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Answer:
12.7m/s
Explanation:
Given parameters:
Mass of diver = 77kg
Height of jump = 8.18m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we apply the motion equation below:
v² = u² + 2gH
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
H is the height
Now insert the parameters and solve;
v² = 0² + 2 x 9.8 x 8.18
v = 12.7m/s
Answer:
A: The acceleration is 7.7 m/s up the inclined plane.
B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane
Explanation:
Let us work with variables and set

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.
Part A:
From the free body diagram we see that the total force along the x-axis is:

Now the force of friction is
where
is the normal force and from the diagram it is 
Thus
Therefore,

Substituting the value for
we get:

Now acceleration is simply

The negative sign indicates that the acceleration is directed up the incline.
Part B:

Which can be rearranged to solve for t:

Substitute the value of
and
and we get:
which is our answer.
Notice that in using the formula to calculate time we used the positive value of
, because for this formula absolute value is needed.
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =
Class 1 lever
Explanation:
In a class 1 lever, the fulcrum is placed between the effort and the load. This lever systems is the most common.
- The effort is the force input and the load is the force output
- The fulcrum is a hinge between the load and effort.
- Movement of the effort and load are in opposite directions.
- There are other classes of lever like the class 2 and 3.
- They all have different load, fulcrum and effort configurations
learn more:
Load related problems brainly.com/question/9202964
Torque brainly.com/question/5352966
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