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Travka [436]
4 years ago
7

Starting friction is the force caused by tiny bonds forming between two surfaces.

Physics
1 answer:
just olya [345]4 years ago
4 0
True because the length along a path between two points of friction forces that resists motion.
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In the heat equation, what does c represent?
padilas [110]

Answer:

specific heat of the substance

4 0
3 years ago
Airplanes moves in blank dimensions through. <br><br> What is Blank
raketka [301]
Blank is something empty lol
6 0
3 years ago
Read 2 more answers
A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi
zzz [600]

Answer:

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

Explanation:

We know that the electric force equation is:

F=k\frac{q_{1}*q_{2}}{r^{2}}

  • k is the electric constant 9*10^{9} Nm^{2}/C^{2}
  • r is the distance between the particles
  • q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}

F_{31}=k\frac{q*2q}{r_{31}^{2}}

F_{31}=k\frac{2q^{2}}{r_{31}^{2}}

r(31) is the distance between 3 and 1

2. Now,  let's find the electric force between the third particle and the second particle.

F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}

F_{32}=k\frac{q*(-q)}{r_{32}^{2}}

F_{32}=-k\frac{q^{2}}{r_{32}^{2}}

r(32) is the distance between 3 and 2.

Now, r_{31}+r_{32}=2a or r_{32}=2a-r_{31}

The net force must be zero so:

F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex]   [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}

We just need to solve it for r(31)

r_{31}^{2}=2(2a-r_{31})^{2}

r_{31}^{2}=2(2a-r_{31})^{2}

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

Therefore the distance from the origin will be:

r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}  

I hope it helps you!        

                 

 

     

4 0
4 years ago
A 0.453 kg pendulum bob passes through the lowest part of its path at a speed of 2.58 m/s. What is the tension in the pendulum c
harkovskaia [24]

Answer with Explanation:

Mass of pendulum bob, m=0.453 kg

Speed, v_1=2.58 m/s

a.r=75.1 cm=75.1\times 10^{-2}m=0.751 m

1cm=10^{-2} m

Tension in the pendulum cable is given  by

Tension=Centripetal force+force due to gravity

T=\frac{mv^2}{r}+mg

Where g=9.8 m/s^2

Substitute the values

T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8

T=8.45 N

b.When the pendulum reaches its highest point,then

Final velocity, v_2=0

According to law of conservation of energy

mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2

gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2

h_1=0

Substitute the values

9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2

3.3282=9.8h_2

h_2=\frac{3.3282}{9.8}=0.34 m

The angle mad  by cable with the vertical=cos\theta=\frac{0.751-0.34}{0.751}=0.55

\theta=cos^{-1}(0.55)=56.6^{\circ}

c.When the pendulum reaches at highest point then

Acceleration, a=0

Therefore, the tension  in the pendulum cable

T=mgcos\theta

Substitute the values

T=0.453\times 9.8cos56.6

T=2.4 N

8 0
3 years ago
A boat moving at 7.8 km/hr relative to the water is crossing a river 3.5 km wide in which the current is flowing at 2.8 km/hr. A
Free_Kalibri [48]

Answer:\theta =21.03^{\circ}

Explanation:

Given

velocity of boat with respect to river =7.8 km/hr

velocity of river is 2.8 km/hr

River is 3.5 km wide

Suppose boat leaves at an angle of \theta  with vertical such that it reaches exactly opposite end of initial Point

Therefore

7.8\sin \thetacomponent will balance the river Flowso that sin component helps to cross the river

7.8\sin \theta =2.8

\sin \theta =0.3589

\theta =sin^{-1}(0.3589)

\theta =21.03^{\circ}

8 0
3 years ago
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