. The magnitudes of two forces are measured to be 120 ± 5 N and 60 ± 3 N. Find the sum

A negatively charged balloon has 4 μC of charge. How many excess electrons are on this bal- loon? The elemental charge is 1.6 ×

bagirrra123 [75]

Data:

The charge of a body depends on the amount of electrons it gains or loses. Q = n * e, where "Q" is charge, "n" is the number of plus or minus electrons, and "e" is the fundamental charge of an electron $1,6 * 10 ^{-19}C$ <span>. To know if the body has gained or lost, we look at the signal of its charge, remembering that the electron is negative. The charge of the body is 4 μC (positive), so there is a lack of electrons!

Q = 4 </span>μC → $Q = 4*10^{-6}$
$e = 1,6 * 10 ^{-19}C$
$n = ?$ <span>

We have:
</span> $Q = n*e$
$n = \frac{Q}{e}$
$n = \frac{4*10^{-6}}{1,6 * 10 ^{-19}}$
$n = 2,5*10^{-6-(-19)}$
$n = 2,5*10^{-6+19}$
$\boxed{n = 2,5*10^{13}electrons}$

7 0

3 years ago

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor

const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is $\tau = 34.3 \ N\cdot m$

Explanation:

From the question we are told that

The mass of the steel ball is $m = 3.0 \ kg$

The length of arm is $l = 70 \ cm = 0.7 \ m$

The mass of the arm is $m_a = 4.0 \ kg$

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

$r = \frac{l}{2}$

=> $r = \frac{ 0.7}{2}$

=> $r = 0.35 \ m$

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

$\tau = m_a * g * r + m * g * L$

=> $\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70$

=> $\tau = 34.3 \ N\cdot m$

5 0

3 years ago

A less massive moving object has an elastic collision with a more massive object that is not moving. Compare the initial velocit

stepladder [879]

Assume that the small-massed particle is $m$ and the heavier mass particle is $M$ .

Now, by momentum conservation and energy conservation:

$mv = mv_{m} + Mv_{M}$

$mv^{2} = mv^{2}_{m} + Mv^{2}_{M}$

Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

$v_{m} = v\frac{m - M}{m + M}\\\\\\v_{M} = v\frac{m + M}{m + M}\\$