Answer:
The acceleration of the mass is 2 meters per square second.
Explanation:
By Newton's second law, we know that force (
), measured in newtons, is the product of mass (
), measured in kilograms, and net acceleration (
), measured in meters per square second. That is:
(1)
The initial force applied in the mass is:
In addition, we know that force is directly proportional to acceleration. If the smaller force is removed, then the initial force is reduced to 
of the initial force. The acceleration of the mass is:
The acceleration of the mass is 2 meters per square second.
Answer:
Explanation:
Given that,
B(t) = B0 cos(ωt) • k
Radius r = a
Inner radius r' = a/2 and resistance R.
Current in the loop as a function of time I(t) =?
Magnetic flux is given as
Φ = BA
And the Area is given as
A = πr², where r = a/2
A = πa²/4
Then,
Φ = ¼ Bπa²
Φ(t) = ¼πa²Bo•Cos(ωt)
Then, the EMF is given as
ε(t) = -dΦ/dt
ε(t) = -¼πa²Bo • -ωSin(ωt)
ε(t) = ¼ωπa²Bo•Sin(ωt)
From ohms law,
ε = iR
Then, i = ε/R
I(t) = ¼ωπa²Bo•Sin(ωt) /R
This is the current induced in the loop.
Check attachment for better understanding
Answer:
P = 33.6 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = forces [N]
m = mass = 14 [kg]
a = acceleration = 6 [m/s²]
![F = 14*6\\F = 84 [N]](https://tex.z-dn.net/?f=F%20%3D%2014%2A6%5C%5CF%20%3D%2084%20%5BN%5D)
In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.
where:
W = work [J]
F = force = 84 [N]
d = displaciment = 40 [m]
![W = 84*40\\W = 3360 [J]](https://tex.z-dn.net/?f=W%20%3D%2084%2A40%5C%5CW%20%3D%203360%20%5BJ%5D)
Finally, the power can be calculated by the relationship between the work performed in a given time interval.
where:
P = power [W]
W = work = 3360 [J]
t = time = 100 [s]
Now replacing:
![P=3360/100\\P=33.6[W]](https://tex.z-dn.net/?f=P%3D3360%2F100%5C%5CP%3D33.6%5BW%5D)
The power is given in watts
The correct answer should be C. Hydroelectric power stations can only produce enough energy for a small town as they do not produce large quantities
Hydroelectric power stations can power even large cities that have millions of people.
since both components, length and time, are measurable
<span>since Rate = length ÷ time </span>
<span>∴ rate is also measurable and ∴ quantitative.
</span>
