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2 years ago

9

A 1.0-c point charge is 15 m from a second point charge, and the electric force on one of them due to the other is 1.0 n. what i

s the magnitude of the second charge?

1 answer:

Fofino [41]2 years ago

4 0

The answer is 25nC !!!

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Question 8 (4 points)

katrin2010 [14]

Answer:

The answer to your question is:

Explanation:

Data

Duane Albert

d = 5 m ; v = 3 m/s v = 4.2 m/s

a) b)

Duane's Albert's

d = 5 + (3)t d = 4.2t

d = 5 + 3t

c) 5 + 3t = 4.2t

4.2t - 3t = 5

1.2t = 5

t = 4.17 s

d)

Duane's

d= 5 + 3(4.17)

d = 17.51 m

Alberts

d = 4.2(4.17)

d = 17.51 m

4 0

2 years ago

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A ball rolls 6.0 meters as its speed changes from 15 meters per second to 10 meters per second. What is the average speed of the

antoniya [11.8K]

Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m

The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
12a = -125
a = -10.4167 m/s²

The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s

The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s

Answer: 12.5 m/s

6 0

2 years ago

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

a.

$I = m*r^2$

$m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg$

$r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m$

$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

$I=2.77x10^{-8} kg*m^2$

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

$T=2\pi *\sqrt{\frac{I}{K}}$

Solve to K'

$K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}$

$K=4.37 x10^{-6} N*m$

3 0

2 years ago

A na+ ion moves from inside a cell, where the electric potential is -72 mv, to outside the cell, where the potential is 0 v. wha

Vlada [557]

The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
$\Delta U = q \Delta V = q (V_f - V_i)$
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge: $q=1.6 \cdot 10^{-19} C$ , and Vf and Vi are the final and initial voltages.

Substituting the numbers, we find:
$\Delta U = (1.6 \cdot 10^{-19}C)(0 V-(-0.072 V))=1.15 \cdot 10^{-20} J$

7 0

2 years ago

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A rabbit is trying to cross the street. Its velocity v as a function of time t is given in the graph below where

Amiraneli [1.4K]

Answer:

2.5

Explanation:

8 0

2 years ago