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Natalka [10]
3 years ago
11

A current of 1.70 A flows in a wire. How many electrons are flowing past any point in the wire per second

Physics
1 answer:
castortr0y [4]3 years ago
8 0

Answer:

1.0625×10¹⁹ electrons

Explanation:

From the question,

Using,

Q = It........................ Equation 1

Where Q = charge, I = current flowing in the wire, t = time (seconds)

Given: I = 1.70 A, t = 1 Seconds.

Substitute these values into equation 1

Q = 1.7(1)

Q = 1.7 C.

But,

1.6×10⁻¹⁹ C = 1 electron

Therefore,

1.7 C = 1.7/(1.6×10⁻¹⁹) electrons

= 1.0625×10¹⁹ electrons

Hence, 1.0625×10¹⁹ electrons flows through the wire per seconds.

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There are four springs stretched by the same mass.
brilliants [131]

Spring C stretches 100 cm.

Explanation:

The spring constant is simply the stiffness of the spring. The higher the spring constant the more stiff the spring is.

Spring constant shows the force needed to stretch a spring from it's equilibrium position. If a material requires more force to cause it to stretch, it will have a high spring constant.

          According to hooke's law "the force needed to extended an elastic material is directly proportional to its extension"

          F = ke

   k is the spring constant

   e is the extension

We see that the spring that stretches by 100 is the less stiff compared to other springs. It has the smallest spring constant.

Learn more;

Force brainly.com/question/8882476

#learnwithBrainly

8 0
3 years ago
When must a psychological researcher debrief human test subjects?
myrzilka [38]
<span>Psychological researchers must debrief human test subjects </span><span>at the end of every experiment.

The current code of ethics in p</span>sychological research states that researchers absolutely must debrief human test subjects at the end of every study regardless or whether or not harm or deception was involved.

Debriefing a subject after a study is an essential opportunity for the researcher to explain the purpose and aim of the study to the subject, make sure the subject is not harmed or mentally disturbed, clarify why deception was used (if deception was involved) and overall, to clarify any questions or doubts the subject might have.
3 0
3 years ago
What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×
xxMikexx [17]

Answer:

9.82 × 10^{-35} Hz

Explanation:

De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:

λ = \frac{h}{mv}

where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.

Given that: h = 6.63 ×10^{-34} Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;

λ = \frac{h}{mv}

  = \frac{6.63*10^{-34} }{2.5*2.7}

 = \frac{6.63 * 10^{-34} }{6.75}

 = 9.8222 × 10^{-35}

The wavelength of the object is 9.82 × 10^{-35} Hz.

4 0
3 years ago
Select all the correct answers.
myrzilka [38]

<em>Anything</em> that's dropped through air is somewhat affected by air resistance.  But, out of that list, the leaf and the balloon are the items that will be affected by air resistance enough so that you can plainly see it.

If you spend some time thinking about it, you can kind of understand why airplane wings and boat propellers are shaped more like leafs and balloons than like bricks and rocks.

3 0
3 years ago
The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

3 0
3 years ago
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