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Delicious77 [7]

3 years ago

5

NEED HELP FAST PLEASE

2 answers:

NNADVOKAT [17]3 years ago

6 0

I’m pretty sure it is expanding

Fittoniya [83]3 years ago

4 0

Answer:

if a star is moving towards earth it is blue shifting

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If a tank filled with water contains a block and the height of the water above point A within the block is 0.6 meter, what's the

garri49 [273]

$Given:\\\rho=1000 \frac{kg}{m^3}\\g=9.8 \frac{m}{s^2} \\h=0.6m\\\\Find:\\p=?\\\\Solution:\\\\p=\rho gh\\\\p=1000 \frac{kg}{m^3}\cdot 9.8 \frac{m}{s^2} \cdot0.6m=5880Pa=5.88kPa\\\\Correct\;is\;answer\;\;D$

6 0

3 years ago

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Why isn't the story of Israel not fully accurate?

maksim [4K]

Answer:

wait what do you mean? And why is this in physics?

Is this about the iron dome or something biblical?

Explanation:

5 0

3 years ago

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A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

7 0

3 years ago

I did questions 2 and 3 but I don't know if they are right. Someone help!

Tems11 [23]

You got it right my friend

8 0

3 years ago

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A dog, with a mass of 10.0 kg, is standing on a flatboat so that he is 22.5 m from the shore. He walks 7.8 m on the boat toward

marissa [1.9K]

Answer:16.096

Explanation:

Given

mass of dog $\left ( m_d\right )=10kg$

mass of boat $\left ( m_b\right )=46kg$

distance moved by dog relative to ground= $x_d$

distance moved by boat relative to ground= $x_b$

Distance moved by dog relative to boat=7.8m

There no net force on the system therefore centre of mass of system remains at its position

0= $m_d\times x_d+m_b\dot x_b$

0= $10\times x_d+46\dot x_b$

$x_d=-4.6x_b$

i.e. boat will move opposite to the direction of dog

Now

$|x_d|+|x_b|=7.8$

substituting $x_d$ value

$5.6|x_b|=7.8$

$|x_b|=1.392m$

$|x_d|=6.4032m$

now the dog is 22.5-6.403=16.096m from shore

4 0

3 years ago