Answer:ok yes
Explanation:yes of course
The skater's final angular speed is equal to 12 rad/s.
When implemented to angular momentum, the regulation of conservation means that the momentum of a rotating item is no longer exchanged until some form of external torque is carried out. Torque, in this sense, can check with any outside pressure that acts upon the object for the purpose to twist or rotate.
The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.
In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.
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Answer:
7.9
Explanation:
When we put the metal piece in the liquid (which is in the graduated cylinder), how much it goes up is equal to the volume of the piece we inserted.
So now we know that the volume of that piece of unknown metal is 7mL (which is the same as 7
).
Density is
.
So the density of that piece of metal is 
Which leaves us with a final density of 7.9
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
Answer:
20 pig callers
Explanation:
Given that:
A pig caller produced intensity level of a sound = 107 dB
To find how many pig callers required to generate an intensity level of 120 dB;
we have:
120 dB - 107 dB = 13 dB
Taking the logarithm function;

where;
= initial intensity


I = 19.95
I ≅ 20 pig callers