<span>The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever.Through this arrangement, a small weight can balance a massive object. If x=450 mm,determine the required mass of the counterweight S required to balance a 90-kg load, l.</span>
We know that the electric field is equal to 1 E 6 V/m
The distance between the thundercloud and the ground is 1.6km = 1600m
Electric field = Voltage/distance
This means that the breakdown voltage must be equal to
V = 1 E 6 V/m * 1600m = 1.6 E 9 V = 1.6 GV
Answer:the Forces cancel out each other
Explanation:the forces cancel out each other
Answer:
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Explanation:
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