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2 years ago

11

List three ways in which decreasing the need to mine gold and reducing its harmful environmental effects. (Core Case Study) coul

d benefit you.__________-

1 answer:

RoseWind [281]2 years ago

4 0

It’s decreasing as environmental cause osmosis

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Help!Yall get brainliest

Alex_Xolod [135]

Answer:

Explanation:

Good sources include barley, oatmeal, beans, nuts, and fruits such as apples, berries, citrus fruits, and pears. Many foods contain both soluble and insoluble fiber. In general, the more natural and unprocessed the food, the higher it is in fiber. There is no fiber in meat, dairy, or sugar.

7 0

2 years ago

Read 2 more answers

If the third charge (–| q 3 |) is placed at point P, but not held fixed, it will experience a force and accelerate away from the

Mazyrski [523]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

The potential of this system is $U=6.75*10^{-7}J$

B

The electric potential at point p is $V_p= -900V$

C

The work required is $W= 9*10^{-7}J$

D

The speed of the charge is $v=600m/s$

Explanation:

A sketch to explain the question is shown on the second uploaded image

Generally the potential energy for a system of two charges is mathematically represented as

$U = \frac{kq_1 q_2}{d}$

where k is the electrostatic constant with a value of $k = 9*10^9 N m^2 /C^2$

q is the charge with a value of $q = 1*10^{-9}C$

d is the distance given as $d =5m$

Now we are given that $q_1 = q$ and $q_2 = 3q$ and

Now substituting values

$U = \frac{9*10^9 *1*10^{-9} * 3*10^{-9}}{5}$

$U=6.75*10^{-7}J$

The electric potential at point P is mathematically obtained with the formula

$V_p = V_{-q} + V_{-3q}$

I.e the potential at $q_1$ plus the potential at $q_2$

Now potential at $q_1$ is mathematically represented as

$V_{-q} = \frac{-kq}{s}$

and the potential at $q_2$ is mathematically represented as

$V_{-3q} = \frac{-3kq}{s}$

Now substituting into formula for potential at P

$V_p = \frac{-kq}{s} + \frac{-3kq}{s} = -\frac{4kq}{s}$

$= \frac{4*9*10^9 *1*10^{-9}}{4*10^{-2}}$

$V_p= -900V$

The Workdone to bring the third negative charge is mathematically evaluated as

$W =\Delta U = \frac{kq_1q_3}{s} + \frac{kq_2q_3}{s}$

$= \frac{kq*q}{s} + \frac{kq*3q}{s}$

$= \frac{4kq^2}{s}$

$= \frac{4* 9*10^9 * (1*10^{-9})^2}{4*10^{-2}}$

$W= 9*10^{-7}J$

From the Question are told that the charge $q_3$ would a force and an acceleration which implies that all its potential energy would be converted to kinetic energy.This can be mathematically represented as

$\Delta U = W = \frac{1}{2} m_{q_3} v^2$

$9*10^{-7} = \frac{1}{2} m_{q_3} v^2$

Where $m_{q_3} = 5.0*10^{-12}kg$

Now making v the subject we have

$v = \sqrt{\frac{9*10^{-12}}{5*10^{-12}*0.5} }$

$v=600m/s$

8 0

3 years ago

What’s 3 times 10 to the 8th power divided by 2.45 times 10 to the 9th power

Virty [35]

0.12244898 is the value obtained when solving the given.

<u>Explanation:</u>

Given:

3 times 10 to the 8th power can be expressed in equation format as $3 \times 10^{8}$ (3 times symbolizes ‘multiplication’ then it is to the tenth power of 8)

2.45 times 10 to the 9th power can be expressed in equation format as $2.45 \times 10^{9}$ (2.45 times symbolizes ‘multiplication’, then it is to the tenth power of 9)

Asked to find the solution when dividing the above,

$\frac{3 \times 10^{8}}{2.45 \times 10^{9}}=\frac{3}{2.45} \times 10^{8-9}$

When the tenth power of any value goes from denominator to numerator and vice-versa, presents in opposite sign to that it possess (like when ‘ $10^{9}$ ' goes to numerator changed as ' $10^{-9}$ ').

Now by solving the above equation, we get

$\frac{3}{2.45} \times 10^{8-9}=1.22 \times 10^{-1}=\frac{1.22}{10}=0.12244898$

5 0

2 years ago

If an airplane is flying directly north at 300.0 km/h, and a crosswind is hitting the airplane at 50.0 km/h from the east, what

Rashid [163]

Answer:

magnitude = 304.14 km/h

direction: $9.46^o$ West of North

Explanation:

The final plane's vector velocity will be the result of the vector addition of one pointing North of length 300 km/h, another one pointing West of length 50 km/h.

To find the magnitude of the final velocity vector (speed) we need to apply the Pythagorean theorem in a right angle triangle with sides: 300 and 50, and find its hypotenuse:

$|v|=\sqrt{300^2+50^2}=\sqrt{92500} = 304.14$ km/h

The actual direction of the plane is calculated using trigonometry, in particular with the arctan function, since the tangent of the angle can be written as:

$tan(\theta)=\frac{50}{300} = \frac{1}{6} \\\theta = arctan(\frac{1}{6} ) = 9.46^o$

So the resultant velocity vector of the plane has magnitude = 304.14 km/h,

and it points $9.46^o$ West of the North direction.

3 0

3 years ago

The Old Faithful geyser found in Yellowstone Park is an example of _____.

katen-ka-za [31]

Answer:

cone geyser

Explanation:

The Old Faithful geyser is the oldest discovered geyser in the Yellowstone national park. The eruptions of the geyser are particularly predictable. It is a cone type geyser.

Cone geysers generally have a spout through which the water ejects out. When super heated water in the tube then the water starts to boil and form bubbles of steam, after this process the eruption takes place.

5 0

2 years ago