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Zarrin [17]
4 years ago
7

The time that is required for a vibrating object to complete one full cycle is called : Group of answer choices

Physics
1 answer:
disa [49]4 years ago
4 0

Answer:

Correct option : (c) period.                        

Explanation:

The time that is required for a vibrating object to complete one full cycle is called the time period. If f is the frequency of a wave, then the relation between the frequency and the time period is given by :

T=\dfrac{1}{f}

These are the characteristics of a wave. Some other characteristics are wavelength, amplitude, intensity etc. So, the correct option is (c) "period".

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I need help with this question. Anyone?
cluponka [151]
The closest answer i would go with is b, I'm not completely sure because I'm not sure what the question actually means because there could be any meaning 
5 0
3 years ago
Read 2 more answers
A low-luminosity star has a small and narrow ________, whereas a high-luminosity star has a large and wide one.
vova2212 [387]

A low-luminosity star has a small and narrow ​<u>habitable zone</u>, whereas a high-luminosity star has a large and wide one.

<h3>What is luminosity of a star?</h3>

The radiant power emitted by a light-emitting item over time is measured as luminosity, which is an absolute measure of radiated electromagnetic power (light).

The total quantity of electromagnetic energy released per unit of time by a star, galaxy, or other celestial object is referred to as luminosity in astronomy.

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4 0
2 years ago
Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a
seropon [69]

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, d=24.44\times 2=48.88\ m

(a) Acceleration, a=-4\ m/s^2

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -4}

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, a=-8\ m/s^2

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -8}

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

4 0
4 years ago
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it wi
Lelu [443]

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s

a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49

Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.

To learn more about oscillations Please click on the given link:

brainly.com/question/26146375

#SPJ4

This is incomplete question Complete Question is:

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?

4 0
1 year ago
A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is oriented
oee [108]

Answer:

Magnetic flux, \phi=2.22\times 10^{-3}\ Wb

Explanation:

It is given that,

Magnitude of the magnetic field, B = 0.078 T

Radius of circular loop, r = 0.1 m

The field is oriented at an angle of θ = 25° with respect to the normal to the surface. The magnetic flux through the surface is given by :

\phi=BA\ cos\theta

\phi=0.078\times \pi \times (0.1)^2\ cos(25)

\phi=0.00222\ Wb

or

\phi=2.22\times 10^{-3}\ Wb

So, the magnitude of magnetic flux through the surface is 2.22\times 10^{-3}\ Wb. Hence, this is the required solution.

4 0
4 years ago
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