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olasank [31]
3 years ago
12

In most cases, magma differentiation (a.k.a. fractional crystallization produces magma with higher ___________ content than the

parent magma.
Physics
1 answer:
Delicious77 [7]3 years ago
7 0
<span>In most cases, magma differentiation (a.k.a. fractional crystallization produces magma with higher silica content than the parent magma. Fractional crystallization removes early formed minerals in magma. The liquid that does not react to the process remains in the magma. </span>
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A caterpillar climbs up a one-meter a wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the
valkas [14]

Answer:

1.5 m is ur answer

Explanation:

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3 years ago
Sam is riding his bike in the park when he sees a soccer ball rolling toward him. He applies his brakes to slow down. Once he st
nlexa [21]

When Sam presses the brake lever, a pair of rubber shoes  clamps onto the metal inner rim of the front and back wheels. As the brake shoes rub  against the wheels, friction is caused and  the kinetic energy possessed by the vehicle is converted into heat which slows down the vehicle.

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3 years ago
Amber asked her roommate to turn down the radio because she was trying to study. Her roommate had increased the volume from a vo
Blizzard [7]

Answer:

Difference threshold or also Just Noticeable Difference

Explanation:

The above mentioned case between room mates, where one room mate was able to detect a minute change in volume shows an instance of the difference threshold.

Difference threshold can be defined as stimulation at its minimum level that can be detected by an individual almost 50 % of the times.

It is the lowest possible level of sound that is detectable by a person.

That is what happened in the mentioned case that when the volume was increased from 14 to 15, Amber was able to detect it.

7 0
3 years ago
Consider a cloudless day on which the sun shines down across the United States. If 2073 kJ of energy reaches a square meter ( m
Mnenie [13.5K]

The total amount of energy per hour is 2.039\cdot 10^{16} kJ

Explanation:

In this problem we are told that the amount of energy reaching a square meter in the United States per hour is

E_1 = 2073 kJ

The total surface area of the United States is

A=9.834\cdot 10^6 km^2

And converting into squared metres,

A=9.834\cdot 10^6 \cdot 10^6 = 9.834\cdot 10^{12} m^2

Therefore, the total energy reaching the entire United States per hour is given by:

E=AE_1 = (9.834\cdot 10^{12})(2073)=2.039\cdot 10^{16} kJ

Learn more about energy and power:

brainly.com/question/7956557

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5 0
3 years ago
An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the
Zarrin [17]

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

x = v_0 t \frac{1}{2} at^2

Where,

x= Displacement

v_0 = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

L = \frac{1}{2} a t_1 ^2

For the second cart

2L \frac{1}{2} at_2^2

When the tenth car is aligned the length will be 9 times the initial therefore:

9L = \frac{1}{2} at_3^2

When the tenth car has passed the length will be 10 times the initial therefore:

10L = \frac{1}{2}at_4^2

The difference in time taken from the second car to pass it is 5 seconds, therefore:

t_2-t_1 = 5s

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

\frac{1}{2} = (\frac{t_1}{t_2})^2

t_1 = \frac{t_2}{\sqrt{2}}

From the relationship when the car has passed and the time difference we will have to:

(t_2-\frac{t_2}{\sqrt{2}}) = 5

t_2 (\sqrt{2}-1) = 3\sqrt{2}

t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2

Replacing the value found in the equation given for the second car equation we have to:

\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2

Finally we will have the time when the cars are aligned is

18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2

t_3 = 36.213s

The time when you have passed it would be:

20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2

t_4 = 38.172

The difference between the two times would be:

t_4-t_3 = 38.172-36.213 \approx 2s

Therefore the correct answer is C.

4 0
3 years ago
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