how can i run my car heater at maximum setting to cool down the engine if the car is overheated and the engine is stalled and just cranks without starting?

Explanation:

Given that,

Mass of a truck = 15,000 kg

Speed of the truck = 25 m/s

Mass of car = 1,100 kg

Speed of the car = 15 m/s

As they stick together after the collision. Let V be the speed after the collision.

Initial momentum of the system,

[tex]P_i=15000\times25+1100\times15\\\\=391500\ kg-m/s[/tex]

Final momentum of the system,

[tex]P_f=(15000+1100)V\\\\=16100V[/tex]

Hence, this is the required solution.

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:

A

Explanation:

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

Answer:

Explanation:

Yes she is doing work. With or without the groceries, she is still doing work. She does more work with the groceries than without because Work is defined by F which is defined by mass. The mass increases with the groceries.

The work done is against the force of gravity.

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:

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]

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:

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:

The momentum, ΔP, and therefore, kinetic energy given to the ball in a serve is the result of the product of the tension force, 'F', in the string and the time of contact, Δt, between the ball and the string

ΔP = F × Δt

Explanation:

The impulse, ΔP, is the produce of the force, 'F', applied to a body for a given period of time, Δt', that gives motion to the body, and it is equal to the change of momentum of the body

ΔP = F × Δt

The momentum, 'P', of a body is the product of the mass, 'm', of the body and its velocity, 'v'

P = m × v

Tension is the axial pulling force of a string

T = Axial Force, F[tex]_{axial}[/tex]

The tension used in Tennis Racket strings is between 40 to 65 lbs.

When high tension is used in the string, the string is taut, and the contact duration between the Racket string and the ball is minimal, and the player needs to use more force to obtain a high momentum, and therefore, energy in the ball, which reduces control, and increase stress, as force is more emphasized

When low tension is used in the string, the Tennis Racket strings are more elastic. During a serve, the ball pushes the strings further back into the racket, such that the ball spends more time in contact with the string, (Δt is larger), and therefore, the impulse, F·Δt = ΔP, given to the ball is larger, therefore, the ball has a larger change in momentum, and therefore more energy in the intended direction.

However, a very slackened string will increase the increase area and time (large Δt) of contact of the ball and the racket such that the force given to the ball, F = ΔP/(large Δt) is reduced and therefore reduce the likelihood of gaining points from a serve against an opponent with a much forceful return of a serve.

Answer:

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

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

The eccentricity is 0.5

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:

yes

Explanation:

Answer:

It converts to other energy

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:

19 km.

..............

Explanation:

.....

Answer:The Distance is 43 and displacement is 5km south

Explanation: To find the distance you add up all the numbers which was 43 and to fine the displacement subtract all the numbers and since all the other numbers were out expect for -5 which was south you would put 5 km south , if the number has a negative on it , it doesnt matter.

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:

Explanation:

Wave motion

ex...longitudinal wave.

Answer:96 W

Explanation:

Answer:

The right solution is "2.5 m/s²". A further explanation is given below.

Explanation:

The given values are:

Initial velocity,

= 20 m/s

Final velocity,

= 0 m/s

Time,

= 8 s

As we know,

⇒  [tex]Acceleration=\frac{Final \ velocity-Initial \ velocity}{2}[/tex]

On substituting the values, we get

⇒                       [tex]=\frac{0-20}{8}[/tex]

⇒                       [tex]=\frac{-20}{8}[/tex]

⇒                       [tex]=-2.5 \ m/s^2[/tex]

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:

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:

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.

The resistance is 15 ohms.

Also, it's dissipating 960 watts whenever it's turned on. So it's almost as hot as a toaster, and Minh no longer has hands.

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.

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.