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Whitepunk [10]
3 years ago
5

The formula shown below is used to calculate the energy released when a specific quantity of fuel is burned. Calculate the energ

y released by a burning sample of fuel if the temperature of 100g of water was raised by 20 °C. (Specific heat capacity of water = 4200J/kg °C)
Physics
1 answer:
olya-2409 [2.1K]3 years ago
8 0

Answer: 8400 J

Explanation:

The formula referenced in the question is:

Q=m. c. \Delta T  

Where:

Q  is the thermal energy

m=100g \frac{1 kg}{1000 g}=0.1 kg is the mass  of the water sample

c=4200 \frac{J}{kg\°C}  is the specific heat capacity of  water

\Delta T=20\°C  is the variation in temperature

Solving:

Q=(0.1 kg)(4200 \frac{J}{kg\°C})(20\°C)  

Q=8400 J  This is the thermal energy released

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As viewed from above in this picture, what direction will the current be in the coil of wire that will cause the loop to rotate
Gala2k [10]

Answer:

When viewed from above, the current in the coil should point towards the top-right corner of the picture.

Explanation:

The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.

Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.

With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.

Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current I. Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field B. The direction of the right thumb should now point in the direction of the force on the wire F.)

Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.

It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.

7 0
3 years ago
An 1800 kg helicopter rises with an upward acceleration of 2.0 m/s?. What lifting force is supplied by its rotating blades?
Viktor [21]

Answer:

Lifting force, F = 21240 N

Explanation:

It is given that,

Mass of the helicopter, m = 1800 kg

It rises with an upward acceleration of 2 m/s². We need to find the lifting force  supplied by its rotating blades. It is given by :

F = mg + ma

Where

mg is its weight

and "ma" is an additional acceleration when it is moving upwards.

So, F=1800\ kg(9.8\ m/s^2+2\ m/s^2)

F = 21240 N

So, the lifting force supplied by its rotating blades is 21240 N. Hence, this is the required solution.

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3 years ago
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AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0
3 years ago
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