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cluponka [151]
3 years ago
15

True or False. Can Metalloids conduct electricity under certain conditions?

Physics
1 answer:
EleoNora [17]3 years ago
4 0

Answer:

Conduction: Some metalloids, such as silicon and germanium, can act as electrical conductors under the right conditions, thus they are called semi-conductors. Luster: Silicon for example appears lustrous, but is not malleable or ductile (it is brittle - a characteristic of some nonmetals).

Explanation:

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In what sense is energy from coal actually solar energy
Jobisdone [24]

Explanation:

The energy stored in coal is directly form the sun which is solar energy. This energy are preserved in carbon atoms and they originate from the life processes of a plant.

  • Plants carry out photosynthesis and produce their energy by this process.
  • Green plants have special adapted structures by which they trap solar energy from the sun and use it to produce their own food.
  • The food is used by plant for nutrition and life activities.
  • In essence the solar energy is converted to chemical energy in plants.
  • To form coal, when plants are rapidly buried their organic matter is preserved from decay by combination with oxygen.
  • More burial preserves the complex molecules that have stored during photosynthesis.
  • The carbon eventually aggregates for coal.

learn more:

Coal brainly.com/question/10055728

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5 0
3 years ago
Explain why objects moving in fluids must have special shapes ?
Alexus [3.1K]

Explanation:

Fluids exert both drag and lift forces on moving objects.  Drag is the frictional force opposing motion.  Lift is the force perpendicular to motion.

Some objects, like parachutes, are designed with large cross sectional areas to increase drag force.  Usually though, objects are designed to minimize drag force.  It's why cars, planes, and boats have sleek shapes.

Airplane wings have shapes called airfoils that generate lift.  It's what makes them fly.  The same shape is found in racecar spoilers.  These spoilers use lift force to push down on the rear tires, increasing traction.

8 0
3 years ago
A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. 11-48).A toy train
masya89 [10]

Answer:

0.166 rad/s

Explanation:

See attachment for calculations

5 0
3 years ago
Read 2 more answers
select the correct scientific notation form of this numeral 541 5 * 10^2 5.4 * 10^2 5.4 * 10^2 5.0 * 10^2
USPshnik [31]
To find the scientific notation, you need to divide at the decimal by the power of 10. So since there are 2 powers of 10, what you want to do is move the decimal 2 places to the left which will give you: .054
7 0
3 years ago
A spaceship hovering over the surface of Venus drops an object from a height of 17 m. How much longer does it take to reach the
Paraphin [41]

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion  that relates the height as follow:

h=v_{0} t+\frac{gt^{2}}{2}

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

h=\frac{gt^{2}}{2}

We know the height h of the spaceship hovering, and the gravity of Venus is g=8.87\frac{m}{s^{2}}. Substituting this values in the equation h=\frac{gt^{2}}{2}:

17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth g=9.81\frac{m}{s^{2}}.

t=\sqrt{\frac{2(17m)}{9.81\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{9.81\frac{m}{s^{2} } } }=1.86s

8 0
3 years ago
Read 2 more answers
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