75 kilometers per hour is an example of: A. speed. B. velocity C. acceleration

Physics

2 answers:

kherson [118]3 years ago

5 0

The answer is A. speed
hope this helps! :D

ozzi3 years ago

4 0

It is A. Speed as speed is how fast you are traveling and 75 kilometers per hour is how fast you are traveling.

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8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer

Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

$M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \ by \ the \ effort}{disatnce \ moved \ by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m$

Therefore, the distance that the object is raised above its initial position is 5.625 m.

3 0

3 years ago

Consider two wind turbines, each located in sites with the same wind speed, and each with the same efficiency. Turbine 1 has a r

Alex787 [66]

Answer:

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Exption:

7 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:

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