what is the efficiency of a machine that uses 102kj of enegery to do 98 kJ of work?

Answer:

The value is  [tex]p = 0.7556 c[/tex]

Explanation:

From the question we are told that

   The speed at which galaxy B moves away from galaxy A is  [tex]v = 0.577c[/tex]

Here c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

     The speed at which galaxy C moves away from galaxy B is  [tex]u = 0.731 c[/tex]

Generally from the equation of  relative speed we have that  

     [tex]u = \frac{p - v}{ 1 - \frac{ p * v}{c^2} }[/tex]

Here p is the velocity at which galaxy C recede from galaxy A so

     [tex]0.731c = \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }[/tex]

=>   [tex]0.731c [1 - \frac{ p * 0.577}{c}] = p - 0.577c[/tex]

=>   [tex]0.731c - 0.4218 p = p - 0.577c[/tex]

=>   [tex]0.731c + 0.577c = p + 0.4218 p[/tex]

=>   [tex]1.308 c = 1.731 p[/tex]

=>    [tex]p = 0.7556 c[/tex]

Answer:

It is called an ion.

Explanation:

Atoms of elements can lose or gain electrons making them no longer neutral, they become charged. A charged atom is called an ion. When an atom loses electron(s) it will lose some of its negative charge and so becomes positively charged. A positive ion is formed where an atom has more protons than electrons.

A C and D

C because ofc when electric charge flow electric field is produced

Answer:

A & D

Explanation:

Answer:

the air is being stopped in the pocket of the parachutes theys why the parachute has a certain shape so that the air that gets caught inside of it as the skydiver goes down slows down the landing

Answer:

Density

Explanation:

Answer:

The location of the center of gravity of the fourth mass is [tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex].

Explanation:

Vectorially speaking, the center of gravity with respect to origin ([tex]\vec r_{cg}[/tex]), measured in meters, is defined by the following formula:

[tex]\vec r_{cg} = \frac{m_{1}\cdot \vec r_{1}+m_{2}\cdot \vec r_{2}+m_{3}\cdot \vec r_{3}+m_{4}\cdot \vec r_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}[/tex] (1)

Where:

[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex], [tex]m_{4}[/tex] - Masses of the objects, measured in kilograms.

[tex]\vec r_{1}[/tex], [tex]\vec r_{2}[/tex], [tex]\vec r_{3}[/tex], [tex]\vec r_{4}[/tex] - Location of the center of mass of each object with respect to origin, measured in meters.

If we know that [tex]\vec r_{cg} = (0,0)\,[m][/tex], [tex]\vec r_{1} = (0,0)\,[m][/tex], [tex]\vec r_{2} = (0, 4.1)\,[m][/tex], [tex]\vec r_{3} = (1.9,0.0)\,[m][/tex], [tex]m_{1} = 6\,kg[/tex], [tex]m_{2} = 1.5\,kg[/tex], [tex]m_{3} = 4\,kg[/tex] and [tex]m_{4} = 7.9\,kg[/tex], then the equation is reduced into this:

[tex](0,0) = \frac{(6\,kg)\cdot (0,0)\,[m]+(1.5\,kg)\cdot (0,4.1)\,[m]+(4.0\,kg)\cdot (1.9,0)\,[m]+(7.9\,kg)\cdot \vec r_{4}}{6\,kg+1.5\,kg+4\,kg+7.9\,kg}[/tex]

[tex](6\,kg)\cdot (0,0)\,[m]+(1.5\,kg)\cdot (0,4.1)\,[m]+(4\,kg)\cdot (1.9,0)\,[m]+(7.9\,kg)\cdot \vec r_{4} = (0,0)\,[kg\cdot m][/tex]

[tex](7.9\,kg)\cdot \vec r_{4} = -(6\,kg)\cdot (0,0)\,[m]-(1.5\,kg)\cdot (0,4.1)\,[m]-(4\,kg)\cdot (1.9,0)\,[m][/tex]

[tex]\vec r_{4} = -0.759\cdot (0,0)\,[m]-0.190\cdot (0,4.1)\,[m]-0.506\cdot (1.9,0)\,[m][/tex]

[tex]\vec r_{4} = (0, 0)\,[m] -(0, 0.779)\,[m]-(0.961,0)\,[m][/tex]

[tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex]

The location of the center of gravity of the fourth mass is [tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex].

