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4 years ago

What is the x-component of the force on the charge located at x = 8 cm given that q = 1.1 μC in N?

Physics

1 answer:

jonny [76]4 years ago

6 0

Answer:

Answer is = 12.3 x 10^6 N

Explanation:

We need to calculate force on the charge which is in Newtons:

Formula is = K q1q2/r^2

k is constant depeneds on the permitivity of the medium and its value is 9x10^9 Nm^2/C^2

q1 and q2 are charges and r is the distance between these two charges so just by putting values we get F= 9x10^9 x 1.1x10^-6/(8x10^-2)^2 (we get the above answer in newton)

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What was the hiker's total displacement?<br>A.24 km B.4 km north C.14 km northeast D.0 km

kipiarov [429]

We can see it as two movements: one on the north-south axis and one on the west-east axis.

On the north-south Axis, the hiker went 8 km north and 4 km south, so in total the hiker went 4 km north (we can subtract one for the other and the direction remains that which was bigger)

On the west-easth Axis, the hiker went 6 km east and 6 km west, so in total the hiker didn't move on this axis! or, he/she moved but then "came back"

So in total, the hiker only moved4 km north - answer B.

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7 0

3 years ago

Read 2 more answers

A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di

navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

$X(van)=5.65t+154$

$X(driver)=34.4t+\frac{(-2)t^{2} }{2}$

or by rearanging the drivers equation.

$X(driver)=34.4t+t^{2}$

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

$X(van)=X(driver)$

$5.65t+154=34.4t-t^{2}$

$0=t^{2} -(34.4-5.65)t+154$ $0=t^{2} -28.75t+154$

To solve this equation we use the following formulas

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

Where a=1; b=-28.75; c=154

So we get:

$t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s$ $t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s$

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

$V(driver)=V_{0} +at$ }

$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}$ $V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}$

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).

Best of luck

8 0

3 years ago

11. You sit at the outer rim of a Ferris wheel that rotates 2 revolutions per minute (RPM). What would your rotational speed be

aalyn [17]

If you are instead clinging to a position halfway from the center to the outer rim of the Ferris wheel, YOUR ROTATIONAL SPEED WILL STILL BE 2 REVOLUTION PER MINUTE.
This is because, every part of the wheel is moving with the same speed, so it does not matter where you sit on the wheel, the rotation per minute will still be the same. It is just like travelling inside a motor car, it does not matter whether you are sitting in the front passenger seat or at the back, the speed of the car remains the same.

4 0

3 years ago

A ball is thrown at a launching angle of 52 o above the horizontal from one meter above the ground, with a velocity of 18 m/s.

Vladimir79 [104]

i dont really know, im sorry

5 0

3 years ago

If a wave is traveling at a constant speed, and the frequency increases, what would happen to the wavelength?

dsp73

Answer:

It would decrease.

Explanation: