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3 years ago

B. what does it mean to release a chemical?|

Physics

1 answer:

blsea [12.9K]3 years ago

3 0

(A "release" of a chemical means that it is emitted to the air or water, or placed in some type of land disposal.)

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Three small spheres, having masses m1 = 1 kg, m2 = 3 kg, and m3 = 4 kg, are held fixed on the x axis in deep space where the eff

Vlad [161]

Answer: $F_{net} = 0.7411 N$ towards the mass $m_{3}$

Explanation: We know that the gravitational force is a long range force which is always attractive in nature.

Given that:

mass $m_{1} = 1 kg$

mass $m_{2} = 3kg$

mass $m_{3} = 4kg$

The masses are positioned on X-axis at the following points:

Position of mass $m_{1}$ $x_{1} = 0$

Position of mass $m_{2}$ $x_{2} = 3$

Position of mass $m_{3}$ $x_{3} = 6$

Mathematically:

Gravitational force on mass $m_{2}$ due to mass $m_{1}$ is given by

$F_{21} = G \frac{m_{1}.m_{2}}{(r_{21})^2}$ ...................(1)

where: $(r_{21})^2$ = the radial distance between masses $m_{2}$ & $m_{1}$ =3

Similarly, gravitational force on mass $m_{2}$ due to mass $m_{3}$ is given by

$F_{23} = G \frac{m_{3}.m_{2}}{(r_{23})^2}$ ............................(2)

where: $(r_{23})^2$ = the radial distance between masses $m_{2}$ & $m_{3}$ =3

Now, put the respective values in the above equations.

$F_{21} = 6.67 \times 10^{-11 }\times \frac{1\times 3}{3^2}$

$F_{21} = 2.2233\times 10^{-11} N$

Again,

$F_{23} = 6.67 \times 10^{-11 }\times \frac{1\times 4}{3^2}$

$F_{23} = 2.9644\times 10^{-11} N$

∵Mass $m_{2}$ is in the middle of the masses $m_{3}$ & $m_{1}$ therefore the forces $F_{23}$ & $F_{21}$ will attract them in radially opposite direction.

∴ $F_{net} = F_{23} -F_{21} \\\\F_{net} = 2.9644-2.2233\\\\F_{net} = 0.7411 N$ towards the mass $m_{3}$

6 0

3 years ago

Read 2 more answers

1. Compare and Contrast microwaves with visible light using wavelength, frequency and energy

coldgirl [10]

Visible light has more energy than microwaves but it has a smaller wavelength than microwave

4 0

3 years ago

If a car takes a banked curve at less than a given speed, friction is needed to keep it from sliding toward the inside of the cu

zubka84 [21]

Answer:

minimum speed is 15.35 m/s

frictional coefficient is 0.26

Explanation:

given data

radius = 84 m

angle = 16°

speed = 16 km/h = 4.43 m/s

to find out

minimum speed and minimum coefficient

solution

we will apply here formula for velocity that is

velocity² = radius × g × tanθ

v² = 84 × 9.8 × tan16

v² = 236.04

v = 15.35 m/s

and

we find first friction force here

friction force 1 = m v² /r

friction force 1 = m (15.35)² / 84 = 2.80 m

and

friction force 2 = m v² /r

friction force 2 = m (4.43)² / 84 = 0.245 m

so total friction force = f1 - f2

total friction force = 2.80 - 0.245 = 2.55 m

so frictional coefficient = friction force /g

frictional coefficient = 2.55 / 9.8

so frictional coefficient is 0.26

4 0

3 years ago

Read 2 more answers

What are 3 ways you can save energy on earth day or any day?

Ber [7]

By unplugging unused devices, by turning off any unused lights, and by switching your lightbulbs to something more energy efficient.

8 0

4 years ago

2/25/20 or 2/28/20 Dispatch #53

mixer [17]

Answer:

Power = 21[W]

Explanation:

Initial data:

F = 35[N]

d = 18[m]

In order to solve this problem we must remember the definition of work, which tells us that it is equal to the product of a force for a distance.

Therefore:

Work = W = F*d = 35*18 = 630 [J]

And power is defined as the amount of work performed in a time interval.

Power = Work / time

Time = t = 30[s]

Power = 630/30

Power = 21 [W]

3 0

3 years ago