Answer:
The angle between the diagonal and edge = 55 degrees
Explanation:
We will find it by finding the angle between two vectors (a and b)
We will assume it to be a unit cube
Vector a = (1,1,1) (defines the diagonal vector)
Vector b = (1,0,0) (defines the edge vector)
cos (theta) = (a.b)/(|a|*|b|)
theta = 54.74 degrees
theta = 55 degrees (Rounded to the nearest degree)
Answer:
The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
Explanation:
Mechanical energy (Potential energy, PE) of the oscillator is calculated as;
PE = ¹/₂KA²
During the first oscillation;
PE₁ = ¹/₂KA₁²
During the second oscillation;
A₂ = A₁ - 0.0342A₁ = 0.9658A₁
PE₂ = ¹/₂KA₂²
PE₂ = ¹/₂K (0.9658A₁)²
PE₂ = (0.9658²)¹/₂KA₁²
PE₂ = (0.9328)¹/₂KA₁²
PE₂ = 0.9328PE₁
Percentage of the mechanical energy of the oscillator lost in each cycle;
Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%
Making a drawing of the system, we will have two forces which are tension and the weight of the object. Balancing the forces present, we do as follows:
T = W
W = 30 N
Therefore, weight is equal to 30 N. Hope this answers the question. Have a nice day. Feel free to ask more questions.
Answer:
where is the graph I can't see it how can I solve the problem if I don't see the graph can you show the graph please
Answer:
Explanation:
To solve this problem we have to take into account that the energy consumed per second is the power. Hence, by multipling the power and the time spent to arrive to the lab we obtain the total energy consumed.
But first we have to calculate the time
Now we use E=W*t for both times
A. Hence, by running the energy consumed is lower.
B.
E1=1008000J
E2=1392000J
C. Because the more intense exercise is made in a lower time in comparison with the less intense exercise, and higher the time, more energy is consumed.