Answer:
v = 4.4 m / s
Explanation:
Unfortunately, the exercise scheme does not appear. Let's analyze the problem the marble leaves point A with an initial velocity, goes down and then rises to a given height where its velocity is zero, in the whole trajectory they tell us that the resistance is zero, so we can use the conservation relations of the enegy.
Starting point. Point A
Em₀ = K + U = ½ m v2 + mg y_a
point B.
Em_f = U = m g y
the energy is conserved
Em₀ = Em_f
½ m v² + mg y_a = m g y
½ m v² = m g (y -y_a)
v =
In the exercise the diagram is not seen, but the height of point A must be known, suppose that y_a = 4 m
v =
v = 4.4 m / s
Answer:
book speed is 3.99 m/s
Explanation:
given data
mass m = 490 g = 0.490 kg
compressing x = 7.10 cm = 0.0710 m
spring constant k = 1550 N/m
to find out
book speed
solution
we know energy is conserve so
we can say
loss in spring energy is equal to gain in kinetic energy
so
..................1
put here value
v = 3.99 m/s
so book speed is 3.99 m/s
Answer:
Explanation:
From the question we are told that:
Period
Trial 1
Spring constant
Period
Mass
Trial 2
Period
Generally the equation for Spring Constant is mathematically given by
\mu=\frac{4 \pi^2 M}{T^2}
Since
Therefore
Answer:
The value is
Explanation:
From the question we are told that
The diameter is
The length of the wire is
Generally the cross sectional area of the copper wire is mathematically represented as
=>
=>
Generally the resistance is mathematically represented as
Here is the resistivity of copper with the value
=>
=>