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iogann1982 [59]
3 years ago
13

Problem 1: A positively charged particle Q1 = +35 nC is held fixed at the origin. A second charge Q2 of mass m = 8.5 micrograms

is floating a distance d = 25 cm above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth. A) What is the sign of charge on Q2? B) Calculate the magnitude of Q2 in units of nanocoulombs.
Physics
1 answer:
tigry1 [53]3 years ago
3 0

Answer:This problem has been solved! See the answer. Two charged particles, Q 1 = + 5.10 nC and Q 2 = -3.40 nC, are separated by 40.0 cm. (a) What is the ... has not been graded yet. (b) What is the electric potential at a point midway between the charged particles? (Note: Assume a reference level of potential V = 0 at r = infinity .)

Explanation:

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Alligators and other reptiles don't use enough metabolic energy to keep their body temperatures constant. they cool off at night
drek231 [11]
To compute for the heat, Q needed to be absorbed or released,  we need

Q = mc\Delta T

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Read 2 more answers
A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
alexgriva [62]
The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
S= \frac{1}{2} a t^2
where a is the acceleration. Using the data of the problem, we can find a:
a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W

b) power output at t=4.0s 
The velocity at t=4.0 s is
v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J

and so the power output is
P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W

c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J

and so the power output is
P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W
8 0
3 years ago
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