PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST The actual subject is Science but they dont have that as a option in pick a subject

Answer:

[tex]Q_t= 8.3 * 10^3 C[/tex]

Explanation:

From the question we are told that:

Resistor [tex]R=1000ohms[/tex]

Voltage [tex]v=120_V[/tex]

Capacitance of c_1 [tex]c_1=20 \mu F[/tex]

Capacitance of c_2 [tex]c_2=60 \mu F[/tex]

Time [tex]t=0[/tex]

Generally the equation for charges is mathematically given by

[tex]For C_1\\Charge\ on\ C_1 = CV = 20*120 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2.4 x 10^-3 C\\[/tex]

[tex]ForC_2\\Charge on C_2 = 60*120 =7200 μC = 7.2 x 10^-3\\Charge on C_2 = 7.2 x 10^-3[/tex]

Generally the equation for voltage across capacitors is mathematically given by

[tex]V_c(t)=V(1-e^{-t/RC})[/tex]

[tex]C=C_1+C_2=80 \mu f\\t=2RC=>160000s[/tex]

[tex]V_c(t)=120(1-e^{-(160000)/1000*(80)})[/tex]

[tex]V_c(t)=103.7598[/tex]

Generally the equation for charges is mathematically given by

[tex]Q1(t) = C1Vc(t)\\Q1(t) = 20*103.7598\\Q1(t) = 2075.196\\\\Q2(t) = 60*103.7598\\Q2(t) = 6225.6\\[/tex]

Generally the equation for total charges [tex]Q_t[/tex] is mathematically given by

[tex]Q_t=Q1(t)+Q2(t)[/tex]

[tex]Q_t= 8.3 * 10^3 C[/tex]

Explanation:

Wave motion

ex...longitudinal wave.

Answer:96 W

Explanation:

Answer:

0.17 kg

Explanation:

mc = mass of cart

mb = mass of ball

vc = speed of cart before collision

vb = speed of ball before collision (vb = 0)

v' = speed of cart and ball after collision

mc . vc + mb . vb = (mc + mb) . v'

(1.5)(1.7) + mb(0) = (1.5 + mb) . 1.53

2.55 + 0 = (1.5 + mb) . 1.53

2.55/1.53 = 1.5 + mb

5/3 = 3/2 + mb ( I change it to fraction)

mb = 5/3 - 3/2 = 10/6 - 9/6 = 1/6 kg

mb = 1/6 kg = 0.17 kg

Answer:

A workout injury can happen to anyone, no matter your experience or fitness level. People hurt themselves in all kinds of ways when they work out. Even walking can cause an injury. But you can significantly cut your risk of getting hurt by following certain workout precautions. There are simple steps that can help keep you injury-free during your workout.

Every workout should begin with a warm-up and end with a cool-down period. A warm-up helps your body get ready for exercise. It gradually increases your heart rate and loosens your muscles and joints. Cool-down after you work out so, it brings your heart rate back to normal. Walking for 5 to 10 minutes after you work out is one way to cool down. Do dynamic stretching before and after you work out. This will help increase flexibility. Don't push yourself too hard. As your fitness abilities increase and you will be able to challenge yourself more.

Hope this helps

Maintaining physical fitness is very important in today's world. Each and everyone should be involved in physical fitness.nowaday people are so busy with their work that they have forgotten health is wealth.

Even if we get engaged in physical fitness activity we can some time get insuries . some of tips that can help prevent from getting physical fitness injury, warm up exercises should be done before exercise to avoid muscle strength.safety precautions should be taken if you are dealing with heavy objects .

hence we should maintain physical fitness with getting rid of injury that may occur

mark me

Answer:

When a right-handed golfer’s tee shot curves to far to the left, the shot is described as a “hook.” This shot typically occurs when the club head moves across the ball from left to right, imparting side-spin on the ball

Answer:

A

Explanation:

Answer: [tex]18.22\ cm[/tex] from [tex]3.5\ nC[/tex] charge.

Explanation:

Given

The magnitude of the first charge is  [tex]Q_1=3.5\ nC[/tex]

The magnitude of the second charge is [tex]Q_2=5\ nC[/tex]

[tex]-6\ nC[/tex] charge must be placed in between the two charges to establish equilibrium

The electrostatic force is given by

[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]

Equilibrium will be established when force by both the charges balance out each other. Suppose [tex]-6\ nC[/tex] is placed at a distance of [tex]x[/tex] cm from [tex]3.5\ nC[/tex] . So, we can write

[tex]\Rightarrow \dfrac{k(3.5)(-6)}{x^2}=\dfrac{k(5)(-6)}{(40-x)^2}[/tex]

Canceling similar terms

[tex]\Rightarrow \left [ \dfrac{40-x}{x}\right ]^2=\dfrac{10}{7}\\\\\Rightarrow \dfrac{40-x}{x}=\sqrt{\dfrac{10}{7}}=1.195\\\\\Rightarrow 40-x=1.195x\\\Rightarrow 40=2.195x\\\Rightarrow x=18.22\ cm[/tex]

Thus, the equilibrium position is [tex]18.22\ cm[/tex] from [tex]3.5\ nC[/tex] charge.

