An example of a product made from a renewable resource is
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Due to equilibrium of moments:
1) The weight of the person hanging on the left is 250 N
2) The 400 N person is 3 m from the fulcrum
3) The weight of the board is 200 N
Explanation:
1)
To solve the problem, we use the principle of equilibrium of moments.
In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.
The moment of a force is defined as:
where
F is the magnitude of the force
d is the perpendicular distance of the force from the fulcrum
In the first diagram:
- The clockwise moment is due to the person on the right is
where is the weight of the person and
is its distance from the fulcrum
- The anticlockwise moment due to the person hanging on the left is
where is his weight and
is the distance from the fulcrum
Since the seesaw is in equilibrium,
So we can find the weight of the person on the left:
2)
Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.
- The clockwise moment due to the person on the right is
where is the weight of the person and
is its distance from the fulcrum
- The anticlockwise moment due to the person on the left is
where is his weight and
is the distance from the fulcrum.
Since the seesaw is in equilibrium,
So we can find the distance of the person on the right:
3)
As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.
- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:
where is the weight of the seesaw and
is the distance of its centre of mass from the fulcrum
- The anticlockwise moment due to the person on the left is
where is his weight and
is the distance from the fulcrum
Since the seesaw is in equilibrium,
So we can find the weight of the seesaw:
#LearnwithBrainly
We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v =
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