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3 years ago

which has a greater speed a heron hat travels 600m in 60 seconds or a duck that travels 60 m in 5 seconds? Explain

Physics

2 answers:

erma4kov [3.2K]3 years ago

4 0

The duck at 12m/s has a greater speed than the heron which travels at 10m/s

anastassius [24]3 years ago

4 0

Answer:

The speed of duck is greater than the speed of heron.

Explanation:

$Speed(S)=\frac{Distance(D)}{Time(T)}$

Speed of heron : S

Distance covered by herons in 60 seconds = 600 m

D = 600 m . T = 60 seconds

$S=\frac{600 m}{60 s}=10 m/s$

Speed of duck : S'

Distance covered by herons in 5 seconds = 60 m

D' = 60 m . T' = 5 seconds

$S'=\frac{60 m}{5 s}=12 m/s$

12 m/s > 10 m/s

S' > S

The speed of duck is greater than the speed of heron.

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3 years ago

James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s

liraira [26]

Answer:

Both of them reach the lake at the same time.

Explanation:

We have equation of motion s = ut + 0.5at²

Vertical motion of James : -

Initial velocity, u = 0 m/s

Acceleration, a = g

Displacement, s = h

Substituting,

s = ut + 0.5 at²

h = 0 x t + 0.5 x g x t²

$t_{James}=\sqrt{\frac{2h}{g}}$

Vertical motion of John : -

Initial velocity, u = 0 m/s

Acceleration, a = g

Displacement, s = h

Substituting,

s = ut + 0.5 at²

h = 0 x t + 0.5 x g x t²

$t_{John}=\sqrt{\frac{2h}{g}}$

So both times are same.

Both of them reach the lake at the same time.

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3 years ago

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s

riadik2000 [5.3K]

Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

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4 years ago

Except for the nodes on a standing wave, what is the frequency f of the points executing simple harmonic motion?

Katen [24]

Take into account that in a standing wave, the frequency f of the points executing simple harmonic motion, is simply a multiple of the fundamental harmonic fo, that is:

f = n·fo

where n is an integer and fo is the first harmonic or fundamental.

fo is given by the length L of a string, in the following way:

fo = v/λ = v/(L/2) = 2v/L

becasue in the fundamental harmonic, the length of th string coincides with one hal of the wavelength of the wave.

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1 year ago

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ratelena [41]

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3 years ago