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g100num [7]
3 years ago
8

A person is riding a motorized tricycle. They weigh 180kg and are moving at 3 m/s over a distance of 300 m. How much work is don

e by the person riding the tricycle?
Physics
1 answer:
agasfer [191]3 years ago
3 0

If I am to understand this question correctly this is what asks you:

If a person is riding a motorized tricycle how much work do they do?

You may ask yourself, why did I only use part of the question. Simple, the rest is not relevant to what is being asked. The weight, speed, and distance wont affect the person riding any <em><u>motorized vehicle</u></em> other than the time it takes to get from one place to another.

So to answer this question I would say:

Not much, all they really have to do is to steer and set the motorized tricycle to cruise control. Just like any rode certified vehicle.

If you have any questions about my answer please let me know and I will be happy to clarify any misunderstandings. Thanks and have a great day!

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KIM [24]

Answer rain gauge measures rain shadow units millimetres

7 0
3 years ago
Read 2 more answers
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
3 years ago
When the potential energy is high, the kinetic energy is _____.
kramer

Answer:

low then it gets high

Explanation:

none

4 0
4 years ago
A turn signal lamp must be visible in normal sunlight at a distance of at least ______ feet from the front and rear of the vehic
kaheart [24]

Answer:

500; 300 feet.

Explanation:

A turn signal lamp can be defined as an amber blinking lamp which indicates a driver's intent to change direction. It is extremely important that drivers gives a turn signal (flashing light) on the side toward which he or she is turning either left or right.

Simply stated, the turn signal lamp indicate the driver's intent to turn either leftward or rightward by displaying flashing lights to the front and rear of his or her vehicle.

A turn signal lamp must be visible in normal sunlight at a distance of at least 500 feet from the front and rear of the vehicle if the vehicle is at least 80 inches wide, and at least 300 feet from front and rear of the vehicle if the vehicle is less than 80 inches wide according to the transportation traffics code.

8 0
3 years ago
Ehbwiuvuyevuhvwihbwijnwiubbs
skad [1K]

Answer:

what this please be clear

5 0
3 years ago
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