Some of the principal minerals commonly found in igneous rocks are

A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 2 × 10−8 C at location m. At locat

liraira [26]

Answer: E = $1.8 *10 ^{4} N$

Explanation: The formulae for intensity of an electric field of a solid metal sphere relative to a point is given below

$E =\frac{Kq}{r^{2} }$ r

where $k=9* 10^{9}N/m^{2}$ , $q=2 *10 ^{-8} c$ , r = 0.1m r = is the position vector of the charge.

it has been stated in the question that the charge is placed at the center thus it has no position vector.

$E=\frac{9 * 10^{9}* 2* 10^{-8} }{0.1^{2} }\\ =\frac{18* 10^{1} }{0.01} \\=\frac{18* 10^{1} }{1 *10^{-2} } = 1.8*10^{4} N$

6 0

2 years ago

The first confirmed detections of extrasolar planets occurred in ____________. The first confirmed detections of extrasolar plan

nirvana33 [79]

Answer:

1992 (Early 1990s)

Explanation:

First of all, i would like to define an extrasolar planet as a planet that orbits a start that is not our own.

The first confirmed detections of extrasolar planets occured in the early 1990s (specifically 1992, some say 1995). The name of the first extrasolar planet is widely believed to be called Dimidium or 51 Pegasi b.

Extrasolar were searched by monitoring stars for slight dimming that might occur as unseen planets pass in front of them.

4 0

3 years ago

Please help step by step 15 points

BartSMP [9]

3.87 i think but if its not correct then let me know.

3 0

3 years ago

Marvin Martian is standing on planet Potatoine.

prisoha [69]

Answer:

W = 113.98 N

Explanation:

Given that,

Radius of Potatoine $R=6.4\times 10^6\ m$

Mass of Potatoine, $M=2\times 10^{24}\ kg$

Mass of Marvin, m = 35 kg

We need to find his weight on Potatoine. Weight of an object is given by :

W = mg

g is acceleration due to gravity, $g=\dfrac{GM}{R^2}$

So,

$W=\dfrac{GMm}{R^2}\\\\W=\dfrac{6.67\times 10^{-11}\times 35\times 2\times 10^{24}}{(6.4\times 10^6)^2}\\\\W=113.98\ N$

So, his weight on Potatoine is 113.98 N.

4 0

2 years ago

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that

ra1l [238]

Answer:

a) $I=0.012\ kg.m^2$

b) $I=0.012\ kg.m^2$

Explanation:

Given that:

mass of rod, $m=0.4\ kg$

length of the rod, $l=0.6\ kg$

(a)

Moment of inertia of rod about its center and perpendicular to the rod is given as:

$I=\frac{1}{12} m.l^2$

$I=\frac{1}{12} 0.4\times 0.6^2$

$I=0.012\ kg.m^2$

(b)

Moment of inertia on bending the rod to V-shape of 60 degree angle and axis being perpendicular to the plane of V at the vertex.

We treat it as two rod with axis of rotation at the end and perpendicular to the plane of rotation. 

So, the mass and the length of the rod will become half of initial value.

$I=I_1+I_2$

$I=\frac{1}{3} \frac{m}{2} (\frac{l}{2})^2 +\frac{1}{3} \frac{m}{2} (\frac{l}{2})^2$

$I=2[ \frac{1}{3}\times 0.2\times 0.3^2]$

$I=0.012\ kg.m^2$

7 0

2 years ago