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3 years ago

What is a heterogeneous mixture and what is a homogeneous mixture?

2 answers:

5 0

A heterogeneous mixture is a mixture that contains two or more distinct substances that you can see. You can see the different part if a heterogeneous mixture. An example of this is a salad. You can see all of the parts. A homogeneous mixture is a mixture that is uniform and you cannot see the different parts. It is still a mixture though. An example of that would be salt water. The water and salt are not chemically combined but you cannot see the salt AND water. It is just one solution.

Katena32 [7]3 years ago

3 0

Answer:

heterogeneous- Is not evenly mixed throughout.

homogeneous- Is evenly mixed throughout.

Explanation:

(APEX)

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When it is summer at the South Pole,

Elza [17]

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The northern hemisphere is experiencing winter because it is tilted away from the sun whereas the south experiences summer because it is tilted towards the sun

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2 years ago

In a binary star system, an unseen component is found to have 8 solar masses. It would be visible if the system were a normal st

maksim [4K]

Answer:

Black Hole

Explanation:

A black hole is a very dense and massive stellar object, which has a field of gravity so large that not even light can escape it.

Since it does not emit light, we cannot see them directly, hence the name of black hole.

So in this case, if the object has a mass of 8 solar masses that is enough to form a black hole, and also cannot be seen, all of this indicates that the object we are talking about is a black hole.

It should be mentioned that although these objects do not emit light, because it cannot escape due to the immense force of gravity, black holes can be detected by a type of radiation emitted on their event horizon due to quantum effects called Hawking radiation .

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2 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$

velocity is constant, a = 0

$m g sin\theta - F_{rope} =0$

$F_{rope} = m g sin\theta$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0$

$F_{rope}= 102.73\ N$

b) now, when acceleration, a = 0.135 m/s²

$F_{rope}-m g sin\theta = ma$

velocity is constant, a = 0.135 m/s₂

$F_{rope} = m g sin\theta+ma$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135$

$F_{rope}= 112.19\ N$

7 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$