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ivanzaharov [21]

3 years ago

11

Why should a rain gauge be raised 30cm above the ground

2 answers:

Mrrafil [7]3 years ago

5 0

Explanation:

The rain gauge is normally placed in the ground, leaving the top of the funnel above ground – about 30 cm above the ground so that it can collect water into the container or jar. ... Rain gauges should be placed in an open area where there are no buildings, trees, or other obstacles to block the rain.

Mrrafil [7]3 years ago

5 0

It is an instrument used for determining the depth of precipitation that occurs over a unit area. ... A graduated measuring glass is used for determining the amount of precipitation. For the measurement of rainfall, the rain gauze should be placed 30 cm above the ground to prevent the splashing of rainwater

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It will have 100 radioactive particles.

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If the work function of a material is such that red light of wavelength 700 nm just barely initiates the photoelectric effect, w

marishachu [46]

Answer:

2.13 x 10^-19 J or 0.53 eV

Explanation:

cut off wavelength, λo = 700 nm = 700 x 10^-9 m

λ = 400 nm = 400 x 10^-9 m

Use the energy equation

$E = \frac{h c}{\lambda _{o}}+K$

Where, K be the work function

$\frac{h c}{\lambda} = \frac{h c}{\lambda _{o}}+K$

$K =hc\left ( \frac{1}{\lambda } -\frac{1}{\lambda _{0}}\right )$

$K =6.63\times 10^{-34}\times 3\times 10^{8}\left ( \frac{1}{4\times 10^{-7} } -\frac{1}{7\times 10^{-7}}\right )$

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3 0

3 years ago

Use the work energy theorem to solve each of these problems and neglect air resistance in all cases. a) A branch falls from the

Sav [38]

Answer:

a)43.8 m/s

b)103.54 m/s

Explanation:

Work energy theorem

Δ $E_{k}=$ Δ $E_{g}$

$E_{k2} - E_{k1}=-(E_{g2} - E_{g1})\\\\ \frac{1}{2} mv^{2}_{2}- \frac{1}{2} mv^{2}_{1}+ mgh_{2} - mgh_{1} = 0$

a) In this part v1=0 and h2=0

$\frac{1}{2} mv^{2}_{2}- \frac{1}{2} mv^{2}_{1}+ mgh_{2}- mgh_{1} = 0\\\\\\\\\v_{2} =\sqrt{2gh_{1}} = \sqrt{2*9.8*98} =43.8 m/s$ =0

b) in this part v2=0, h1= 0, and h2= 545m

$\frac{1}{2} v^{2}_{1}= gh_{2} \\v_{1} =\sqrt{2gh_{2}}=103.54m/s$

5 0

3 years ago

Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (

Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

$x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}$

$0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}$

$10 + 8.3 x = 0$

$x = -1.20 m$

Similarly for y direction we have

$y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}$

$0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}$

$9.28 + 8.3 x = 0$

$x = -1.12 m$

5 0

3 years ago