Answer:

F = 200 [N]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which can be calculated by means of the following equation.

[tex]P=m*v[/tex]

where:

P = lineal momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

Now we must understand that the momentum is conserved before and after the firing of the cannon. Before firing the cannon we have the mass of the cannon and mass of the cannonball together at rest (speed = 0). After firing the cannon the cannonball moves forward with positive speed, while the cannon moves back (negative), in this way knowing the masses of each one we can determine the speed of the cannon.

[tex](m_{cannon}+m_{ball})*v_{1}=-(m_{cannon}*v_{2})+(m_{ball}*v_{3})[/tex]

where:

m_cannon = 2500 [kg]

m_ball = 2.5 [kg]

v₁ = 0 (velocity of the group before firing)

v₂ = velocity of the cannon after firing [m/s]

v₃ =  160 [m/s] (velocity of the cannonball after firing)

[tex](2500+2.5)*0 = -(2500*v_{2})+(2.5*160)\\v_{2}=0.16[m/s][/tex]

Now using the following equation of kinematics, we can calculate the acceleration.

[tex]v_{f}=v_{o}-a*t[/tex]

where:

Vf = final velocity = 0 (cannon comes to rest)

Vo = initial velocity = 0.16 [m/s]

a = acceleration [m/s²]

t = time = 2 [s]

[tex]0 = 0.16 - a*2\\2*a=0.16\\a = 0.08 [m/s^{2}][/tex]

With the value of acceleration, we can use Newton's second law which tells us that the forces acting on a body is equal to the product of mass by acceleration.

ΣF = m*a

where:

F = force [N] (units of Newtons)

m = mass = 2500 [kg]

a = acceleration = 0.08 [m/s²]

[tex]F = 2500*0.08\\F = 200 [N][/tex]

Answer:

in a spontaneous process, the entropy of the universe increases.

Explanation:

Entropy is a measure of of the degree of  randomness or disorderliness in a system.

The second law of thermodynamics can be stated as follows; "in any spontaneous process, the entropy of the universe increases."  

The universe here refers to the system's disorder and the disorder of the surroundings.  Therefore, a spontaneous process can occur, in which the entropy of the system decreases, only if the entropy increases in the surroundings.

For instance, when ice freezes, the entropy of liquid water decreases, that is, the entropy of the system decreases. However, heat is given off to the surroundings and the entropy of the surroundings increases. This is an obvious expression of this law.

Explanation:

It's a transistor. Hope that helps!

Answer:

D. Transistor

Explanation:

Edge 2021

Answer:

8

Explanation:

Atomic number in the Bohr model given is 8.

For every atom, the atomic number is the number of protons in such an atom.

The protons are the positively charged particles found in an atom. They are called the atomic number.

Answer:

Explanation:

Step one:

given data

speed = 123m/s

radius of circle= 4330m

Step two:

We need to find the circumference of the circle, it represents the distance traveled

C=2πr

C= 2*3.142*4330

C= 27209.72m

Step three:

We know that velocity= distance/time

time= distance/velocity

time= 27209.72/123

time=221.2 seconds

in minute = 221.2/60

time= 3.7 min

Answer:

[tex]\mathbf{\beta = 123.75 \ dB}[/tex]

Explanation:

From the question, using the expression:

[tex]125 \ dB = 10 \ log (\dfrac{I}{I_o})[/tex]

where;

[tex]I_o = 10^{-12} \ W/m^2[/tex]

[tex]I = 10^{12.5} \times 10^{-12} \ W/m^2[/tex]

[tex]I = 3.162 \ W/m^2[/tex]

This is a combined intensity of 4 speakers.

Thus, the intensity of 3 speakers = [tex]\dfrac{3.162\times 3}{4}[/tex]

= 2.372 W/m²

Thus;

[tex]\beta = 10 \ log ( \dfrac{2.372}{10^{-12}} ) \ W/m^2[/tex]

[tex]\mathbf{\beta = 123.75 \ dB}[/tex]

Since vector B was not specified, I'll assume one at random. You can later answer your own question.

