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1 year ago

A pendulum has a mass of 3kg and is lifted to a height of .3m. What is the maximum speed of the pendulum

Physics

1 answer:

Kryger [21]1 year ago

7 0

Given data

*The given mass of the pendulum is m = 3 kg

*The given height is h = 0.3 m

The formula for the maximum speed of the pendulum is given as

$v_{\max }=\sqrt[]{2gh}$

*Here g is the acceleration due to the gravity

Substitute the values in the above expression as

$\begin{gathered} v_{\max }=\sqrt[]{2\times9.8\times0.3} \\ =2.42\text{ m/s} \end{gathered}$

Hence, the maximum speed of the pendulum is 2.42 m/s

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Radial acceleration is given by

$a_{rad}= \frac{v^2}{r}$
where

$v=r \omega$
then

$a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2$

Now

$70\omega^2=3.90 \frac{m}{s^2} \\ \\ \omega= \sqrt{ \frac{3.9}{70} }$

Using the relation

$\omega=2 \pi f$

$2 \pi f= \sqrt{ \frac{3.9}{70} }\\ \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz$

Putting into rpm

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4 years ago

All of the following elements produce magnetic fields except

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3 years ago

Read 2 more answers

Compare the catching of two different water balloons.

Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = $F_{average}$ × Δt = mΔV

∴ $F_{average}$ = m·ΔV/Δt

∴ For Case A $F_{average}$ = 149.55×8/Δt = 1196.4/Δt N

For Case B $F_{average}$ = 598.2×8/Δt = 4785.6/Δt

Where Δt is the same for Case A and Case B, $F_{average}$ for Case B >> $F_{average}$ for Case B

Therefore, Case B involves the greatest force.

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3 years ago

A transformer is designed to convert the voltage from 240 V to 12 V, and has 4000 turns on its primary coil. The turns on its se

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3 years ago

If a 20 kg green fish swimming at 2 m/s swallows a 1 kg orange fish at rest, in what direction, and how fast

krok68 [10]

Answer: 1.9 m/s

Explanation:

The question should be:

If a 20 kg green fish swimming at 2 m/s swallows a 1 kg orange fish at rest, in what direction, and how fast will the green fish swim after eating the orange fish?

Ok, here we have conservation of momentum.

At the beginning, the total momentum is equal to the sum between the momentum of the green fish and the momentum of the orange fish.

Where the momentum is written as:

P = m*v

m = mass

v = velocity.

The momentum of the green fish is:

Pg = 20kg*2m/s = 40 kg*m/s.

The momentum of the orange fish is:

Po = 1kg*0m/s = 0

The total initial momentum is:

Pi = Pg + Po = 40 kg*m/s.

After the green fish eats the orange fish, we do not have an orange fish anymore, and the mass of the green fish will be equal to it's initial mass, plus the mass of the fish that it ate, this will be:

M = 20kg + 1kg = 21kg.

Then the momentum will be:

Pf = 21kg*V

Where V is the final velocity.

For conservation of momentum, the initial momentum is equal to the final momentum, then:

Pi = Pf

40 kg*m/s = 21kg*V

(40/21) m/s = 1.9 m/s = V

The fish's final velocity is 1.9 m/s

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3 years ago