Answer:

What school do you go to?

Explanation:

Answer:

like I commented it yesterday can i have brainiest

Explanation: oh and HI

Answer:

Like

Explanation:

I put it before but my answer got deleted

Answer:

B

Explanation:

Answer:

The main function of the circulatory system is to carry nutrients to cells and waste away from cells.

Explanation:

Answer:

125

Explanation:

use the formula e=1/2mv² and use m=10 and v=5

this produces the equation e=1/2x10x25=125

dont forget the units (J) ;)

Answer:

125J

Explanation:

K.e=1/2mv^2

=1/2(10)(5)^2

=125J

Answer:

w = 1,662 rad / s,    v = 33.25 m / s

Explanation:

This this exercise indicates that the acceleration felt by the astronauts is

         a = 5.64 g

in a circular motion the centripetal acceleration is

        a_c = v²/ r

angular and linear variables are related

         v = w r

we substitute

        a_c = w² r

we finally have

         5.64 g = w² r

         w² = 5.64 g / r

let's calculate

         w = [tex]\sqrt{ \frac{5.64 \ 9.8}{20.0} }[/tex]

         w = 1,662 rad / s

linear velocity is

        v = w r

        v = 1,662 20

         v = 33.25 m / s

Answer:

a = 2 m/s²

Explanation:

Given: 20 g, 40N

To find: Acceleration (a)      

Solution: To find the acceleration (A), divide the force by the weight    

A = F ÷ m

=  40 ÷  20

= 2 m/s²

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

Answer:

A

Explanation:

The answer is A because proton number is the same as atomic number

Answer:

Rae Young at Bell Labs. They diagrammed a network of wireless towers into what they called a cellular layout. Cellular was the chosen term because each tower and its coverage map looked like a biological cell. Eventually, phones that operated on this type of wireless network were called cellular phones.

Explanation:

Answer:

The correct option is;

B) No, the Navy vessel is slower

Explanation:

The speed of some torpedoes can be as high as 370 km/h. The average speed of a fast Navy vessel is approximately 110 km/h

Therefore, the torpedoes travel approximately 3 times as fast as the (slower) Navy vessel, such that the torpedo covers three times the distance of the Navy vessel in the same time and therefore, if the Navy vessel and the torpedo continue in a straight line (in the same direction) due north the vessel can not outrun the torpedo

Therefore, no the Navy vessel travels slower than a torpedo.

Answer: It would be the dependent variable.

Explanation:

The dependent variable is limited and can be affected by the changes and manipulation of the independent variable.

Answer:

Yes, but its magnetic power is very weak so its not that noticeable to the average person whos not paying attention.

Answer:

When the sum of all forces applied to the object is 0.

Explanation:

N2L states that net force equals an object's mass times its acceleration (F = ma). If the net force is 0, the acceleration of the object is 0.

On the second note, when an object has 0 acceleration (which means it has a constant velocity), it is in equilibrium (N1L).

Answer:

It converts to other energy

Answer:

The sound level of the point source at a distance of 10 meters is approximately 110.294 decibels.

Explanation:

First, we calculate the intensity ([tex]I[/tex]), in watts per square meter, by the Inverse-Square Law:

[tex]I= \frac{\dot W}{4\pi\cdot r^{2}}[/tex] (1)

Where:

[tex]\dot W[/tex] - Power, in watts.

[tex]r[/tex] - Radius, in meters.

And the sound intenity level ([tex]L[/tex]), in decibels, is expressed by:

[tex]L = 10\cdot \log_{10} \frac{I}{I_{o}}[/tex] (2)

Where [tex]I_{o}[/tex] is the reference sound intensity, in watts per square meter.

If we know that [tex]\dot W = 135\,W[/tex], [tex]r = 10\,m[/tex] and [tex]I_{o} = 10^{-12}\,\frac{W}{m^{2}}[/tex], then we find that sound level at a distance of 10 meters is:

[tex]I= \frac{\dot W}{4\pi\cdot r^{2}}[/tex]

[tex]I = 0.107\,\frac{W}{m^{2}}[/tex]

[tex]L = 10\cdot \log_{10} \frac{I}{I_{o}}[/tex]

[tex]L \approx 110.294\,dB[/tex]

The sound level of the point source at a distance of 10 meters is approximately 110.294 decibels.

Answer:

it is b on edge not d, well atleast for me it was b

Explanation:

Answer:

B) one battery, two switches, and three light bulbs

Explanation:

The eccentricity is 0.5

Answer:

yes

Explanation:

Many of the symbols are the first one or two letters of the element’s name in English. The first letter of a symbol is always written as a capital letter (uppercase) and the second letter as a small letter (lowercase).

For example

(i) hydrogen, H

(ii) aluminium, Al and not AL

(iii) cobalt, Co and not CO.

Symbols of some elements are formed from the first letter of the name and a letter, appearing later in the name.

Examples are: (i)chlorine, Cl, (ii) zinc, Zn etc.

Other symbols have been taken from the

names of elements in Latin, German or Greek.

For example, the symbol of iron is Fe from its Latin name ferrum, sodium is Na from natrium, potassium is K from kalium. Therefore, each element has a name and a unique chemical symbol.