Answer:
196
Explanation:
subtract 24 from 220 to get your answer.
Power can be defined as the rate at which work is accomplished.
Option D is the correct answer.
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Power </h3>
The work done by an object in a given time interval is called the power of that object.
Suppose an external force F is applied to any object for the time interval T seconds. Due to this external force, the object will perform some amount of work for the time T seconds. This work W done by the object for the time interval T seconds is called the power of that object.
Power can be defined in mathematical term which is given below.

Thus the power can also be defined as the work done by the object per unit time interval.
Hence we can conclude that option D is the correct answer.
To know more about power, follow the link given below.
brainly.com/question/1618040.
A. 0.5kg
To get this answer you need to follow the equation of KE=0.5*mv^2
But we don't have the m part in the equation. So just plug in the numbers to see which works best, though I can tell you before we do that the answer would be a.
As you may know, gravity, is a force of 9.8 m/s. And we want to get 9.8 Joules. So if we take a half a kg stone, release it at one meter, we get half of the normal gravity pull, 4.90 Joules. That means if we take half a kg stone and drop it at a doubled height, we get 9.8 Joules.
That is also to say that if we have a 1kg stone and drop it at one meter you will get the normal pull of gravity in Joules, 9.8J.
Be careful though, this does not mean if you drop a 1kg stone and a .5 kg stone the 1kg will hit first. This simply means that the 1kg stone will have twice the Joules that the .5kg stone has.
Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J