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Misha Larkins [42]

1 year ago

13

You have found that the net force exerted on the cylinder depends on several different independent variables in different ways.

how can you combine those separate dependencies into one equation relating the net force to the relevant independent variables

1 answer:

postnew [5]1 year ago

4 0

The net force on the cylinder depends on several independent variables such as mass, acceleration, velocity, time, etc, and it is given as F(net) = ma.

<h3>Net force exerted on the cylinder</h3>

The net force on the cylinder is calculated from Newton's second law of motion.

F(net) = ma

F(net) = m(v/t)

where;

m is mass of the cylinder
a is acceleration of the cylinder
v is velocity of the cylinder
t is time of motion of the cylinder

Thus, the net force on the cylinder depends on several independent variables such as mass, acceleration, velocity, time, etc, and it is given as F(net) = ma.

Learn more about net force here: brainly.com/question/11556949

#SPJ1

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Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

a)

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago

When an object moves away from Earth, its light waves are stretched to a lower frequency, or longer wavelength, and thus we say

Gelneren [198K]

Your answer will be False because wavelengths are usually are use as sound waves.

3 0

3 years ago

A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f

MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

$m_T = 25(7.7*10^4)Kg$

$m_T = 1.925*10^6Kg$

Therefore the friction force at 29Km / h would be,

$f=250v$

$f= 250*29Km/h$

$f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})$

$f = 2013.889N$

In this way the tension exerts between first car and locomotive is,

$T=m_Ta+f$

$T=(1.925*10^6)(0.2)+2013.889$

$T= 3.8701*10^5N$

Therefore the tension in the coupling between the car and the locomotive is $3.87*10^5N$

6 0

3 years ago

In-s [12.5K]

Answer:

Explanation:

1) acceleration is the change in velocity of a body with respect to time.

acceleration = velocity/time

Given

velocity = 139m/s

time = 20secs

acceleration = 139/20

acceleration = 6.95m/s²

Hence its average acceleration during the first 20 seconds of the launch is 6.95m/s²

2) Speed is the rate of change of distance with respect to time.

average speed = distance/time

Time = distance/speed

Time = 100/7823

Time = 0.013s

3) Using the equation of motion v² = u²+2as

v² = 0²+2(15)(60)

v² = 30*60

v² =1800

v = √1800

v = 42.43m/s

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3 years ago

How does the number of valence electrons affect the reactivity?

ololo11 [35]

The less valence electrons the more reactive the element is.

6 0

3 years ago