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2 years ago

How are the components of a heterogeneous mixture distributed?

Physics

1 answer:

Lena [83]2 years ago

3 0

Answer:

heterogeneous mixture has components that are not evenly distributed. This means that you can easily distinguish between the different components.

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A 140 N block rests on a table. The suspended mass has a weight of 77 N.

Basile [38]

I'm assuming that this is what your diagram looks like because these types of problems always look like this (lol).

Comment if you need clarification!

5 0

3 years ago

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At the equator, the earth spins a distance of

valentina_108 [34]

Answer:

About 1080 miles per hour

Explanation:

If the earth spins 25,922 miles everyday (24 hours) then is would spin about 1080 miles per hour.

25 922 miles / 24

24 divided by 24 to get 1 ( 1 hour)

*if we divided 24 by 24 then we do the same to 25 922 (to make it equivalent)

25 922 divided by 24 = 1080.083333333333.

So, 1080 mph or 1080 miles / 1 hour

I hope this helped :)

3 0

3 years ago

This type of stretch keeps your heart rate elevated and muscles warm.

Illusion [34]

Answer:

dynamic and sometimes ballistic

4 0

2 years ago

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If an electronic circuit experiences a loss of 3 decibels with an input power of 6 watts, what would its output power be, to the

agasfer [191]

Answer:

Output power of the circuit is 3 Watt.

Given:

loss in decibles = 3 dB

Input power = 6 Watt

To find:

Output power = ?

Formula used:

Output power = Input power × loss in ratio

Solution:

3 dB loss = 0.5 ratio

Output power is given by,

Output power = Input power × loss in ratio

Output power = 6 × 0.5

Output power = 3 Watt

Thus, output power of the circuit is 3 Watt.

4 0

2 years ago

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Two small insulating spheres with radius 3.50×10^−2m are separated by a large center-to-center distance of

Liula [17]

Answer:

Part (i) the magnitude E of the electric field midway between the spheres is 8.71 × 10⁵ N/C

Part (ii) the direction of the electric field is towards the negative charge

Explanation:

Given;

+q = 3.93.90μC, r = 3.50×10⁻²m

-q = −2.40μC, r = 3.50×10⁻²m

magnitude of the electric field is experienced, midway between the spheres at a distance r, r = ¹/₂ × 0.51 = 0.255 m

Electric field due to point charge is given as;

$E = \frac{F}{q} = \frac{Kq^2}{qr^2} = \frac{kq}{r^2}$

K is coulomb's constant = 8.99 x 10⁹ Nm²/C²

The positive charge on positive x-axis and the negative charge is on negative x-axis.

part (a)

The electric field due to positive charge; +q = 3.93.90μC

$E_+ = \frac{kq}{r^2}\\\\E_+ = \frac{8.99 X10^9*3.9X0^{-6}}{0.255^2}\\\\E_+= 5.3919 X10^5\frac{N}{C}$

The electric field due to negative charge; -q = −2.40μC

$E_- = \frac{kq}{r^2}\\\\E_- = \frac{8.99 X10^9*2.4X0^{-6}}{0.255^2}\\\\E_-= 3.3181 X10^5\frac{N}{C}$

From superimposition theorem

The magnitude of the electric field is;

E = E₊ + E₋

E = (5.3919 × 10⁵ + 3.3181 × 10⁵) N/C

E = 8.71 × 10⁵ N/C

Therefore, the magnitude E of the electric field midway between the spheres is 8.71 × 10⁵ N/C

Part (b)

The direction of the electric field is towards the negative charge.

7 0

3 years ago