a)]An expression for the magnitude of charge transferred Q is Q = W/V.
b) The magnitude of the charge, in coulombs is 1.02 x 10⁻⁹ Coulomb.
c) Number of electrons in this are 63.75 x 10⁸ electrons.
<h3>What are electrons?</h3>
The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.
A balloon is charged until it has a voltage of v=9800V. A student reaches out and is shocked by a spark. The electric force does W= 1x10⁻⁶ J of work transferring the charge.
a) Work done = Charge x potential
W = Q x V
Q = W/V
b)Substitute the values into the expression, we have
|Q| = 1x10⁻⁶/9800
|Q| = 1.02 x 10⁻⁹ Coulomb
c) No of electrons = total charge /charge on electron
n = 1.02 x 10⁻⁹/1.6 x 10⁻¹⁹
n = 63.75 x 10⁸ electrons
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best and most correct answer among the choices provided by the question is<span> A.
If the uncharged object is a conductor, the charged object can attract opposite
charges. </span></span>
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Answer:
H = 109.14 cm
Explanation:
given,
Assume ,
Total energy be equal to 1 unit
Balance of energy after first collision = 0.78 x 1 unit
= 0.78 unit
Balance after second collision = 0.78 ^2 unit
= 0.6084 unit
Balance after third collision = 0.78 ^3 unit
= 0.475 unit
height achieved by the third collision will be equal to energy remained
H be the height achieved after 3 collision
0.475 ( m g h) = m g H
H = 0.475 x h
H = 0.475 x 2.3 m
H = 1.0914 m
H = 109.14 cm
Answer:
See the answers below
Explanation:
We can solve this problem using the principle of energy conservation. That is, the energy is conserved before and after dropping the bag.
For this case we have mechanical energy, which is the sum of the kinetic and potential energies.
where:
Ek = kinetic energy [J] (units of Joules)
Ep = potential energy [J]
In the final Energy (2), there is only potential energy. since when the balloon reaches the maximum height its velocity is zero, that is, there is no kinetic energy.
A)
![m*g*h+\frac{1}{2}*m*v^{2} =m*g*h_{1} \\9.81*50+0.5*(15)^{2}=9.81*h_{1}\\h_{1} = 61.46 [m]](https://tex.z-dn.net/?f=m%2Ag%2Ah%2B%5Cfrac%7B1%7D%7B2%7D%2Am%2Av%5E%7B2%7D%20%3Dm%2Ag%2Ah_%7B1%7D%20%5C%5C9.81%2A50%2B0.5%2A%2815%29%5E%7B2%7D%3D9.81%2Ah_%7B1%7D%5C%5Ch_%7B1%7D%20%3D%2061.46%20%5Bm%5D)
B)
With the value calculated above we can find the acceleration of the balloon.
The distance traveled is the difference between the maximum height and 50 meters.
![x = 61.46-50\\x = 11.46[m]](https://tex.z-dn.net/?f=x%20%3D%2061.46-50%5C%5Cx%20%3D%2011.46%5Bm%5D)
With the following equation of kinematics.
![0 = 15^{2} +2*a*11.46\\a = - 9.816 [m/s^{2} ]](https://tex.z-dn.net/?f=0%20%3D%2015%5E%7B2%7D%20%2B2%2Aa%2A11.46%5C%5Ca%20%3D%20-%209.816%20%5Bm%2Fs%5E%7B2%7D%20%5D)
The negative sign indicates that the acceleration acts downward. That is, in the opposite direction to the movement.
We can use the following equation of kinematics to find the final velocity after 4 seconds.
![v_{f}=v_{o}-a*t\\v_{f}=15-9.816*(4)\\v_{f}=-24.24 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D-a%2At%5C%5Cv_%7Bf%7D%3D15-9.816%2A%284%29%5C%5Cv_%7Bf%7D%3D-24.24%20%5Bm%2Fs%5D)
Now the distance:
![v_{f}^{2} =v_{o}^{2}-2*a*x\\(24.24)^{2} =(15)^{2} -2*9.81*x\\x = 18.48 [m]\\x_{f}=50+18.48 = 68.48 [m]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%20%3Dv_%7Bo%7D%5E%7B2%7D-2%2Aa%2Ax%5C%5C%2824.24%29%5E%7B2%7D%20%3D%2815%29%5E%7B2%7D%20-2%2A9.81%2Ax%5C%5Cx%20%3D%2018.48%20%5Bm%5D%5C%5Cx_%7Bf%7D%3D50%2B18.48%20%3D%2068.48%20%5Bm%5D)
c) Using the following equation of kinematics.
![v_{f}=v_{o}-a*t\\0 = 15-9.81*t\\15=9.81*t\\t = 1.52 [s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D-a%2At%5C%5C0%20%3D%2015-9.81%2At%5C%5C15%3D9.81%2At%5C%5Ct%20%3D%201.52%20%5Bs%5D)
