Answer:
PE = 58.8,J
Explanation:
The potential energy at the top of ramp1 can be found by the application of the formula PE = mgh
Where PE =potential energy
m = mass of block = 2kg
g = acceleration due to gravity = 9.80m/s²
h = height of the top of the ramp from its bottom where the block is initially positioned = 3m
PE = 2×9.8×3 = 58.8J
Answer:
A. Endothermic reaction.
B. +150KJ.
C. 250KJ.
Explanation:
A. The graph represents endothermic reaction because the heat of the product is higher than the heat of the reactant.
B. Determination of the enthalpy change, ΔH for the reaction. This can be obtained as follow:
Heat of reactant (Hr) = 50KJ
Heat of product (Hp) = 200KJ
Enthalphy change (ΔH) =..?
Enthalphy change = Heat of product – Heat of reactant.
ΔH = Hp – Hr
ΔH = 200 – 50
ΔH = +150KJ
Therefore, the enthalphy change for the reaction is +150KJ
C. The activation energy for the reaction is the energy at the peak of the diagram.
From the diagram, the activation energy is 250KJ.
To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.
Since the propagation occurs in an area of spherical figure we will have to
Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that
The relation between intensity I and 
Here,
= Permeability constant
c = Speed of light
Rearranging for the Maximum Energy and substituting we have then,
Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,
Therefore the maximum value of the magnetic field is
Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.
- 1. Ball A will have the greater density
- 2. Ball C and Ball D have the same density.
- 3. Ball Q will have the greater density.
- 4. Ball X and Y will have the same density
The density of an object is given as its mass per unit volume of the object.
Mathematically;.
For Case 1:
- Va = Vb and Ma = 2Mb
- D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
- Therefore, the density of ball A,
- D(a) = 2D(b).
- Therefore, ball A has the greater density.
For Case 2:
- D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd
- Therefore, ball C and D have the same density
For Case 3:
- Vp = 2Vq and Mp = Mq
- D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
- Therefore, the density of ball P is half the density of ball Q
- Therefore, ball Q has the greater density.
For case 4:
Therefore, Ball X and Ball Y have the same density.
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