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yuradex [85]
3 years ago
11

While sitting motionless in a 5 kg friction free wagon, an 80kg clown catches a 15 kg cannonball traveling horizontally at 20 m/

s. After the cannonball is caught, at what speed do the clown, wagon, and ball move?
Physics
1 answer:
posledela3 years ago
4 0
Momentum is conserved
15*20= initial momentum =300
total mass after being caught = 100 
momentum=mass*velocity
thus dividing the momentum by total mass we get the velocity of 3 m/s
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Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o
tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D  is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

f =\dfrac{1}{P}

Put the value into the formula

f=\dfrac{100}{2.55}

f=39.21\ cm

We need to calculate the image distance

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}

\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}

\dfrac{1}{v}=-\dfrac{1421}{98025}

v=-68.98\ cm

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

f =\dfrac{1}{P}

Put the value into the formula

f=-\dfrac{100}{3.00}

f=-33.33\ cm

We need to calculate the image distance

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}

-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}

\dfrac{1}{v}=-\dfrac{1}{33.33}

v=-33.33\ cm

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

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The acceleration of the first block (4 kg) is -9.8 m/s².

The given parameters:

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  • <em>Mass of the second block, m₂ = 2.0 kg</em>

The net force on the system of the two blocks is calculated as follows;

m_2 g - T = m_1 a

where;

  • <em>T </em><em>is the tension in the connecting string due weight of the first block</em>

m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2

Thus, the acceleration of the first block (4 kg) is -9.8 m/s².

Learn more about net force on two connected blocks here: brainly.com/question/13539944

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Answer:

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Xelga [282]

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7 0
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Lapatulllka [165]

Answer:

X-rays travel through space faster than radio waves.

Explanation:

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All electromagnetic waves travel in a vacuum always at the same speed, the speed of light, whose value is:

c=3.0\cdot 10^8 m/s

Electromagnetic waves are classified into 7 different types, according to their wavelength/frequency. From shortest to longest wavelength (and so, from highest to lowest frequency), we have:

Gamma rays

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Ultraviolet

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Now we can analyze the 4 statements:

X-rays and radio waves are both forms of light, or electromagnetic radiation --> TRUE. They are both types of electromagnetic waves.

X-rays have higher frequency than radio waves. --> TRUE, as we can see from the table above.

X-rays have shorter wavelengths than radio waves. --> TRUE, as we can see from the table above.

X-rays travel through space faster than radio waves. --> FALSE: all electromagnetic waves travel in space at the same speed, the speed of light.

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