If the ball rolls out of bounds on the sideline in soccer, what is the result?

The speeds of 22 particles are as follows: two at 5.30 cm/s, four at 1.40 cm/s, six at 7.14 cm/s, eight at 1.52 cm/s, two at 7.6

Leona [35]

Answer:

Average velocity is 3.93 cm/s

Root mean square velocity is 4.79 cm/s

Velocity peak to peak is 6.28 cm/s

Explanation:

Speed of 2 particles = 5.3 cm/s

Speed of 4 particles = 1.4 cm/s

Speed of 6 particles = 7.14 cm/s

Speed of 8 particles = 1.52 cm/s

Speed of 2 particles = 7.68 cm/s

$v_{avg}=\frac{2\times 5.3+4\times 1.4+6\times 7.14+8\times 1.52+2\times 7.68}{22}\\\Rightarrow v_{avg}=\frac{86.56}{22}\\\Rightarrow v_{avg}=3.93\ cm/s$

Average velocity is 3.93 cm/s

$v_{rms}=\sqrt{\frac{2\times 5.3^2+4\times 1.4^2+6\times 7.14^2+8\times 1.52^2+2\times 7.68^2}{22}}\\\Rightarrow v_{rms}=\sqrt{\frac{507.34}{22}}\\\Rightarrow v_{rms}=4.79\ cm/s$

Root mean square velocity is 4.79 cm/s

$v_p=7.68-1.4=6.28\ cm/s$

Velocity peak to peak is 6.28 cm/s

8 0

3 years ago

What is the relation between force and acceleration

Galina-37 [17]

The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

6 0

3 years ago

Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

6 0

3 years ago

Two engines create SIL values of 75 dB and 85 dB respectively. What dB reading would you expect when both are run simultaneously

stepladder [879]

Answer:

Explanation:

In a quiet forest, you can sometimes hear a single leaf fall to the ground. ... greater its pressure amplitude, the more the air is compressed in the sound it creates. ... Graphs of the gauge pressures in two sound waves of different intensities. ... The sound intensity level β in decibels of a sound having an intensity I in watts per .

4 0

3 years ago

What is the specific heat for the aluminum wire?

Alisiya [41]

0.902

hope this helps :)

4 0

3 years ago