Match the example to the type of equipment category it fits best in.

A secret agent is locked in a room. He pushes against the door but cannot open it. Finally, he falls to the floor exhausted. Has

Vitek1552 [10]

Answer: There is not work done at the door because the door did not move.

Explanation: Work is defined as the movement done by a force.

So if you move to apply a force F in an object and you move it a distance D, the work applied on the object is

W = F*D

In this case, the secret agent pushes against the door, so there is a force, but the agent does not move the door, so D = 0, so there is no motion of the door, which implies that there is no work done at the door.

8 0

3 years ago

Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim

blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

$x=x_{0}+V t \\ x=\frac{1}{2}+V t$

The distance between the police car and the intersection is,

$y=y_{0}+V t$

$y=\frac{1}{2}-40 t$

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

$z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })$

$z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)$

The rate of change is,

$2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)$

$2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots$

Now finding $z$ when $t=0,$ from (1) we have

$z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}$

$z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071$

The officer's radar gun indicates 25 mph pointed at the other car then, $\frac{d z}{d t}=25$ when $t=0,$ from

From (2) we get

$2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)$

$2(0.7071)(25)=V+2 V^{2}(0)-40$

$35.36=V-40$

$V=35.36+40=75.36$

Hence the speed of the car is $75.36 mph$

7 0

3 years ago

Please answer as fast as possible

aleksley [76]

Electric pumps are not useful they clog

6 0

3 years ago

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Which statement BEST describes how the body moves?

Alex73 [517]

<h3><u>Answer;</u></h3>

A. Skeletal muscles, which are made of fibers, nerves, and blood vessels, contract in order to make the body move.

<h3><u>Explanation</u>;</h3>

<em><u>Muscular tissues functions in movement and locomotion through its direct connection with the skeletal system. </u></em>
<em><u>Skeletal muscles are responsible for moving the body. The skeletal muscle contractions pull on tendons, which as attached to bones. When the skeletal muscle contraction causes the muscle to shorten, the bone and thus the body par will move. </u></em>
<em><u>Skeletal muscles also provide structural support for the entire body. </u></em>

7 0

3 years ago

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A runner moves 2.88 m/s north. She accelerates at 0.350 m/s^2 at a -52.0 angle. At the point in the motion where she is running

MrRissso [65]

The runner has initial velocity vector

$\vec v_0=\left(2.88\dfrac{\rm m}{\rm s}\right)\,\vec\jmath$

and acceleration vector

$\vec a=\left(0.350\dfrac{\rm m}{\mathrm s^2}\right)(\cos(-52.0^\circ)\,\vec\imath+\sin(-52.0^\circ)\,\vec\jmath)$

so that her velocity at time $t$ is

$\vec v=\vec v_0+\vec at$

She runs directly east when the vertical component of $\vec v$ is 0:

$2.88\dfrac{\rm m}{\rm s}+\left(0.350\,\dfrac{\rm m}{\mathrm s^2}\right)\sin(-52.0^\circ)\,t=0\implies t=10.4\,\rm s$

It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time $t$ would be

$\vec x=\vec v_0t+\dfrac12\vec at^2$

so that after 10.4 s, her position would be

$\vec x=(10.1\,\mathrm m)\,\vec\imath+(17.2\,\mathrm m)\,\vec\jmath$

which is 19.9 m away from her starting position.

8 0

3 years ago

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