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Wittaler [7]
3 years ago
5

Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =

2.0 c and q2 = 6.0 c, separated by klein's total driving distance. a third charge, q3 = 4.0 c, is placed on the line connect- ing q1 and q2. how far from q1 should q3 be placed for q3 to be in equilibrium
Physics
1 answer:
Semenov [28]3 years ago
6 0
For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
r_1+r_2=L km
The second equation is going to the total force acting on the charge q3:
F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}
Let's solve this for r_1^2:
0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0
Now we have a quadratic equation with following parameter:
a=2\\ b=2L\\ c=-L^2
We know that two solutions are:
r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\
We need a positive solution. When we plug in all the numbers we get:
r_1=0.915\cdot 10^6$km

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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude
Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

\omega_f = \omega_i + \alpha t

where

\omega_i = 0.300 rev/s is the initial angular velocity

\alpha = 0.895 rev/s^2 is the angular acceleration

Substituting t = 0.200 s, we find

\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s

Let's now convert it into rad/s:

\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

r=\frac{0.800}{2}=0.400 m

And so now we can find the tangential speed at t = 0.200 s:

v=\omega_f r =(3.01)(0.400)=1.2 m/s

2) 2.25 m/s^2

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

a_t = \alpha r

where \alpha is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

a_t = (5.62)(0.400)=2.25 m/s^2

3) 3.6 m/s^2

The centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

a=\frac{1.2^2}{0.400}=3.6 m/s^2

7 0
3 years ago
804 n of force are applied to a 51.7 kg. What is the acceleration that the object experiences?
Andreyy89

We can use Newton II here  (where F=m*a), that F is the net (or resultant) force on the object, m is the mass of the object and a is the acceleration the object experiences.

This means, in this case there would be no friction and absolutely no other force which gives a component in the plane of motion, only then can you assume that F=804N.

Now using F= m*a

804 = 51.7*a

Therefore a = 804/51.7 = 15.55 m/s²


7 0
2 years ago
PLEASE EXPLAIN AND YOU WILL GET BRAINLIST Ms. R is curious if the type of gasoline she uses in her car affects how many miles sh
inessss [21]
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8 0
3 years ago
a 150 N force is used to pull a wooden box across a wooden surface at a constant velocity. what is the mass of the box?
IRINA_888 [86]

Answer:

The mass of the box:

m =  60 kg

Explanation:

Given:

F = 150 N

g = 10 m/s²

_________

m - ?

Coefficient of friction wood on wood:

μ = 0.25

Friction force:

F₁ = μ*m*g

Newton's Third Law:

F = F₁

F = μ*m*g

The mass of the box:

m = F / ( μ*g) = 150 / (0.25*10) =  60 kg

7 0
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