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2 years ago

8

Two spherical objects are separated by a distance that is 9.00 X 10m The objects are initially electrically neutral and are very

small compared to the stance between them. Each objectures the same negative charge due to the addition of electrons. As a result, each experiences an lectr ic force that has a magnitude of 1.137 10 . How many electrons did t o produce the charge on one of the objects?

1 answer:

romanna [79]2 years ago

5 0

Answer: 3.2 * 10^-3 C

Explanation: In order to solve this problem we have to use the Coulomb force given by:

F=k*q^2/ d^2 where he consider the same charge for each point

so, we have

q^2= F*d^2/k= 1.137* (10 C*9*10m)^2/9*10^9 N*m^2/C^2)=3.2 * 10^-3 C.

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After a surface interval my pressure group is D. I want to do a repetitive dive to 38 feet for 32 minutes. What will be my new p

ExtremeBDS [4]

Answer:

New pressure group is N

Explanation:

The Repetitive Dive table has to be checked, the residual nitrogen for the maximum depth of 38 feet as a "D" diver, then add that to the actual dive time to get the total nitrogen time. The total nitrogen time and the maximum depth of 38 feet would put you on a pressure group N.

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2 years ago

Read 2 more answers

A 3.30 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring const

avanturin [10]

Answer:

0.17547 m

Explanation:

m = Mass of block = $3.3\times 10^{-2}\ kg$

v = Velocity of block = 10.8 m/s

k = Spring constant = 125 N/m

A = Amplitude

The kinetic energy of the system is conserved

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow \\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{3.3\times 10^{-2}\times 10.8^2}{125}}\\\Rightarrow A=0.17547\ m$

The amplitude of the resulting simple harmonic motion is 0.17547 m

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3 years ago

A 30.0 g mass of iron at 24.5°C is heated to 45.0°C. The theoretical specific

diamong [38]

Answer:

276.135 J

Explanation:

Given that:

mass of Fe = 30.0 g

initial temperature = 24.5°C

final temperature = 45.0°C

specific heat of Fe = 0.449 J/g°C

We can determine the thermal energy added by using the formula;

Q = mcΔT

Q = 30.0g × 0.449 J/g°C × (45.0 - 24.5)°C

Q = 276.135 J

8 0

3 years ago

Name the 2 types of tissue that form your skin?

QveST [7]

Answer:

Epithelial tissue and Muscle tissue

Explanation:

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2 years ago

1. Bone has a Young’s modulus of about

Blababa [14]

#1

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

#2

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

#3

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

$\rho = 1185.3 kg/m^3$

#4

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

so it is 3.48 cm below the interface

5 0

3 years ago