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3 years ago

Why do iron meteorites represent a much higher percentage of finds than of falls?

Physics

1 answer:

KiRa [710]3 years ago

6 0

Answer:

Explanation:

Meteorites falls are those which are collected just after knowing it is falling, whereas meteorites find are found at later times.

Iron meteorites have higher percentage of find because people have found that they are useful and have great qualities. They collect such meteorites thus removing them from scientific record.

The iron meteorite consist of iron nickle alloy and is a usable iron source for humans. Iron meteorites are more resistance to weathering and can easily survive atmospheric entry hence are found in large pieces.

Because of all such features, people collect them and make their most of use, hence its find percentage is higher.

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two forces act at an angle of 135 degree.the forces are F1 = 50N and F2 = 25N respectively.graphically construct a parallelogram

olya-2409 [2.1K]

Answer:gay

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3 years ago

g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9

Elza [17]

Answer:

The answer is " $\bold{dosage = 0.031 rem}$ "

Explanation:

please find the complete question in the attached file.

Given value:

$m = 79\ kg \\\\n = 3.4 \times 10^9 \\\\E = 5.5 \times 10^{-13} \\\\ RBE = 15$

$\to E = n E\\$

$= 3.4 \times 10^9 \times 5.5 \times 10^{-13} \\\\ = 1.87 \times 10^{-3}$

$\to E(absorbed) = 1.87 \times 10^{-3} \times 0.87 = 1.63 \times 10^{-3}$

calculating the radiation absorbed per kg:

$= \frac{1.63 \times 10^{-3}}{79} \\\\ = 2.06 \times 10^{-5} \\\\ = 0.00206 \ rad$

$\to Dosage = 0.00206 \times 15 \\$

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4 0

2 years ago

A hollow, uniformly charged sphere has an inner radius of r1 = 0.105 m and an outer radius of r2 = 0.31 m. The sphere has a net

Sliva [168]

Answer:

E = 77532.42N/C

Explanation:

In order to find the magnitude of the electric field for a point that is in between the inner radius and outer radius, you take into account the Gauss' law for the electric flux trough a spherical surface with radius r:

$\int E\cdot dS=\frac{Q}{\epsilon_o}$ (1)

Q: net charge of the hollow sphere = 1.9*10-6C

ε0: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

Furthermore, you have that the net charge contained in a sphere of radius r is:

$Q=\rho V=\rho \frac{4\pi (r^3-r_1^3)}{3}$ (2)

with the charge density is:

$\rho=\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}$ (3)

r2: outer radius = 0.31m

r1: inner radius = 0.105m

The electric field trough the Gaussian surface is parallel to the normal to the surface, the, you have in the integral of the equation (1):

$\int E\cdot dS=E(4\pi r^2)$ (4)

where you have used the expression for a surface of a sphere.

Next, you replace the expressions of equations (2), (3) and (4) in the equation (1) and solve for E:

$E(4\pi r^2)=\frac{1}{\epsilon_o}\frac{Q}{\frac{4}{3}\pi(r_2^3-r_1^3)}(\frac{4\pi (r^3-r_1^3)}{3})\\\\E=\frac{1}{\epsilon_o}\frac{Q(r^3-r_1^3)}{4\pi r^2(r_2^3-r_1^3)}$

you replace the values of all parameters, and with r = 0.17m

$E=\frac{(1.6*10^{-6}C)((0.17m)^3-(0.105m)^3)}{4\pi(8.85*10^{-12}C^2/Nm^2)(0.17m)^2((0.31m)^3-(0.105m)^3)}\\\\E=77532.42\frac{N}{C}$

The magnitude of the electric field at a distance r=0.17m to the center of the hollow sphere is 77532.42N/C

5 0

3 years ago