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miv72 [106K]
4 years ago
12

Something that allows electrons to flow easily through it is called

Physics
2 answers:
zaharov [31]4 years ago
6 0
A material that allows electrons to flow easily through it is called a conductor.
SCORPION-xisa [38]4 years ago
4 0
A closed path through<span> which electrons </span>flow<span>. conductor. A material that </span>allows electrons<span> to move </span>easily through<span> it. current. </span>
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A 300g ball and a 100g ball are dropped from a tower. Both balls are the same size and their is no air resistance. Which one hit
kifflom [539]

Answer: The balls would hit the ground at the same time.

Explanation: Since there is no air resistance, we would put Galileo's experiment into motion. Galileo once performed an experiment of dropping two items, with different masses, from the tower of Pisa. Since there was no air resistance, the balls hit the ground at the same time. In this problem, the balls are "free-falling." Freefall is a term used in Physics to describe the motion of a falling object experiencing only the acceleration due to gravity. g=9.8 m/s^2 (acceleration due to gravity.) 9.8 will always be the acceleration in a free fall event like this with no air resistance, since Gravity remains constant.

Newton's second law of motion states that F=ma, (force is equal to mass times acceleration), but since Newton also states that f=mg, we can conclude that ma=mg. Stating that no matter what the mass of the object is, both of the objects will fall at the same time with the same velocity.

3 0
3 years ago
The Great Lakes region has both biotic and abiotic components. Abiotic components include water, temperature, rainfall, and geol
Andrew [12]

the answer is C. all of the biotic and abiotic components of the great lakes region

3 0
4 years ago
Read 2 more answers
A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground. Calculate the velocity of the acorn
AleksAgata [21]

Answer:

The velocity and translational kinetic energy of the acorn when hitting the ground are approximately 19 meters per second and 3 joules, respectively.

Explanation:

Let suppose that the acorn is a conservative system. By Principle of Energy Conservation, we understand that initial potential gravitational potential energy (U_{g}), in joules, which is related to initial height above the ground, is equal to the final translational kinetic energy (K), in joules, related to the instant just before hitting the ground. Let suppose that ground has a height of zero. That is:

U_{g} = K (1)

m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2} (1b)

Where:

m - Mass, in kilograms.

g - Gravitational acceleration, in meters per square second.

h - Height, in meters.

v - Speed, in meters per second.

If we know that m = 0.017\,kg, g = 9.807\,\frac{m}{s^{2}} and h = 18.5\,m, then the velocity and the translational kinetic energy of the acorn just before hitting the ground is:

m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}

v = \sqrt{2\cdot g \cdot h}

v \approx 19.049\,\frac{m}{s}

K = \frac{1}{2}\cdot m\cdot v^{2}

K = 3.084\,J

The velocity and translational kinetic energy of the acorn when hitting the ground are approximately 19 meters per second and 3 joules, respectively.

8 0
3 years ago
On the basis, a given figure, calculate rfficiency of the lever. ​
daser333 [38]

Explanation:

here,load=100N

effort=200N

distance travelled by load=2cm

distance travelled by effort=10cm

Now, output work=L*Ld

=100*2

=200g

Again,

input work =E*Ed

200*10

2000j

again, efficiency = output work/input work*100%

200/2000*100%

=10%

please don't forget to write symbol

4 0
3 years ago
Two stationary point charges of +60.0 uC and +50.0 uC exert a repulsive force
joja [24]

Answer:

The value is r =  0.39279 \ m

Explanation:

From the question we are told that

     The first point charge is q_1 = 60.0 \mu C  =  60  *10^{-6} \ C

      The second point charge is  q_2 = 50 \mu C  =  50 *10^{-6} \  C

     The repulsive force exerted is  F  =  175 \  N

Generally the repulsive force  exerted is mathematically represented as

     F =  \frac{k * q_1 * q_2 }{r^2}

Here k is the coulomb constant with a value  k  = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

So

       r =  \sqrt{\frac{ k *  q_1 * q_2 }{F} }

=>    r =  \sqrt{\frac{ 9*10^9 * 60 *10^{-6} * 50 *10^{-6}}{175} }

=>    r =  0.39279 \ m

8 0
3 years ago
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