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Lunna [17]
3 years ago
6

Microwave ovens have a plate that rotates at a rate of about 7.0 rev/min. What is this in revolutions per second?

Physics
1 answer:
Lera25 [3.4K]3 years ago
5 0

Answer:

The value is  w =  0.1167 \ rev/second

Explanation:

From the question we are told that

    The rate at which the plate rotates is  w =7.0 \ rev/min

Generally the revolution per second is mathematically represented as

       w =  \frac{7.0}{60}

=>    w =  0.1167 \ rev/second

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The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is _____________.
RSB [31]

Answer: the correct option is B ( on the half of the road that the vehicle is traveling).

Explanation: according to Georgia Code About Pedestrians; The driver must stop and remain stopped to let a pedestrian cross at a crosswalk when the pedestrian is

on the half of the road that the vehicle is traveling.

4 0
3 years ago
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A technician at a semiconductor facility is using an oscilloscope to measure the AC voltage across a resistor in a circuit. The
d1i1m1o1n [39]

Answer:

The value to be reported is 5.48V

Explanation:

The RMS (root mean square) is defined as the value of voltage that will produce the same heating effect, or power dissipation, in circuit, as this AC voltage.

The RMS voltage is also called effective voltage because it is just as effective as DC voltage in providing power to an element.

It is expressed as V_{rms} = \frac{V_{m} }{\sqrt{2} }

where Vm is the maximum or peak value of the voltage

In calculating the RMS of the voltage , we simply divide the peak voltage by square root of 2 (√2)

V_{rms} = \frac{7.75}{\sqrt{2} }

= \frac{7.75}{1.414}

= 5.48 V

6 0
3 years ago
A bird lands on a bare copper wire carrying a current of 51
nirvana33 [79]
The resistance of the piece of wire is
R= \frac{\rho L}{A}
where
\rho = 1.68 \cdot 10^{-8}\Omega m is the resistivity of the copper
L=5.1 cm=0.051 m is the length of the piece of wire
A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2 is the cross sectional area of the wire
By substituting these values, we find the value of R:
R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V
7 0
3 years ago
Item 4 Which conditions produce the smallest and largest ocean waves? Choose the two correct answers.
Darya [45]

Answer:

strong winds that blow for a long time over a great distance

weak winds that blow for short periods of time with a short fetch

Explanation:

When the winds are weak and blow for short periods, we experience the smallest ocean waves but when there are strong winds over a longer duration, the largest ocean waves are seen. Therefore, the conditions to produce the smallest and largest ocean waves are strong winds that blow for a long time over a great distance and weak winds that blow for short periods of time with a short fetch.

5 0
3 years ago
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A rifle that shoots bullets at 477 m/s is to be aimed at a target 45.5 m away. If the center of the target is level with the rif
Free_Kalibri [48]

Answer:

The rifle barrel must be pointed at a height of 4.45cm above the target so that the bullet hits dead center.

Explanation:

First, we need to sketch the situation so we can have a better idea of what the problem looks like (Refer to uploaded picture).

So as you may see in the drawing, when pointing the rifle to the target, we can see it as a triangle, but in reality, the bullet will have a parabolic trajectory. Both points of view will help us determine what the height must be. In order to find it, we need to first determine at what angle the bullet should be shot. In order to do so we can use the range formula, which looks like this:

R=\frac{v^{2}sin(2\theta)}{g}

Where R is the range of the bullet (this is how far it goes before it has the

same height it was shot from), v is the original speed of the bullet, θ is the angle at which the bullet is shot and g is the acceleration of gravity.

We can solve this equation for theta, so we get:

gR=v^{2}sin(2\theta)

\frac{gR}{v^{2}}=sin(2\theta)

sin^{-1}(\frac{gR}{v^{2}})=2\theta

\theta=\frac{sin^{-1}(\frac{gR}{v^{2}})}{2}

so now we can substitute the given data:

\theta=\frac{sin^{-1}(\frac{(9.8m/s^{2})(45.5m)}{(477m/s)^{2}})}{2}

so we get:

θ=0.05614°

once we get the angle, we can look at the triangle diagram. From the drawing we can see that we can use the tan function to find the height:

tan \theta = \frac{h}{45.5m}

so we can solve this for h, so we get:

h=45.5m*tan(0.05614^{o})

which yields:

h=0.0445m

or

h=4.45cm

5 0
3 years ago
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