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agasfer [191]
3 years ago
14

The restoring force on a pendulum bob is proportional to sin(θ), where θ is the angle which the pendulum makes with the vertical

. The pendulum performs simple harmonic motion if the approximation sin(θ) ≈ θ is satisfied. Which angle below is the largest θ can be while the approximation holds to within 0.1%?
Physics
1 answer:
anyanavicka [17]3 years ago
8 0

Answer:

The angle is 4.44º

Explanation:

If:

\frac{sin\theta -\theta}{\theta } \leq 0.1o/o\\\frac{sin\theta -\theta}{\theta }\leq 10^{-3}

According the Taylor`s series:

sin\theta =\theta -\frac{\theta ^{3}  }{3!}  ....

\frac{\theta ^{2} }{6} \leq 10^{-3} \\\theta ^{2} \leq 6x10^{-3} \\\theta \leq \sqrt{6x10^{-3} } \\\theta \leq 0.0775\\\theta = 0.0775rad=4.44

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A 5.8 kg bowling ball is rolling down the lane with a velocity of 8.1 m/s . Calculate the momentum of the bowling ball. Provide
zheka24 [161]

Answer:

46.98 kg m/s

Explanation:

P = mv

P = 5.8*8.1 = 46.98 kg m/s

8 0
2 years ago
Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p
Natali5045456 [20]

Answer:

Acceleration, a=3.6\times 10^{16}\ m/s^2

Explanation:

It is given that, two isolated protons are separated by 2 nm. The force due to charged particles is given by :

F_e=\dfrac{kq^2}{d^2}

Force due to mass of proton, F_g=ma

ma=\dfrac{kq^2}{d^2}

a=\dfrac{kq^2}{md^2}

a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.6\times 10^{-27}\times (2\times 10^{-9})^2}

a=3.6\times 10^{16}\ m/s^2

So, the acceleration of two isolated protons is 3.6\times 10^{16}\ m/s^2. Hence, this is the required solution.

5 0
3 years ago
A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
n200080 [17]

Answer:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0
3 years ago
What is most likely the amount of energy available at a trophic level of primary consumers if the amount of energy available to
docker41 [41]

Answer:

200 kilocalories

Explanation:

6 0
2 years ago
Which of the following statements about ycarrier(x,t) is correct?
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Answer: Option D : is traveling rapidly but oscillating slowly.

Explanation:

ycarrier(x,t) is traveling rapidly but oscillating slowly.

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