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FromTheMoon [43]
3 years ago
5

How is an emulsion different from other colloids?

Physics
1 answer:
Usimov [2.4K]3 years ago
8 0

Answer:

The difference between colloid and emulsion is that a colloid can form when any state of matter (solid, liquid or gas) combine with a liquid whereas an emulsion has two liquid components which are immiscible with each other

Explanation:

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depends on wind

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Which equation could not be used to determine straight line acceration?
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the answer for the question is the last option

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The unit for work can be written as __________ or as the nickname ______
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N or joule cuz joule is newton’s
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Read 2 more answers
Cars cross a certain point on the highway in accordance with a Poisson process with rate = 3 per minute. If Al runs across the h
Svetradugi [14.3K]

Answer:

The probability is 0.2212

Solution:

As per the question:

Poisson rate, \lambda = 3/min = \frac{3}{60}\ s = 0.05\ s

Now,

Let

X: time of waiting for the next vehicle on the highway

Now,

To find the probability of the next vehicle to arrive after 10 s

The probability distribution function is given by:

f(x) = \lambda e^{- \lambda x}

Now,

P(X < x) = 1 - e^{- \lambda x}

P(X \leq x) = 1 - e^{- 0.05 x}

For X> 0,

P(X > x) = e^{- \lambda x}

P(X < 5) = 1 - e^{- 5\lambda} = 1 - e^{- 0.25} = 1 - 0.7788 = 0.2212

4 0
3 years ago
A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

6 0
3 years ago
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