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anygoal [31]
2 years ago
8

Two point charges are on the y-axis. A 3.0 µC charge is located at y = 1.15 cm, and a -2.28 µC charge is located at y = -2.00 cm

. (a) Find the total electric potential at the origin. g
Physics
1 answer:
Ghella [55]2 years ago
3 0

Answer:

Total electric potential, V=1.32\times 10^6\ volts

Explanation:

It is given that,

First charge, q_1=3\ \mu C=3\times 10^{-6}\ C

Second charge, q_2=-2.28\ \mu C=-2.28\times 10^{-6}\ C

Distance of first charge from origin, r_1=1.15\ cm=0.0115\ m

Distance of second charge from origin, r_2=2\ cm=0.02\ m

We need to find the total electric potential at the origin. The electric potential at the origin is given by :

V=\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}

V=k(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2})

V=9\times 10^9(\dfrac{3\times 10^{-6}}{0.0115}+\dfrac{-2.28\times 10^{-6}}{0.02})

V = 1321826.08 V

or

V=1.32\times 10^6\ volts

So, the total electric potential at the origin is 1.32\times 10^6\ volts. Hence, this is the required solution.

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After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48
masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

15min*\frac{1hr}{60min}=0.25hr

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

V=\frac{distance}{time}

the plane traveled a distance to the north of 48km so the velocity is:

V=\frac{48km}{0.25hr}

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

V_{x}=42km/hr*cos(-19^{o} )=39.71 i

and

V_{y}=42km/hr*sin(-19^{o} )=-13.67 j

So the velocity of the wind can be expressed as a vector as:

V_{wind}=(39.71i - 13.67j)km/hr

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

V_{plane x}=V_{plane/wind x}+V_{wind x}

V_{plane/wind x}=V_{plane x}-V_{wind x}

V_{plane/wind x}=(0-39.71 i)km/hr

V_{plane/wind x}= -39.71 i km/hr

and

V_{plane y}=V_{plane/wind y}+V_{wind y}

V_{plane/wind y}=V_{plane y}-V_{wind y}

V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr

V_{plane/wind x}= 205.67 j km/hr

So the velocity of the plane with respect to the wind can be rewritten as:

V_{plane/wind x}= (-39.71i + 205.67 j) km/hr

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

speed=\sqrt{(-39.71)^{2}+(205.67)^{2}  }

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0
3 years ago
The angular momentum about the center of the planet and the total mechanical energy will be conserved regardless of whether the
ValentinkaMS [17]

Answer:

True

Explanation:

The angular momentum around the center of the planet and the total mechanical energy will be preserved irrespective of whether the object moves from large R  to small R. But on the other hand the kinetic energy of the planet will not be conserved because it can change from kinetic energy to potential energy.

Therefore the given statement is True.

5 0
3 years ago
If you double the voltage in a circuit and reduce the resistance by a factor of four, what will happen to the current?
Gennadij [26K]

Answer:

It will increase by a factor of 8.

Explanation:

By Ohm's Law, we can relate current, voltage and resistance. It is expressed as V=IR. That is, there is a direct relationship between voltage and resistance and voltage and current.

V = IR

V1/2V1 = I1R1 / I2R1/4

1/2 = 4I1/I2

I2 = 8I1

3 0
3 years ago
WILL MARK BRAINLIEST! The driver of a pickup truck going 120 km/h applies the brakes, giving the truck a uniform deceleration of
Rudik [331]
First put the speed in m/s.  120km/h = 33.33m/s.  Now the position function is the integral of velocity, and velocity is in turn the integral of acceleration.  The velocity is:
v= \int\limits^{} _ {} {-5} \, dt =-5t+33
Now we integrate this expression to get the position. The constant of integration will be the distance the truck travels.
s= \int\limits^{} {-5t+33} \, dx =- \frac{5}{2}t^2+33.33t-35=0
Here we set the distance, 35m as negative because I assumed the stopping point of the truck is the origin.  Putting t=0 shows it starts at -35m. 
Now solve the following equation for the time, t using the quadratic equation:
s=- \frac{5}{2}t^2+33.33t-35=0
and choose the  value t=1.149s
7 0
3 years ago
Conductors can be charged by, while insulators cannot.
ikadub [295]

Answer:

Electrons

Explanation:

Conductors can be charged by electrons, while insulators cannot. Insulators inhibit the flow of electrons; conductors allow free flow of electrons.

7 0
2 years ago
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