Answer:
See the answer below
Explanation:
A poker that will effectively and safely function to move pieces of coal or logs in a burning fire must be fireproof itself. Hence, to be as safe as possible, such <u>poker should be made from a material that is fireproof</u> and that does not conduct a lot of heat. Otherwise, the poker will catch fire/becomes too hot during the course of usage.
Heat required to melt 0.05 kg of aluminum is 28.7 kJ.
<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>
The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.
The formula to be used is given below:
- Heat required = mass * heat capacity * temperature change
Assuming the aluminum sheet was at room temperature initially.;
Room temperature = 25 °C
Melting point of aluminum = 660.3 °C
Temperature difference = (660.3 - 25) = 635.3 903
Heat capacity of aluminum = 903 J/kg/903
Heat required = 0.05 * 903 * 635.3
Heat required = 28.7 kJ
In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.
Learn more about heat capacity at: brainly.com/question/21406849
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Explanation:
Given that,
Electric field = 5750 N/C
Charge 
Distance = 5.50 cm
(a). When the charge is moved in the positive x- direction
We need to calculate the change in electric potential energy
Using formula of electric potential energy
Put the value into the formula
The change in electric potential energy is 
(b). When the charge is moved in the negative x- direction
We need to calculate the change in electric potential energy
Using formula of electric potential energy
Put the value into the formula
The change in electric potential energy is 
Hence, This is the required solution.
Answer:
the gauge pressure at the upper face of the block is 116 Pa
Explanation:
Given the data in the question;
A cubical block of wood, 10.0 cm on a side.
height h = 1.50 cm = ( 1.50 × ( 1 / 100 ) ) m = 0.0150 m
density ρ = 790 kg/m³
Using expression for the gauged pressure;
p-p₀ = ρgh
where, p₀ is atmospheric pressure, ρ is the density of the substance, g is acceleration due to gravity and h is the depth of the fluid.
we know that, acceleration due to gravity g = 9.8 m/s²
so we substitute
p-p₀ = 790 kg/m³gh × 9.8 m/s² × 0.0150 m
= 116.13 ≈ 116 Pa
Therefore, the gauge pressure at the upper face of the block is 116 Pa
<h2>
Answer: 277.777 m</h2>
Explanation:
The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.
In this sense, the movement equations in the Y axis are:
(1)
(2)
Where:
is the rock's final position
is the rock's initial position
is the rock's initial velocity
is the final velocity
is the time the parabolic movement lasts
is the acceleration due to gravity at the surface of the moon
As we know , equation (2) is rewritten as:
(3)
On the other hand, the maximum height is accomplished when 
:
(4)
(5)
Finding :
(6)
Substituting (6) in (3):
(7)
(8) Now we can calculate the maximum height of the rock
(9)
Finally:
