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2 years ago

15

This diagram shows a rubber bung, of mass 200 g, on the end of a length of string being swung in a horizontal circle of radius 4

0 cm. The string makes an angle of 56° with the vertical.
Calculate:
i the tension in the string
ii the angular speed of the bung
iii the time it takes to make one complete revolution

1 answer:

stiks02 [169]2 years ago

4 0

I) Tension in the spring = 2N

ii) Angular speed of the bung=5 rads-¹

iii) The time(T) it takes to make one complete revolution=1.256 seconds

<h3>Circular motion</h3>

Formula used to calculate the time(T) it takes to make one complete revolution = 2π √(l/g)

l = radius= 40cm = 0.4m

g = acceleration due to gravity = 10m/s-²

π =3.14( constant)

T = 2× 3.14 × √0.4/10

T = 6.28 ×√0.04

T = 6.28 × 0.2

T= 1.256 seconds

Formula to calculate angular speed(w)= 2π/T

where T= 1.256 secs

Therefore w = 2×3.14/1.256

w = 6.28/ 1.256

w = 5 rads-¹

Formula to calculate Tension in the spring(F)

= mrw²

where m= 200 g = 0.2kg

r = radius= 40cm = 0.4m

w = angular speed = 5 rads-¹

Therefore F = 0.2 × 0.4 × 5²

= 0.08 ×25

= 2N

Learn more about angular speed here:

brainly.com/question/6860269

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3 years ago

Read 2 more answers

In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p

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given,

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plate length = 6 cm

distance between them = 12 cm

using formula

$v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}$

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

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$v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}$

v_0 = 6496355.63 m/s

$v_0 = \dfrac{E}{B_0}$

$B_0 = \dfrac{E}{v_0}$

$B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}$

$B_0 = 1.69 \times 10^{-4}\ T$

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3 years ago