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3 years ago

Lunar missions have revealed that the moon has:

Physics

1 answer:

myrzilka [38]3 years ago

5 0

That the moon has soil within its shadowy craters rich and useful material

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When a species has no members left that are alive

denpristay [2]

Answer:

it means that species is extinct.

Explanation:

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3 years ago

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The most common oxidation numbers for a given element

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3 years ago

A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is th

vladimir2022 [97]

As we know that friction force on box is given by

$F_s = \mu_s N$

here we know that

$N = mg$

here we have

m = 12 kg

$\mu_s = 0.42$

so now we have

$N = 12(9.8) = 117.6 N$

now we will have

$F_s = 0.42(12)(9.8)$

$F_s = 49.4 N$

so it required minimum 49 N(approx) force to move the block

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4 years ago

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if charge is located at center of spherical volume and electric flux through surface of sphere is Φ what would be flux through s

jonny [76]

Answer:

The flux will be nine times as great.

Explanation:

The electric flux due to a charge Q located in the center of a sphere can be obtained using Gauss's law. Considering a Gaussian surface in the form of a sphere of radius r:

$\Phi_E=\int\limits {Ecos\theta dS} \,$

The electric field (E) is parallel to the surface vector (dS), so $\theta=0$

$\Phi_E=\int\limits{Ecos(0)dS} \,\\\Phi_E=E\int\limits{dS} \,\\\Phi_E=ES\\\Phi_E=E4\pi r^2$

Since the electric flux is proportional to the square of the sphere's radius, if radius of sphere were tripled, the flux will be nine times as great.

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3 years ago

Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of

Pachacha [2.7K]

Answer:

The frequency changes by a factor of 0.27.

Explanation:

The frequency of an object with mass m attached to a spring is given as

$f$ = $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

where $f$ is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

$f_{n}$ = $\frac{1}{2\pi } \sqrt{\frac{k}{14m} }$

simplifying, we have

$f_{n}$ = $\frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }$

$f_{n}$ = $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$

if we divide this final frequency by the original frequency, we'll have

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ ÷ $\frac{1}{2\pi } \sqrt{\frac{k}{m} }$

==> $\frac{1}{3.742*2\pi } \sqrt{\frac{k}{m} }$ x $2\pi \sqrt{\frac{m}{k} }$

==> 1/3.742 = 0.27

7 0

3 years ago