By how many times will the kinetic energy of a body increase if its speed is tripled? Show by calculation .
1 answer:
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a)
Here the crate is moving at constant velocity, so no acceleration:
a = 0
Let's analyze the forces acting along the horizontal and vertical direction.
- Vertical direction: the equation of the forces is
(1)
where
R is the normal reaction of the floor (upward)
is the component of the force F in the vertical direction (downward)
mg is the weight of the crate (downward)
- Horizontal direction: the equation of the forces is
(2)
where
is the horizontal component of the force F (forward)
is the force of friction (backward)
From (1) we get
And substituting into (2)
b)
In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.
The equations of the forces will be:
(1)
(2)
In this second case, we want to find the critical value of such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction,
.
Therefore, using the same procedure as before,
And solving for ,
Now we analyze the expression that we found. We notice that if the force applied F is very large, , therefore we can rewrite the expression as
So, this is the critical value of the coefficient of static friction.
Answer:
The 5-number summary is
1. Median = 0.93 W/kg
2. Lower quartile = 0.69 W/kg
3. Upper quartile = 1.16 W/kg
4. Minimum value = 0.54 W/kg
5. Maximum value = 1.42 W/kg
Explanation:
We are given the measured radiation absorption rates (in W/kg) corresponding to 11 cell phones.
1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57
What is 5-number summary?
A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.
These statistical characteristics are:
1. Median
2. Lower quartile
3. Upper quartile
4. Minimum value
5. Maximum value
1. Median:
Arrange the data in ascending order
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
(n+1)/2 gives the median value of the data set.
(11 + 1)/2 = 6th position
Therefore, 0.93 W/kg is the median of the data set.
2. Lower quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The lower quartile is the median of the lower half of the data set.
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
The median is 6/2 = 3rd position
Therefore, the lower quartile of the data set is 0.69 W/kg
3. Upper quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The upper quartile is the median of the lower half of the data set.
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The median is 6/2 = 3rd position
Therefore, the upper quartile of the data set is 1.16 W/kg
4. Minimum value:
The minimum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the minimum value of the data set is 0.54 W/kg
5. Maximum value
The maximum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the maximum value of the data set is 1.42 W/kg
The box plot is illustrated in the attached diagram.
Answer:
1.35m
Explanation:
At the highest point of the jump, the vertical speed of the skier should be 0. So the 13m/s speed is horizontal, this speed stays the same from the jumping point to the highest point. The 14m/s speed at jumping point is the combination of both vertical and horizontal speeds.
The vertical speed at the jumping point can be computed:
When the skier jumps to the its potential energy is converted to kinetic energy:
where m is the skier mass and h is the vertical distance traveled, is the vertical velocity at jumping point, and h is the highest point.
Let g = 10m/s2
We can divide both sides of the equation by m:
Answer:
The Amount of work is done on the box by the net force is 378 Joules.
Explanation:
Given:
A 42 N net force is applied to a box which slides horizontally across a floor.
Force = 42 N
Force applied for a displacement of 9.0 m
Displacement = 9.0 m
To Find:
Amount of work is done on the box by the net force = ?
i.e Work = ?
Solution:
When a force causes a object to move, work is being done on the object by the force. Work is the measure of energy transfer when a force (F) moves an object through a distance (d).
Formula is given by
Substituting the given values we get
∴ Work = 42 × 9.0
∴ Work = 378 N-m
or
∴ Work = 378 Joules
The Amount of work is done on the box by the net force is 378 Joules.