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3 years ago

9

The current illustrated in the diagram is directed upward in a straight vertical wire. A compass is placed alongside the wire at

the X mark. Which way will the compass point?

1 answer:

Inga [223]3 years ago

6 0

<span>here we have to apply the right hand rule no, since the magnetic field is perpendicular with respect to the direction of the current</span>

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An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to

mafiozo [28]

Answer:

wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

w_f = w₀ - a t

w₀ = w_f + a t

w₀ = 109 + 4.7 1.87

w₀ = 117.8 rad / s

let's reduce to revolutions / s

w₀ = 117.8 rad / s (1 rev / 2pi rad)

w₀ = 18.75 rev / s

8 0

3 years ago

Describe how an electromagnet can be switched on and off.

Phoenix [80]

Answer:The magnetic field around an electromagnet is just the same as the one around a bar magnet. It can, however, be reversed by turning the battery around. Unlike bar magnets, which are permanent magnets, the magnetism of electromagnets can be turned on and off just by closing or opening the switch.

8 0

3 years ago

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

On a trip you travel at a constant speed of 50 kilometers per hour North for 1 hour. Then you turn West and travel at a constant

polet [3.4K]

Answer:27 km per hour West + 17 km per hour North

6 0

3 years ago

if the velocity of a body changes from 13m/s to 30m/s while undergoing constant acceleration,what's the average velocity of the

Bess [88]

   Average speed = (1/2) (beginning speed + ending speed)

   = (1/2) ( 13 m/s + 30 m/s )

   = (1/2) ( 43 m/s )

   = 21.5 m/s

5 0

3 years ago