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Vlad1618 [11]
3 years ago
15

How long does it take an airplane to fly 1200 miles if it maintains a speed of 300 miles per hour

Physics
1 answer:
sdas [7]3 years ago
5 0

Answer:

The answer is

<h2>4 hrs</h2>

Explanation:

To find the time taken we use the formula

t =  \frac{d}{v}  \\

where

d is the distance covered

v is the velocity

t is the time

From the question

d = 1200 miles

v = 300 mi/hr

We have

t =  \frac{1200}{300}  =  \frac{12}{3}  \\

We have the final answer as

<h3>4 hrs</h3>

Hope this helps you

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25  m/s² is the object's acceleration

m1a1=m2a2

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m2=3 kg

a1= 15 m/s²

a2=?

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Acceleration is a vector variable that describes the rate at which an object changes its velocity.

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The difference between an object with a constant acceleration and one with a constant velocity must be understood. Do not be fooled! If an object's velocity changes, whether it does so by a constant amount or a variable amount, then it is accelerating. Furthermore, something that is travelling at a steady speed is not accelerating.

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1 year ago
The very strong source of radio waves at the center of our galaxy is called.
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The answer is Sagittarius A, a very large black hole.
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3 years ago
100 g of Ice at -10°C is added into a
Andrei [34K]

Answer:

The mass of the juice responsible for melting the ice is 949.043 grams.

Explanation:

By the First Law of Thermodynamics, we understand that juice releases heat to the ice, which turns into water under the assumption that interactions between the ice-juice system and surroundings are negligible and energy processes are done in steady-state. Since juice is done with water, its specific heat will be taken as of the water. The process is described by the following formula:

m_{i} \cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})] + m_{w} \cdot  c_{w}\cdot (T_{4}-T_{3}) = 0 (1)

Where:

m_{i} - Mass of ice, in grams.

m_{w} - Mass of the juice, in grams.

c_{i} - Specific heat of ice, in joules per gram-degree Celsius.

c_{w} - Specific heat of water, in joules per gram-degree Celsius.

L_{f} - Latent heat of fusion, in joules per gram.

T_{1} - Initial temperature of ice, in degrees Celsius.

T_{2} - Melting point of water, in degrees Celsius.

T_{3} - Final temperature of the ice-juice system, in degrees Celsius.

T_{4} - Initial temperature of the juice, in degrees Celsius.

If we know that m_{i} = 100\,g, c_{i} = 2.090\,\frac{J}{g\cdot ^{\circ}C}, c_{w} = 4.18\,\frac{J}{g\cdot ^{\circ}C}, L_{f} = 334\,\frac{J}{g}, T_{1} = -10\,^{\circ}C, T_{2} = 0\,^{\circ}C, T_{3} = 10\,^{\circ}C and T_{4} = 20\,^{\circ}C, then the mass of the juice is:

m_{w} = \frac{m_{i}\cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})]}{c_{w} \cdot (T_{3}-T_{4})}

m_{w} = \frac{(100\,g)\cdot  \left[\left(2.090\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) - 334\,\frac{J}{g} +\left(4.18\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C)  \right]}{\left(4.180\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C)}

m_{w} = 949.043\,g

The mass of the juice responsible for melting the ice is 949.043 grams.

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3 years ago
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GarryVolchara [31]
Yes you were correct the answer was c


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Answer:

This question is incomplete as it lacks options, the options are:

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