q = MCΔT
1610 = 0.476 × C × 20
1610 = 9.52 × C
Divide both side by 9.52
C = 1610 / 9.52
C = 169.118 J/KgºC
Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC
Waxing gibbous
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Answer:
819.78 m
Explanation:
<u>Given:</u>
- OA = range of initial position of the airplane from the point of observation = 375 m
- OB = range of the final position of the airplane from the point of observation = 797 m
= angle of the initial position vector from the observation point = 
= angle of the final position vector from the observation point = 
= displacement vector from initial position to the final position
A diagram has been attached with the solution in order to clearly show the position of the plane.
Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:
The magnitude of the displacement vector = 
Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.
