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Kay [80]
1 year ago
11

a tire has a tread pattern with a crevice every 4.00 cm. each crevice makes a single vibration as the tire moves. what is the fr

equency (in hz) of these vibrations if the car moves at 24.2 m/s?
Physics
1 answer:
koban [17]1 year ago
5 0

the frequency (in hz) of these vibrations if the car moves at 24.2 m/s is 605 HZ .

Calculation :

frequency = \frac{number of contacts}{second}

frequency = \frac{24.2 m/s}{1}*100cm*\frac{1}{4cm}

= 605 HZ

Frequency describes the number of waves passing through a particular location in a particular time. So if the wave takes 1/2 second to travel, the frequency is 2 per second. If it takes 1/100th of an hour, the frequency is 100 per hour.

Frequency is the number of occurrences of a repeating event per unit time. ... sometimes called time-frequency for clarity,

Learn more about frequency here : brainly.com/question/254161

#SPJ4

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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
Oxygen is the most abundant gas in Earth’s atmosphere true or false
iogann1982 [59]

Answer: false

Explanation:

Nitrogen is actually the most abundant gas in the atmosphere.

The answer to the question is false. Hope this helps you!

3 0
3 years ago
Read 2 more answers
What is the distance of earths surface from the surface of sun
storchak [24]
Max: 152 million km
min 146 million km
4 0
3 years ago
Read 2 more answers
IF YOU GOOD AT SCIENCE PLEASE ANSWER THIS ASAP
xz_007 [3.2K]

Answer:

I would believe that it would be the last option

Explanation:

Physical science is a type of science that mainly focuses on natural objects that are not alive, such as minerals and rocks.

4 0
2 years ago
Read 2 more answers
A 300 kg car initially travels with a velocity of 20 m/s to the right. A net force F acts on the car for 5 s, which causes the v
hram777 [196]

Answer:

<em>600N.</em>

Explanation:

From the question, we are to calculate the net force acting on the car.

According to Newton's second law of motion:

F = ma

m is the mass of the car

a is the acceleration = change in velocity/Time

a = v-u/t

F = m(v-u)/t

v is the final velocity = 30m/s

u is the initial velocity = 20m/s

t is the time = 5secs

m = 300kg

Get the net force:

Recall that: F = m(v-u)/t

F  = 300(30-20)/5

F = 60(30-20)

F = 60(10)

<em>F = 600N</em>

<em>Hence the net force acting on the car is 600N.</em>

<em></em>

<em></em>

3 0
3 years ago
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