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2 years ago

If you added sand to your coffee, what kind of mixture would your coffee now be?

Physics

1 answer:

Pepsi [2]2 years ago

7 0

Answer:

A heterogenous mixture.

Explanation:

If you added sand to your coffee, the kind of mixture your coffee would now

be is a heterogenous mixture.

A heterogenous mixture can be defined as any mixture which has a different or non-uniform composition and properties throughout any given sample of the mixture. This ultimately implies that, the constituents of a heterogenous mixture always remain separate in the sample and as such comprises of two or more phases.

Hence, a mixture of sand and coffee is simply a heterogenous mixture and can easily be separated into their various components.

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The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud

motikmotik

<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = $\frac{Gm_1m_2}{r^2}$

Here G is the gravitational constant

m₁ is the mass of one student

m₂ is the mass of second student .

and r is the distance between them

Thus r = $\sqrt{\frac{Gm_1m_2}{F} }$

If we substitute the values in the above equation

r = $\sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }$

= 2.46 m

3 0

3 years ago

Read 2 more answers

A boy with a mouse is 80 kg stands on a scale in an ascending elevator with an acceleration of 3 m/s What is the resultant upwar

Rus_ich [418]

Answer:

Upward force on the boy will be 1024 N

Explanation:

We have given mass of the boy m = 80 kg

Acceleration due to gravity $g=9.8m/sec^2$

It is given that elevator is ascending with acceleration of $3m/sec^2$

So net acceleration on the boy = 9.8+3 = 12.8 $m/sec^2$ ( As the car elevator is moving ascending )

So upward force on the boy will be equal to F = m(g+a)

So $F=80\times 12.8=1024N$

So upward force on the boy will be 1024 N

3 0

2 years ago

A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po

aleksandrvk [35]

Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5 / x² = 2 / ( .4 - x )²

Taking square root on both sides

2.236 / x = 1.414 / .4 - x

2.236 ( .4 - x ) = 1.414 x

.8944 - 2.236 x = 1.414 x

.8944 = 3.65 x

x = .245 m

24.5 cm

So the third charge should be at a distance of 24.5 cm from q₁ .

4 0

2 years ago

قوة الجذب المركزي تكون في اتجاه

pochemuha

Answer:

تكون دائمًا متعامدة مع سرعة الجسم وتكون دائمًا في اتجاه مركز انحناء المسار

Explanation:

6 0

2 years ago

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A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius

ahrayia [7]

Answer:

The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.

at r < R:

Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.

E = 0.

at R < r < 2R:

The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.

at 2R < r:

The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.

Explanation:

Gauss’ Law is straightforward when applied to spheres. The area of the sphere is $A = 4\pi r^2$ , and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.

3 0

2 years ago