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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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A spiracle polish metallic ball is uses a diverging mirror with a focal length of -20 cm a bird 25 cm tall standing 50 cm away L

Aleksandr-060686 [28]

Answer:

1/f = 1/i + 1/o thin lens equation

1/i = 1/f - 1/o

i = o * f / ( o - f) = 50 * (-20) / (50 - (-20)) = -14.3 cm

The final image is erect and 14.3 cm behind the curved surface

M = -o / i = 14.3 / 50 = .29 magnificaton of object

S = .29 * 25 cm = 7.1 cm appearance of bird in mirror (height)

5 0

2 years ago

A 20-kg child starts at the center of a playground merry-go-round that has a radius of 1.5 mm and rotational inertia of 500kg⋅m2

Lemur [1.5K]

Answer:

w = 0.189 rad/ s

Explanation:

This exercise we work with the conservation of the moment, the system is made up of the merry-go-round and the child, for which we write the moment of two instants

Initial

L₀ = I₀ w₀

Final

$L_{f}$ = I w

L₀ = L_{f}

I₀ w₀ = I_{f} w

.w = I₀/I_{f} w₀

The initial moment of inertia is

I₀ = 500 kg. m2

The final moment of inertia

$I_{f}$ = 500 + m r²

I_{f} = 500 + 20 1.5

I_{f} = 530 kg m²

Initial angular velocity

w₀ = 0.20 rad / s

Let's calculate

w = 500/530 0.20

w = 0.189 rad / s

6 0

3 years ago

The quantum state of a particle can be specified by giving a complete set of quantum numbers (n,l, ml, ms). How many different q

Delvig [45]

Answer:

number of quantum states = 8

Explanation:

To find the total number of allowed states you take into account the following relations:

$l=n-1\\\\m_l=-l,-(l-1),...,0,,,,(l-1),l$

in this case you have:

$n=2\\\\l=0,1\\\\m_0=0\\\\m_1=-1,0,1$

furthermore, for each n,l,ml quantum state you have two additional states due to the spin of the electrons.

then, you have (n,l,ml) = (2,0,0), (2,1,-1), (2,1,0), (2,1,1) and with the spin:

number of quantum states = 2*(1+3) = 8

4 0

3 years ago

Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far doe

JulsSmile [24]

Answer:

(A) $a=2.0.37m/sec^2$

(B) s = 146.664 m

Explanation:

We have given car starts from the rest so initial velocity u = 0 m /sec

Final velocity v = 88 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $88km/hr=88\times \frac{1000}{3600}=24.444m/sec$

Time is given t = 12 sec

(A) From first equation of motion v = u+at

So $24.444=0+a\times 12$

$a=2.0.37m/sec^2$

So acceleration of the car will be $a=2.0.37m/sec^2$

(b) From third equation of motion $v^2=u^2+2as$

So $24.444^2=0^2+2\times 2.037\times s$

s = 146.664 m

Distance traveled by the car in this interval will be 146.664 m

6 0

3 years ago

What are three type of pedestrians you should pay special attention to while driving

sveticcg [70]

The answer is :

Children

Mature Pedestrians

Runners/Joggers

5 0

3 years ago

Read 2 more answers