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2 years ago

A student mixes two substances in a large beaker. After two hours, the student sees that the substances have separated again.

Physics

1 answer:

S_A_V [24]2 years ago

7 0

The full question is found in the image attached

Answer:

Densities

Explanation:

When we talk about density, one of the things that come to our mind is actually how heavy something is.

If i have substances that possess different densities, i will notice that i can not really mix them because they will separate and the heavier substance will be at the bottom while the lighter substance will be at the top.

Hence the two substances separated due to difference in density.

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List 2 examples of objects demonstrating mechanical energy.

OlgaM077 [116]

Windmills run on the principle of mechanical energy and work. Moving air (wind) possesses some amount of energy in the form of kinetic energy (due to motion). This energy gives the air the ability to do work on the blades of the fan.

6 0

2 years ago

A heater is being designed that uses a coil of 14-gauge nichrome wire to generate 300 W using a voltage of V = 110 V . How long

Mazyrski [523]

Answer:

The length of the wire is 83.2 m.

Explanation:

Given that,

Power of heater, P = 300 W

Voltage, V = 110 V

The resistivity of nichrome wire is, $\rho=100\times 10^{-8}\ \Omega-m$

The electric power of a wire is given by :

$P=\dfrac{V^2}{R}\\\\R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{300}\\\\R=40.34\ \Omega$

Area of cross section of the wire is :

$A=\pi r^2\\\\A=\pi (0.000815)^2\\\\A=2.08\times 10^{-6}\ m^2$

Resistance of a material is given by :

$R=\rho \dfrac{L}{A}\\\\L=\dfrac{RA}{\rho}\\\\L=\dfrac{40\times 2.08\times 10^{-6}}{10^{-6}}\\\\L=83.2\ m$

So, the length of the wire is 83.2 m. Hence, this is the required solution.

5 0

3 years ago

While riding in a moving car, you toss an egg directly upward. Does the egg tend to land behind you, in front of you, or back in

bazaltina [42]

Answer:

Explanation:

The movement of a body can be analyzed using New's first law. In an inertial frame (without acceleration) every body is kept at rest or moving at constant speed until there is an external force that changes this state

Let's analyze these cases in the framework of this first law

a) If the vehicle is going at constant speed the two bodies (the egg and the hands) do not change movement so he had returned to the hands

b) If the vehicle accelerates the passenger goes faster, but the egg that is not subject to anything does not change the movement, so it falls behind the passenger

c) If the vehicle slows down, the passenger reduces its speed and the distance traveled in time, but the egg that is not attached follows its movement and falls in front of the passenger.

4 0

3 years ago

2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

$\\ \\$

So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

$\\ \\$

$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

2 years ago

Coulomb's law states that the force F of attraction between two oppositely charged particles varies jointly as the magnitude of

Mars2501 [29]

Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved

Explanation:

Let the charges be q1 and q2 and the distance between the charges be 'd'

Mathematical representation of coulombs law will be;

F1=kq1q2/d²...(1)

Where k is the electrostatic constant.

If q1 and q2 is doubled and the distance halved, we will have;

F2 = k(2q1)(2q2)/(d/2)²

F2 = 4kq1q2/(d²/4)

F2 = 16kq1q2/d²...(2)

Dividing equation 1 by 2

F1/F2 = kq1q2/d² ÷ 16kq1q2/d²

F1/F2 = kq1q2/d² × d²/16kq1q2

F1/F2 = 1/16

F1 = 1/16F2

This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved

4 0

3 years ago