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Ghella [55]
2 years ago
11

An object attached to one end of a spring makes 20 vibrations in 10 seconds. Its angular frequency is: 0. 79 rad/s 1. 57 rad/s 2

. 0 rad/s 6. 3 rad/s 12. 6 rad/s
Physics
1 answer:
morpeh [17]2 years ago
5 0

Angular frequency in radian per second for 20 vibrations in 10 seconds is 12.6 rad/s

<h3>What is Angular frequency?</h3>

Angular frequency is the number of vibrations in radian per second.

The total number of vibrations n is 20 and the time taken for these vibrations is 10 s

The frequency of the vibrations will be

f = 20 / 10 = 2 Hz

Angular frequency is related to the frequency as

ω = 2πf

ω=2π × 2

ω = 12.6 rad/s

Thus, the angular frequency is 12.6 rad/s.

Learn more about Angular frequency.

brainly.com/question/14244057

#SPJ

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What is the mass of a stone moving at a speed of 15 meters/second and having a momentum of 7.5 kilogram - meters / seconds
Elanso [62]

Answer:

0.5 kg

Explanation:

The momentum of an object is defined as

p = mv

where

m is the mass

v is the velocity

In this problem we have,

v = 15 m/s is the velocity of the stone

p = 7.5 kg m/s is the momentum

Solving for m, we can find the mass of the stone:

m=\frac{p}{v}=\frac{7.5 kg m/s}{15 m/s}=0.5 kg

4 0
2 years ago
Read 2 more answers
Boron coated with SiC (or Borsic) reinforced aluminum containing aligned 20 vol% fibers is an important high-temperature, lightw
Scilla [17]

Answer:

Option C is correct.

Modulus of elasticity of the composite perpendicular to the fibers = (12 × 10⁶) psi

Explanation:

For combination of materials, the properties (especially physical properties) of the resulting composite is a sum of the fractional contribution of each material thay makes up the composite.

In this composite,

The fibres = 20 vol%

Aluminium = 80 vol%

Modulus of elasticity of the composite

= [0.2 × E(fibres)] + [0.8 × E(Al)]

Modulus of elasticity of the fibers = E(fibres) = (55 × 10⁶) psi. =

Modulus of elasticity of aluminum = E(Al) = (10 × 10⁶) psi.

But modulus of elasticity of the composite perpendicular to the fibers is given in the expression.

[1 ÷ E(perpendicular)]

= [0.2 ÷ E(fibres)] + [0.8 ÷ E(Al)]

[1 ÷ E(perpendicular)]

= [0.2 ÷ (55 × 10⁶)] + [0.8 ÷ (10 × 10⁶)]

= (3.636 × 10⁻⁹) + (8.00 × 10⁻⁸)

= (8.3636 × 10⁻⁸)

E(perpendicular) = 1 ÷ (8.3636 × 10⁻⁸)

= 11,961,722.5 psi = (11.96 × 10⁶) psi

= (12 × 10⁶) psi

Hope this Helps!!!

6 0
2 years ago
A surveyor is using a magnetic compass 5.6 m below a power line in which there is a steady current of 140 A. (a) What is the mag
ArbitrLikvidat [17]

Answer:

(a) B = 5.6 micro Tesla

Explanation:

Current in the wire, i = 140 A

distance, r = 5 m

The formula for the magnetic field at a distance r due to the current carrying wire

B=\frac{\mu _{0}}{4\pi }\times \frac{2i}{r}

B=10^{-7}\times \frac{2\times140}{5}

B = 5.6 x 10^-6 Tesla

B = 5.6 micro Tesla

(b) As the magnetic field of earth at this site is 20 micro tesla so the magnetic field due to current carrying wire interfere the magnetic compass.

4 0
2 years ago
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

5 0
2 years ago
What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg
scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

F_c = F_g

\frac{mv^2}{r} = \frac{GmM}{r^2}

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

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

At the same time we know that the period is equivalent in terms of the linear velocity to,

T = \frac{2\pi}{\frac{v}{r}}

v = \frac{2\pi r}{T}

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

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

( \frac{2\pi r}{T})^2 =  \frac{GM}{r}

T = \sqrt{\frac{4\pi^2 r^3}{GM}}

Replacing we have,

T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}

T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})

T= 1.74hour

Therefore the correct answer is C.

7 0
3 years ago
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