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2 years ago

5

J.J. Thomson theorized that, if an atom had all of its negatively charged

1 answer:

Bumek [7]2 years ago

6 0

Answer:

C. Positively charged

Explanation:

The plum pudding model of the atom proposes by J. J. Thomson consisted of electrons which lay embedded as the raisins within a dough or soup that was positively charged. The electron was discovered by J. J. Thomson in 1897 through cathode ray tube experiments.

Based on the plum pudding model, if all the negatively charge electrons contained in an atom are removed, the material remaining will be the <em>positively charged</em> soup

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An amoeba has 1.00 x 1016 protons and a net charge of 0.300 pC. Assuming there are 1.88 x 106 fewer electrons than protons, If y

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Answer:

The fraction of the protons would have no electrons $=1.88\times 10^{-10}$

Explanation:

We are given that

Amoeba has total number of protons= $1.00\times 10^{16}$

Net charge, Q=0.300pC

Electrons are fewer than protons= $1.88\times 10^6$

We have to find the fraction of protons would have no electrons.

The fraction of the protons would have no electrons

= $\frac{Fewer\;electrons}{Total\;protons}$

The fraction of the protons would have no electrons

= $\frac{1.88\times 10^{6}}{1.00\times 10^{16}}$

$=1.88\times 10^{-10}$

Hence, the fraction of the protons would have no electrons $=1.88\times 10^{-10}$

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3 years ago

Light with a wavelength of 725 nm passes through the slits of a double-slit experiment. The slits of the experiment are a distan

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Answer:

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Explanation:

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2 years ago

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3 years ago

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The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

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Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

$b = 1795.4 kg/s$

7 0

3 years ago