Answer:
Explanation:
Solution is in the picture attached
To practice Tactics Box 14.1 Identifying and analyzing simple harmonic motion. A complete description of simple harmonic motion
must take into account several physical quantities and various mathematical relations among them. This Tactics Box summarizes the essential information needed to solve oscillation problems of this type. The position of a 60 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. Determine the velocity at t=0.40s. Assume that the oscillating mass described in Part A is attached to a spring. What would the spring constant k of this spring be? What is the total energy E of the mass described in the previous parts?
1 answer:
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Answer:
Focal Length = 38.61cm, Power = 2.59 Diopter, Converging lens.
Explanation:
When an object is placed 25cm from Gbenga's eye, the glasses lens must produce an image 61cm away (Gbenga's eye near point).
An image 61cm from the eye will be (61cm - 1.6cm) from the glasses.
i.e.
and
note will be negative because the image is formed on the same side as the object.
finally,
the formula for finding the focal length is given as
The focal length is positive which indicates converging lens
power
but must be in metres
Therefore,
Answer:
The current in the rods is 171.26 A.
Explanation:
Given that,
Length of rod = 0.85 m
Mass of rod = 0.073 kg
Distance
The rods carry the same current in the same direction.
We need to calculate the current
I is the current through each of the wires then the force per unit length on each of them is
Using formula of force
Where, m = mass of rod
l = length of rod
Put the value into the formula
Hence, The current in the rods is 171.26 A.
Answer:
26.2 m/s
Explanation:
We can find the speed of the cannonball just by analyzing its vertical motion. In fact, the initial vertical velocity is
(1)
where u is the initial speed and is the angle of projection.
We can therefore use the following suvat equation for the vertical motion of the ball:
where is the vertical velocity at time t, and
is the acceleration of gravity. The time of flight is 3.78 s, so we know that the ball reaches its maximum height at half this time:
And at the maximum height, the vertical velocity is zero:
Substituting these values, we find the initial vertical velocity:
And using eq.(1) we now find the initial speed: