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Tanya [424]
3 years ago
14

A particle with charge 3.20×10−19 c is placed on the x axis in a region where the electric potential due to other charges increa

ses in the +x direction but does not change in the y or z direction.part a the particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, increasing its kinetic energy by 1.60×10−18 j . in what direction and through what potential difference vb−va does the particle move?
Physics
1 answer:
lys-0071 [83]3 years ago
8 0

Answer:

-5 V

Explanation:

The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value

vb−va

is negative.

We can calculate the potential difference between the two points by using the law of conservation of energy:

\Delta K+ \Delta U=0\\\Delta K + q\Delta V=0

where:

\Delta K=+1.6\cdot 10^{-18} J is the change in kinetic energy of the particle

q=3.2\cdot 10^{-19} C is the charge of the particle

\Delta V =V_b-V_a is the potential difference

Re-arranging the equation, we can find the value of the potential difference:

\Delta V=V_b-V_a = -\frac{\Delta K}{q}=-\frac{1.6\cdot 10^{-18} J}{3.2\cdot 10^{-19} C}=-5 V

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On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed som
k0ka [10]

Answers:

a) 222.22 m/s

b) 800.00 km/h

Explanation:

The speed of a wave is given by the following equation:

v=f \lambda

Where:

v is the speed

f=\frac{1}{T} is the frequency, which has an inverse relation with the period T=1 h

\lambda=800 km is the wavelength

Solving with the given units:

v=\frac{1}{T}\lambda

v=\frac{1}{1 h}800 km

v=800.00 km/h This is the speed of the wave in km/h

Transforming this speed to m/s:

v=800.00 \frac{km}{h} \frac{1 h}{3600 s} \frac{1000 m}{1 km}

v=222.22 m/s This is the speed of the wave in m/s

5 0
3 years ago
I can fly but have no wings. I can cry but I have no eyes. Wherever I go, darkness follows me. What am I?
nlexa [21]

cloud?? thats prob wrong lol

7 0
3 years ago
Read 2 more answers
Describe a real-world example of the law of conservation of momentum.
mario62 [17]
Imagine a car crash. A car coming at a high speed has a head on collision with a car at rest. When the car makes impact, it will move the other car with it at a slower speed then it was travelling at. In this case, the velocity decreased since the car slowed down, but the mass increased since there are now two cars moving. Momentum was conserved because the change in mass accounts for the loss of velocity.
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3 years ago
Which of the following has the most momentum?
IrinaK [193]

Answer:you riding your bike at 12m/s

Explanation: this is because momentum P = mass x velocity. With a bigger mass and a velocity of about 12m/s, you really have a great momentum.

6 0
3 years ago
Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

8 0
3 years ago
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