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BigorU [14]
3 years ago
13

The electron drift speed in a 1.2-mm-diameter gold wire is 5.00 10-5 m/s. how long does it take 1 mole of electrons to flow thro

ugh a cross section of the wire?
Physics
1 answer:
Vlada [557]3 years ago
7 0
Twenty four minutes per second

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The ability of atoms to attract electrons is referred to as
dexar [7]

Answer: electronegativity

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself.  

The size of an atom decreases as we move across the period because the electrons get added to the same shell and the nuclear charge keeps on increasing. Thus the electrons get more tightly held by the nucleus.

As, the size of an element decreases, the valence electrons come near to the nucleus. So, the attraction between the nucleus and the shared pair of electrons increases and thus the electronegativity increases.

8 0
4 years ago
Read 2 more answers
17. A 120 m long train is moving in a direction with speed 20 m/s. A train B moving with 30
nydimaria [60]
Answer is D.

Speed:
Use relative speed to simplify the situation. Since the trains are moving in opposite directions, you can add the speeds and pretend the first train is stationary (moving at 0m/s) and the second train is moving at 50m/s.

Distance:
The front of the second train needs to travel 120m to get from the front to the back of the first train. When the front of the second train is at the back of the first train, the back of the second train is still 10m in front of the first train. The back therefore has to travel 130m to clear the first train. The total distance over which the trains are overlapping in this scenario is therefore 120 + 130 = 250m.

You have speed and you have distance so now just calculate time:

v = d / t
50 = 250 / t
t = 5s
4 0
3 years ago
A rubber band has potential energy of 5 J. If the spring constant of the rubber band is 50 N/m, what is the displacement of the
Valentin [98]
To determine the displacement, since we are given the potential energy, we use the equation for potential energy. For a spring, it is one-half the product of the spring constant and the square of the displacement. We do as follows:

PE = kx^2/2
5 Nm = 50N/m (x^2)
x = 0.32 m

Therefore, the displacement would be 0.32 m.
7 0
4 years ago
Read 2 more answers
A drag racer crosses the finish line doing 212 mi/h and promptly deploys her drag chute. (A.) what force must the drag chute exe
Svet_ta [14]

initial speed of the racer is given as

v_i = 212 mi/h

v_i = 212*\frac{1609}{3600} = 94.75 m/s

after applied force the final speed is given as

v_f = 45 mi/h

v_f = 45 * \frac{1609}{3600} = 20.11 m/s

now during this speed change the racer will cover total distance 185 m

so here we will use kinematics

v_f^2 - v_i^2 = 2 a d

20.11^2 - 94.75^2 = 2*a*185

a = -23.2 m/s^2

now the force that chute will exert on the racer will be given as

F = ma

F = 898* 23.2

F = 2.1* 10^4 N

B) here following is the strategy for solving it

1. first we used kinematics to find the acceleration of the car

2. then we used Newton's II law (F = ma) to find the force

4 0
3 years ago
F the mass of both vehicles were doubled, how would the final velocity and the change in kinetic energy be affected? (Select all
Ira Lisetskai [31]

Answer:

a) True. Speeds are reduced by half .  

e) True. kinety energy  is reduced by half

Explanation:

In order to analyze the statements you have we have to raise the problem solution.

If we fear a constant force when double the mass we can see the result in acceleration with Newton's second law

    F = m a₁

    F = (2m) a₂

    m a1 = 2m a₂

    a₂ = ½ a₁

We see that when I double the mass the acceleration is reduced by half

With kinematics we see that there is a linear relationship between speed and acceleration

    v_{f} = v₀ + at

For the same initial speed and the same time if the acceleration is reduced by half the speed is reduced by half.

Therefore, by doubling the mass the speed reduces by half

kinetic energy is

     K = ½ m v²

     K = ½ (2m) (v/2)²

     K = (½ m v²) /2

     K = Ko / 2

Now we can review the claims

a) True. Speeds are reduced by half

b) False. We saw that they change their value

c) False. They reduce not increase

d) False. We saw that the kinetic energy is reduced

e) True. It is reduced by half

f) False. We saw that change

4 0
3 years ago
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