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mart [117]
3 years ago
10

Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magni

tude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.5 μC at x = 0.40 m, y = 0?
Physics
1 answer:
IgorLugansk [536]3 years ago
3 0
PDF attached.....................
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If a town was 90 miles away and you travel at 45 mph how long would it take to get there
notka56 [123]

Answer:

t = 2 hours

Explanation:

Given that,

Distance of the town, d = 90 miles

Speed, v = 45 mph

We need to find the time to get there. The speed of an object is given by :

v=\dfrac{d}{t}

Where

t is time

t=\dfrac{d}{v}\\\\t=\dfrac{90}{45}\\\\t=2\ h

So, the required time is 2 hours.

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3 years ago
During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel
BigorU [14]

Answer:

47.4 m

Explanation:

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In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

t=\frac{6.22}{2}=3.11 s

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

s=ut+\frac{1}{2}gt^2

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

g=9.8 m/s^2 is the acceleration of gravity (taking downward as positive direction)

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s=\frac{1}{2}(9.8)(3.11)^2=47.4 m

7 0
3 years ago
At which location would you expect the LOWEST TEMPERATURE?
MissTica

Answer:

c

Explanation:

3 0
2 years ago
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trapecia [35]
It holds the atoms together (aka your last option)
8 0
3 years ago
Using a force of 28.0 Newton, a student pulls a 70.0 Newton weight along the tabletop for a distance of 15.0 meters in 3.0 secon
Musya8 [376]

Answer:

140 watt

Explanation:

We are given that

Force applied by student ,F=28 N

Weight pulled  by students=70 N

Displacement,s=15 m

Time=3 s

We have to find the power developed  by the student.

Work done=w=F\times s

Work done by the student=28\times 15=420 J

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Using the formula

Power=\frac{420}{3}=140watt

Hence, the power developed by the students=140 watt

8 0
3 years ago
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