How much heat energy must be added to 52kg Of water at 68°F to raise the temperature to 212°F? The specific heat capacity for wa
1 answer:
Answer:
The amount of energy added to rise the temperature Q = 17413.76 KJ
Explanation:
Mass of water = 52 kg
Initial temperature = 68 °F = 20° c
Final temperature = 212 °F = 100° c
Specific heat of water
Now heat transfer Q = m × C × ( -
)
⇒ Q = 52 × 4.186 × ( 100 - 20 )
⇒ Q = 17413.76 KJ
This is the amount of energy added to rise the temperature.
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Answer:
e.)At twice the distance, the strength of the field is E/4.
Explanation:
The strength of the electric field at a certain distance from a point charge is given by:
where
k is the Coulomb's constant
Q is the charge
r is the distance from the point charge
In this problem, the distance from the point charge is doubled:
r' = 2r
So the new electric field strength is
so, at twice the distance the strength of the field is E/4.
The object's speed will not change.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.