Can any one help me with 45 and 46 please
1 answer:
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Answer:
9.9 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
If the body has started from rest then the initial velocity is 0. In order to find the velocity just before hitting the water then the distance at which the downward motion stops is irrelevant.
Hence, the speed of the diver just before striking the water is 9.9 m/s
Answer:
<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>
Explanation:
<u>Physical Power </u>
It measures the amount of work W an object does in certain time t. The formula needed to compute power is
Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F
The second Newton's law gives us the net force as
being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:
Solving for a
The distance traveled in the interval is given by
Since vo=0
The force is given by
We don't know the value of m, so the force is
Computing the work done by the sprinter
The power is finally computed
During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration
The distance is
The net force is
The work done by the sprinter is now computed as
At last, the output power is
By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s
Answer:
Length = 2.32 m
Explanation:
Let the length required be 'L'.
Given:
Resistance of the resistor (R) = 3.7 Ω
Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]
Resistivity of the material of rod (ρ) =
First, let us find the area of the circular rod.
Area is given as:
Now, the resistance of the material is given by the formula:
Express this in terms of 'L'. This gives,
Now, plug in the given values and solve for length 'L'. This gives,
Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.