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3 years ago

9

The mining-resource cycle determines the use of ore materials. Which step is not part of the cycle?

1 answer:

ExtremeBDS [4]3 years ago

8 0

Answer:

c) marketing and selling products that use ore materials

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What is the maximum m2 value that the system can be stationary at?

Ne4ueva [31]

Answer: 37.5 kg in 3 s.f.

Explanation:

7 0

3 years ago

2. An object's weight is proportional to its _ or _ from another object

leva [86]

Answer:

mass

gravitational pull

5 0

2 years ago

Two loudspeakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from the other. (a) What is the lowest fre

VMariaS [17]

Answer:

Explanation:

Given

Distance between two loud speakers $d=1.6\ m$

Distance of person from one speaker $x_1=3\ m$

Distance of person from second speaker $x_2=3.5\ m$

Path difference between the waves is given by

$x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}$

for destructive interference m=0 I.e.

$x_2-x_1=\frac{\lambda }{2}$

$3.5-3=\frac{\lambda }{2}$

$\lambda =0.5\times 2$

$\lambda =1\ m$

frequency is given by

$f=\frac{v}{\lambda }$

where $v=velocity\ of\ sound\ (v=343\ m/s)$

$f=\frac{343}{1}=343\ Hz$

For next frequency which will cause destructive interference is

i.e. $m=1$ and $m=2$

$3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda$

$\lambda =\frac{1}{3}\ m$

frequency corresponding to this is

$f_2=\frac{343}{\frac{1}{3}}=1029\ Hz$

for $m=2$

$3.5-3=\frac{5}{2}\cdot \lambda$

$\lambda =\frac{1}{5}\ m$

Frequency corresponding to this wavelength

$f_3=\frac{343}{\frac{1}{5}}$

$f_3=1715\ Hz$

8 0

3 years ago

A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves

expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

$q_1=3.4\mu C=3.4\times 10^{-6} C$

$1\mu C=10^{-6} C$

$q_2=-4.9\mu C=-4.9\times 10^{-6} C$

$x_1=0.125,y_1=0$

$x_2=0.280,y_2=0.235$

We have to find the work done by the electric force on the moving point charge.

$r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125$

$r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366$

Work done, $W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})$

Where $k=9\times 10^9$

Using the formula

$W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})$

$W=-0.79 J$

5 0

3 years ago

A 57 kg person in a rollercoaster moving through the bottom of a curved track of radius 42.7 m feels a normal force of 995 N. Ho

natita [175]

Answer:

Use Fc centripetal force as positive and W the weight as negative

N = m v^2 / R + m g

v^2 = (N - m g) R / m

v^2 = (995 - 57 * 9.8) 42.7 / 57 = 327 m^2/s^2

v = 18.1 m/s

Note: N - m g is the net force producing the centripetal force

5 0

2 years ago