In general, the quantity of heat energy, Q, required to raise a mass m kg of a substance with a specific heat capacity of <span>c </span>J/(kg °C), from temperature t1 °C to t2 °C is given by:
<span>Q </span>= <span>mc(t</span><span>2 </span><span>– t</span>1<span>) joules</span>
<span>So:</span>
(t2-t1) =Q / mc
<span>As we know:
Q = 500 J </span>
<span>m = 0.4 kg</span>
<span>c = 4180 J/Kg </span>°c
<span>We can take t1 to be 0</span>°c
t2 - 0 = 500 / ( 0.4 * 4180 )
t2 - 0 = 0.30°c
Answer:
The object will move to Xfinal = 7.5m
Explanation:
By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:
Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s
With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:
Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =
= 3.9m + 3.6m = 7.5m
Answer:
Kinetic energy, E = 40 Joules
Explanation:
Given that,
Mass of the object, m = 2 kg
It is released from rest from a height of 3 m above Earth's surface, ![h_1=3\ m](https://tex.z-dn.net/?f=h_1%3D3%5C%20m)
We need to find the kinetic energy the object have when it reaches a height of 1 m, ![h_2=1\ m](https://tex.z-dn.net/?f=h_2%3D1%5C%20m)
It is a case of conservation of mechanical energy. It states that the total energy of the system remains constant. The kinetic energy is given by :
![E=mg(h_1-h_2)](https://tex.z-dn.net/?f=E%3Dmg%28h_1-h_2%29)
![E=2\times 10\times (3-1)](https://tex.z-dn.net/?f=E%3D2%5Ctimes%2010%5Ctimes%20%283-1%29)
E = 40 Joules
So, the kinetic energy of the object when it reaches a height of 1 m is 40 joules. Hence, this is the required solution.
The hint is wrong.
Volume of a cylinder is π r² h .
For the cylinder in the question,
Volume = ( π ) · (1.5 cm)² · (5.3 cm)
= ( π ) · (2.25 cm²) · (5.3 cm)
= ( π ) · (11.925 cm³)
= 37.4 cm³