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3 years ago

A tree bumps into a tree with a force of 4500 N. What force does the tree exert on the car?

Physics

1 answer:

madreJ [45]3 years ago

4 0

If one tree bumps into another tree, it's entirely possible
that a car could get away without a scratch.

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Find the resistance of a 0.03 m long copper wire with a radius of .005 m (Area of a circle: π r²).

Bas_tet [7]

Answer:

578feet ygti

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3 years ago

Which of the following is NOT true regarding the far side of the moon:

frutty [35]

It contains no large maria

8 0

3 years ago

when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe

Alinara [238K]

Answer:

Rate of change of area will be $9.796cm^2/min$

Explanation:

We have given rate of change of radius $\frac{dr}{dt}=0.03cm/min$

Radius of the circular plate r = 52 cm

Area is given by $A=\pi r^2$

So $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$

Puting the value of r and $\frac{dr}{dt}$

$\frac{dA}{dt}=2\times 3.14\times 52\times 0.03=9.796cm^2/min$

So rate of change of area will be $9.796cm^2/min$

6 0

3 years ago

Chapter 21, Problem 009 Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.12

PilotLPTM [1.2K]

Answer:

a) -1.325 μC

b) 4.17μC

Explanation:

First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:

$q_1 + q_2 = constant\\q_1_f = q_2_f |Then\\q_1_f + q_2_f = 2q_1_f = q_1_o+q_2_o\\q_1_f = q_2_f = \frac{q_1_o+q_2_o}{2}$

We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.

The electrostatic force is equal to:

$F_e = k\frac{q_1q_2}{r^2}$

K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2

Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:

$F_e_f = k \frac{q_1_fq_2_f}{r^2} = k\frac{\frac{q_1_o+q_2_o}{2}\frac{q_1_o+q_2_o}{2}}{r^2} = k\frac{(q_1_o+q_2_o)^2}{4r^2} \\(q_1_o+q_2_o) = \sqrt{\frac{F_e_f*4r^2}{k}} = \sqrt{\frac{0.0443N *4(0.641m)^2}{9*10^9Nm^2/C^2} } = 2.844 *10^{-6}C \\ q_1_o = 2.844*10^{-6}C - q_2_o$

Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:

$F_e_o = k\frac{q_1_oq_2_o}{r^2}\\ q_1_oq_2_o = \frac{F_e_o*r^2}{k} = \frac{-0.121N(0.641m)^2}{9*10^9Nm^2/C^2} = -5.524*10^{-12}$