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Ilya [14]
3 years ago
14

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

ately 57.0 m57.0 m . If the track is completely flat and the race car is traveling at a constant 24.5 m/s24.5 m/s (about 55 mph55 mph ) around the turn, what is the race car's centripetal (radial) acceleration
Physics
1 answer:
sergiy2304 [10]3 years ago
6 0

Answer:

10.53m/s²

Explanation:

Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:

a = \frac{v^2}{r}

v is the velocity of the car = 24.5m/s

r is the radius of the track = 57.0m

Substitute the given values into the formula:

a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}

Hence the centripetal acceleration of the race car is 10.53m/s²

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A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.6 A runs through the wire
lana66690 [7]

Answer:

the magnitude of the magnetic force on the wire is 0.2298 N

Explanation:

Given the data in the question;

we know that, the magnitude of magnetic force is given as;

|F_{mg}^> | = I(B^> × L^> )

given that

I = 2.6 A

B^> = 0.17

L^> = 0.52

so we substitute

|F_{mg}^> | = 2.6( 0.17i" × 0.52j" )

|F_{mg}^> | = 0.2298 N

Therefore, the magnitude of the magnetic force on the wire is 0.2298 N

4 0
3 years ago
Sandra takes a class field trip to a geology museum. She sees an interesting sample of igneous rock. She wonders if the material
Aleonysh [2.5K]

Answer:

Yes

Explanation:

It is possible for sedimentary rocks to be converted to igneous rocks. Under conditions of high temperature and pressure, sedimentary rocks can be broken down into igneous rock by melting this rock type.

When the rock is broken down, it forms melt which when cooled and solidifies will form igneous rocks.

Sedimentary rocks are formed from the breaking down of pre-existing rocks through the action of weathering, erosion and sediment transportation. Within a basin, the sediments are compacted and lithified.

When this is subjected to intense pressure and temperature, the rock hardens and might further break down to melt.

6 0
2 years ago
Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone o
Sholpan [36]

Answer:

Explanation:

We shall apply Doppler's effect of sound .

speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at  6 m/s .

apparent frequency = f_o\times\frac{V+v_o}{ V-v_s}

V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .

Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f

apparent frequency =  f\times \frac{340+6}{340-10}

= f\times \frac{346}{330}

So m = 346 , n = 330 .

8 0
3 years ago
A light bulb can be all of the following except
Dennis_Churaev [7]

Answer: A light bulb can be all of the following except option C (a consumer product if it is used to light the office of the board of directors.)

Explanation:

Products are classified as being BUSINESS or CONSUMER products according to the buyer's intended use of the product.

-Consumer products: these are sold goods that are used for personal, family, or household use. The intention of the buyer is for the products to satisfy his personal needs and desires. Example of some of the consumer products include: toothpaste, eatables and clothes.

Business products: products that are not for personal use but for the manufacturing of other goods are called business products.

Therefore a bulb is not serving as a personal use when used to light the office of the board of directors rather it's serving as a business product .

3 0
3 years ago
Bottlenose dolphins use echolocation pulses with a frequency of about 100 kHz, higher than the frequencies used by most bats. Wh
Alinara [238K]

Answer:

1. greater

2. direct

3. smaller

4. inverse

Explanation:

The speed of sound in water is greater than in air; hence for the same frequency the sound wavelength in water is <u>greater </u>than in air (for the given frequency the wavelength is in the <u>direct </u>proportion with the speed of sound).

To "see" an object via the echolocation creature needs to use sound with the wavelength <u>smaller </u>than the size of an object viewed.

That means to "see" objects of the same size dolphin and bat need to use ultrasound of the same wavelength, hence dolphin needs to use higher frequency (for the given speed of sound the wavelength is in <u>inverse </u>proportion with the frequency).

8 0
2 years ago
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