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loris [4]
3 years ago
8

Newton´s laws are based on which two concepts?

Physics
1 answer:
ioda3 years ago
5 0
Inertia and force

PLEASE RATE THIS THE BRAINLIEST ANSWER! THANK YOU! :)
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What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

3 0
3 years ago
Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co
patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of ^{3890}Sr\, t_{\frac{1}{2}} = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where

N = No. of nuclei left after time t

N_{o} = No. of nuclei initially started with

\frac{N}{N_{o}} = 1\times 10^{- 4}

(Since, 100% - 99.99% = 0.01%)

Thus

1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}

Taking log on both the sides:

- 4 = \frac{t}{28.5}log\frac{1}{2}

t = \frac{-4\times 28.5}{log\frac{1}{2}}

t = 387.69 yrs

5 0
3 years ago
Dr. Spock would have supported which of the following actions by parents? (2 points)
vampirchik [111]
<span>A parent develops a set of rules collaboratively with her child.

Dr. Benjamin Spock believed that children should be treated as individuals.</span>
7 0
3 years ago
Read 2 more answers
A mixture of nitrogen and xenon gases, at a total pressure of 836 mm Hg, contains 2.80 grams of nitrogen and 24.9 grams of xenon
larisa86 [58]

Answer: Partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

p_A=p_T\times \chi_A

where,

p_A = partial pressure of substance A

p_T = total pressure

\chi_A = mole fraction of substance A

We are given:

m_{N_2}=2.80g

m_{Xe}=24.9g

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

And,

n_A=\frac{m_A}{M_A}

Mole fraction of nitrogen is given as:

\chi_{N_2}=\frac{\frac{m_{N_2}}{M_{N_2}}}{(\frac{m_{N_2}}{M_{N_2}}+\frac{m_{Xe}}{M_{Xe}})}

Molar mass of N_2 = 28 g/mol

Molar mass of Xe =  g/mol

Putting values in above equation, we get:

\chi_{N_2}=\frac{\frac{2.80}{28}}{\frac{2.80}{28}+\frac{24.9}{131}}

\chi_{N_2}=\frac{0.100}{0.100+0.190}=0.345

To calculate the mole fraction of xenon, we use the equation:

\chi_{Xe}+\chi_{N_2}=1\\\\\chi_{Xe}=1-0.345=0.655

p_{N_2}=836mmHg\times 0.345=288mmHg

p_{Xe}=836mmHg\times 0.655=548mmHg

Thus partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

6 0
3 years ago
A horizontal 649 N merry-go-round of radius 1.05 m is started from rest by a constant horizontal force of 61.3 N applied tangent
nexus9112 [7]

Answer:

The kinetic energy of the merry-go-round is 632.82 J

Explanation:

Given;

weight of the merry-go-round, W = 649 N

radius of the merry-go-round, r = 1.05 m

applied horizontal force, F =  61.3 N

acceleration due to gravity, g = 9.8 m/s²

mass of  merry-go-round, m = W/g

                                               = 649/9.8  = 66.225 kg

moment of inertia of merry-go-round, I = ¹/₂mr²

                                                                 = ¹/₂ x 66.225 x (1.05)²

                                                                 = 36.507 kg.m²

Angular acceleration of the merry-go-round, α

τ = Iα = Fr

α = Fr / I

Where;

α is angular acceleration

α = (61.3 x 1.05) / 36.507

α = 1.763 rad/s²

Angular velocity of the merry-go-round, ω

ω = αt

ω = 1.763 x 3.34

ω = 5.888 rad/s

Finally, the kinetic energy of the merry-go-round, K.E

K.E = ¹/₂Iω²

K.E = ¹/₂ x 36.507 x (5.888)²

K.E = 632.82 J

4 0
2 years ago
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