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yKpoI14uk [10]
3 years ago
6

Your car's speedometer works in much the same way as its odometer, except that it converts the angular speed of the wheels to a

linear speed of the car, assuming standard-size tires and no slipping on the pavement. Part A If your car has standard 24-in.-diameter tires, how fast are your wheels turning when you are driving at a freeway speed of 35 mph
Physics
1 answer:
brilliants [131]3 years ago
3 0

Answer:

Speed will be 30810 rpm      

Explanation:

We have given diameter of the tire d = 24 inch

So radius r=\frac{d}{2}=\frac{24}{2}=12imch

We have given linear velocity v = 35 mph

We know that linear velocity is given by v=\omega r

35=\omega \times 12

\omega =\frac{35}{12}\times \frac{63360}{60}=3080rad/min

As we know that 1 mile = 63360 inch and 1 hour = 60 min

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A 0.45-m metal rod moves 0.11 m in a direction that is perpendicular to a 0.80-T magnetic field in an elapsed time of 0.036 s. A
Gwar [14]

Answer:

1.1 V

Explanation:

L = 0.45 m

d = 0.11 m

B = 0.80 T

t = 0.036 s

Let e be the emf.

e = B v L

e = 0.80 x 0.11 x 0.45 / 0.036 = 1.1 V

3 0
3 years ago
Read 2 more answers
A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

5 0
3 years ago
10 rectangular object's mass from greatest to least. You can choose books, sandwiches, phones, pictures - as long as the shape i
Greeley [361]

Answer: 1 is phone 2 is sandwich, Last is picture.

Explanation: I hoped That Helped !!

7 0
2 years ago
How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated
guapka [62]

Answer:

The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate

Explanation:

This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error

5 0
3 years ago
An object is placed perpendicular to the the principal axis of convex lens of focal length 8,the distance of the object from the
Digiron [165]

Answer:

v = 24 cm and inverted image

Explanation:

Given that,

The focal length of the object, f = +8 cm

Object distance, u = -12 cm

We need to find the position &nature of the image​. Let v be the image distance. Using lens formula to find it :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

Put all the values,

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{8}+\dfrac{1}{(-12)}\\\\v=24\ cm

So, the image distance from the lens is 24 cm.

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{24}{-12}\\\\m=-2

The negative sign of magnification shows that the formed image is inverted.

7 0
2 years ago
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