Answer:
The rate of change of the shadow length of a person is 2.692 ft/s
Solution:
As per the question:
Height of a person, H = 20 ft
Height of a person, h = 7 ft
Rate = 5 ft/s
Now,
From Fig.1:
b = person's distance from the lamp post
a = shadow length
Also, from the similarity of the triangles, we can write:
Differentiating the above eqn w.r.t t:
Now, we know that:
Rate = 
Thus
Answer:
3.67 N
Explanation:
From the question given above, the following data were obtained:
Charge of 1st object (q₁) = +15.5 μC
Charge of 2nd object (q₂) = –7.25 μC
Distance apart (r) = 0.525 m
Force (F) =?
Next, we shall convert micro coulomb (μC) to coulomb (C). This can be obtained as follow:
For the 1st object
1 μC = 1×10¯⁶ C
Therefore,
15.5 μC = 15.5 × 1×10¯⁶
15.5 μC = 15.5×10¯⁶ C
For the 2nd object:
1 μC = 1×10¯⁶ C
Therefore,
–7.25 μC = –7.25 × 1×10¯⁶
–7.25 μC = –7.25×10¯⁶ C
Finally, we shall determine the force. This can be obtained as follow:
Charge of 1st object (q₁) = +15.5×10¯⁶ C
Charge of 2nd object (q₂) = –7.25×10¯⁶ C
Distance apart (r) = 0.525 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 15.5×10¯⁶ × 7.25×10¯⁶ / 0.525²
F = 3.67 N
Therefore, the force on the object is 3.67 N
Answer:
The energy that the truck lose to air resistance per hour is 87.47MJ
Explanation:
To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by
Our values are:
Replacing,
We need calculate now the energy lost through a time T, then,
But we know that d is equal to
Where
v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,
(per hour)
Therefore the energy that the truck lose to air resistance per hour is 87.47MJ
Answer:
because they are same and their properties
Answer:
d. 87,500 J
e. 49,600 J
Explanation:
The total energy is the heat absorbed by the copper plus the heat absorbed by the water.
d)
E = m₁C₁ΔT + m₂C₂ΔT
E = (1 kg) (390 J/kg/°C) (10 °C) + (2 kg) (4180 J/kg/°C) (10 °C)
E = 87,500 J
e)
E = m₁C₁ΔT + m₂C₂ΔT
E = (2 kg) (390 J/kg/°C) (10 °C) + (1 kg) (4180 J/kg/°C) (10 °C)
E = 49,600 J
