Answer:
the octate rule says that electron after completing all the valency of the element doesnot goes as 9 but continues next round after completing the next octat then the first shell completes the all electron valency .thanks if you dont understand then please comment l will try to solve your confusion
Answer:
a) 1.3 rad/s
b) 0.722 s
Explanation:
Given
Initial velocity, ω = 0 rad/s
Angular acceleration of the wheel, α = 1.8 rad/s²
using equations of angular motion, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
where
θ2 - θ1 = 53.2 rad
t2 - t1 = 7s
substituting these in the equation, we have
θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²
53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²
53.2 = 7.ω(0) + 1/2 * 1.8 * 49
53.2 = 7.ω(0) + 44.1
7.ω(0) = 53.2 - 44.1
ω(0) = 9.1 / 7
ω(0) = 1.3 rad/s
Using another of the equations of angular motion, we have
ω(0) = ω(i) + α*t1
1.3 = 0 + 1.8 * t1
1.3 = 1.8 * t1
t1 = 1.3/1.8
t1 = 0.722 s
The amount of heat required is B) 150 J
Explanation:
The amount of heat energy required to increase the temperature of a substance is given by the equation:
where:
m is the mass of the substance
C is the specific heat capacity of the substance
is the change in temperature of the substance
For the sample of copper in this problem, we have:
m = 25 g (mass)
C = 0.39 J/gºC (specific heat capacity of copper)
(change in temperature)
Substituting, we find:
So, the closest answer is B) 150 J.
Learn more about specific heat capacity:
brainly.com/question/3032746
brainly.com/question/4759369
#LearnwithBrainly
Answer:
Explanation:
The mass of the block is 0.5kg
m = 0.5kg.
The spring constant is 50N/m
k =50N/m.
When the spring is stretch to 0.3m
e=0.3m
The spring oscillates from -0.3 to 0.3m
Therefore, amplitude is A=0.3m
Magnitude of acceleration and the direction of the force
The angular frequency (ω) is given as
ω = √(k/m)
ω = √(50/0.5)
ω = √100
ω = 10rad/s
The acceleration of a SHM is given as
a = -ω²A
a = -10²×0.3
a = -30m/s²
Since we need the magnitude of the acceleration,
Then, a = 30m/s²
To know the direction of net force let apply newtons second law
ΣFnet = ma
Fnet = 0.5 × -30
Fnet = -15N
Fnet = -15•i N
The net force is directed to the negative direction of the x -axis
