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Elenna [48]
4 years ago
14

A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m twice every second, what is the tension i

n the string
Physics
1 answer:
Svetradugi [14.3K]4 years ago
3 0

Answer:

T = 19.75 N

Explanation:

given,

mass of ball = 0.25 Kg

radius = 0.5 m

frequency = 2 s⁻¹

tension in the string = ?

angular velocity

ω = 2 π f

ω = 2 π x 2

ω = 12.57 rad/s

tension on the string is equal to the centripetal force

T = m ω² r

T = 0.25 x 12.57² x 0.5

T = 19.75 N

Tension in the string is equal to T = 19.75 N

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A fixed 15.3-cm-diameter wire coil is perpendicular to a magnetic field 0.77 T pointing up. In 0.20 s , the field is changed to
maksim [4K]

Answer:

Induced emf will be 0.468 volt

Explanation:

We have given diameter of wire d = 15.3 cm

So radius r=\frac{d}{2}=\frac{15.3}{2}=7.65cm

So area A=\pi r^2=3.14\times (7.65)^2=183.76\times 10^{-4}m^2

Change in magnetic field dB = 0.26 - 0.77 = -0.51 T

Time for change in magnetic field dt = 0.26 sec

We know that emf is given by e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}=-183.76\times 10^{-4}\times \frac{-0.51}{0.2}=0.0468volt

5 0
3 years ago
A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the
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Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating  system = 6.74 J

We know that energy is given by

(a) Ke=\frac{1}{2}mv_{max}^2

6.74=\frac{1}{2}\times 0.267\times v_{max}^2

v_{max}=7.1m/sec

(b) Now \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec

We know that \omega =\sqrt{\frac{k}{m}}

35.68=\sqrt{\frac{k}{0.267}}

k=339.9N/m

(c) We know that energy is given by

E=\frac{1}{2}KA^2

6.74=\frac{1}{2}\times 339.9\times A^2

A=19.91cm

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3 years ago
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it can make it stronger!

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