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3 years ago

Of the more than 2,000 known minerals approximately 95% of them belong to what mineral group?

Physics

1 answer:

Temka [501]3 years ago

5 0

Your answer is Silicates.

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Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5

pishuonlain [190]

Answer

given,

force per unit length = 350 µN/m

current, I = 22.5 A

y = y = 0.420 m

$\dfrac{F}{L}= \dfrac{KI_1I_2}{d}$

$I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}$

$I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}$

I₂ = 32.67 A

distance where the magnetic field is zero

$\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}$

$y_1 = 0.248\ m$

there the distance at which the magnetic field is zero in the two wire is at 0.248 m.

3 0

3 years ago

What measurement is the quantity of mass per unit volume

ANTONII [103]

That's the definition of the property called "density".

6 0

2 years ago

Read 2 more answers

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

6 0

3 years ago

Two boats start together and race across a 48-km-wide lake and back. boat a goes across at 48 km/h and returns at 48 km/h. boat

jolli1 [7]

Answer:

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

Explanation:

Case 1: Boat 1

Speed of boat = $\frac{distance of river}{time}$

time = $\frac{distance of river}{speed of boat}$

While going to another end

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{48}$

time = 1 hour

While going back,

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{48}$

time = 1 hour

Total time taken by boat 1 is,

Total time by boat 1 = 1 hour + 1 hour = 2 hour

Total time by boat 1 = 2 hour

Total time taken by boat 1 for the round trip is 2 hour.

Case 2: Boat 2

Speed of boat = $\frac{distance of river}{time}$

time = $\frac{distance of river}{speed of boat}$

While going to another end

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{24}$

time = 2 hour

While going back,

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{72}$

time = 0.66 hour

Total time taken by boat 2 is,

Total time by boat 1 = 2 hour + 0.66 hour

Total time by boat 1 = 2.66 hour

Total time taken by boat 2 for the round trip is 2.66 hour.

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

5 0

3 years ago

You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel wi

Elenna [48]

Answer:

$\frac{g_{2}}{g_{1}} = \frac{1}{4}$

Explanation:

The period of the simple pendulum is:

$T = 2\pi\cdot \sqrt{\frac{l}{g} }$

Where:

$l$ - Cord length, in m.

$g$ - Gravity constant, in $\frac{m}{s^{2}}$ .

Given that the same pendulum is test on each planet, the following relation is formed:

$T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}$

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:

$\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}$

$\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}$

$\frac{g_{2}}{g_{1}} = \frac{1}{4}$

5 0

3 years ago