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nlexa [21]
3 years ago
6

n the metric system the acceleration due to gravity near Earth's surface is about 9.81 meters per second per second. What is the

acceleration due to gravity as measured in the English system, that is, in feet per second per second? (Round your answer to one decimal place.)
Physics
1 answer:
Anna007 [38]3 years ago
3 0

Answer:

32.2\frac{ft}{s^2}

Explanation:

To do this conversion we have to know that:

1ft=0.3048m

(It is easy to find this value on physics textbooks or internet physics webpages)

So, we should convert the gravitational acceleration in the next way:

9.81\frac{m}{s^{2}}=9.81\frac{m}{s^{2}}*\frac{1ft}{0.3048m}

So, the meters cancel out:

9.81\frac{m}{s^{2}}=9.81\frac{1}{s^{2}}*\frac{1ft}{0.3048}=32.18\frac{ft}{s^2}

9.81\frac{m}{s^{2}}=32.2\frac{ft}{s}

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Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

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