Answer:

[tex]\mid\mid \vec A+\vec B \mid \mid=\sqrt{105}[/tex]

[tex]\mid\mid \vec A-\vec B \mid \mid=\sqrt{149}[/tex]

Explanation:

Given:

[tex]\vec A=3\hat i +2\hat j+3\hat k[/tex]

And (assumed):

[tex]\vec B=-5\hat i +8\hat j-4\hat k[/tex]

Find the magnitude of

[tex]\vec A+\vec B[/tex]

[tex]\vec A-\vec B[/tex]

Given a vector

[tex]\vec P=x\hat i +y\hat j+z\hat k[/tex]

The magnitude of the vector is:

[tex]\mid\mid \vec P\mid \mid=\sqrt{x^2+y^2+z^2}[/tex]

[tex]\vec A+\vec B =3\hat i +2\hat j+3\hat k-5\hat i +8\hat j-4\hat k[/tex]

[tex]\vec A+\vec B =-2\hat i +10\hat j-\hat k[/tex]

The magnitude of the sum is:

[tex]\mid\mid \vec A+\vec B \mid \mid=\sqrt{(-2)^2+10^2+(-1)^2}=\sqrt{4+100+1}[/tex]

[tex]\mathbf{\mid\mid \vec A+\vec B \mid \mid=\sqrt{105}}[/tex]

[tex]\vec A-\vec B =3\hat i +2\hat j+3\hat k-(-5\hat i +8\hat j-4\hat k)[/tex]

[tex]\vec A-\vec B =3\hat i +2\hat j+3\hat k+5\hat i -8\hat j+4\hat k[/tex]

[tex]\vec A-\vec B =8\hat i -6\hat j+7\hat k[/tex]

The magnitude of the difference is:

[tex]\mid\mid \vec A-\vec B \mid \mid=\sqrt{8^2+(-6)^2+7^2}=\sqrt{64+36+49}[/tex]

[tex]\mathbf{\mid\mid \vec A-\vec B \mid \mid=\sqrt{149}}[/tex]

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

[tex]W = F\cdot \Delta s[/tex] (1)

Where:

[tex]F[/tex] - Force applied by the cart, measured in newtons.

[tex]\Delta s[/tex] - Distance travelled by the car, measured in meters.

[tex]W[/tex] - Work done on the cart, measured in joules.

If we know that [tex]F = 888\,N[/tex] and [tex]\Delta s = 22\,m[/tex], then the work done on the cart is:

[tex]W =(888\,N)\cdot (22\,m)[/tex]

[tex]W = 19536\,J[/tex]

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) If we know that [tex]F = 1111\,N[/tex] and [tex]\Delta s = 22\,m[/tex] , then the work on the cart is:

[tex]W = (1111\,N)\cdot (22\,m)[/tex]

[tex]W = 24442\,N[/tex]

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

a. The amount of work done by the go cart engine is 19,536 Nm.

b. Work is being done by the go cart engine i.e the engine installed on the go cart.

c. The amount of work will increase if the driver wants to go further because work done is directly proportional to the distance covered. No, you don't need a bigger engine to go further.

d. The amount of work done when a bigger engine is used is 24,442 Nm.

e. The reason why a bigger engine is used is, so that the engine can easily do more work in the same amount of distance.

f. Since we know that work requires a change in energy, the bigger engine would use more gas to go 22 meters.

g. An empty semi truck uses much more gas than a car because it has more weight (mass) and as such requires more energy, in accordance with the law of inertia.

Given the following data:

a. To determine the amount of work done by the go cart engine:

Mathematically, the work done by an object is given by the formula;

[tex]Work\;done = Force \times distance[/tex]

Substituting the given parameters into the formula, we have;

[tex]Work\;done = 888 \times 22[/tex]

Work done = 19,536 Nm.

b. Work is being done by the go cart engine i.e the engine installed on the go cart.

c. The amount of work will increase if the driver wants to go further because work done is directly proportional to the distance covered. No, you don't need a bigger engine to go further.

d. To determine the amount of work done when a bigger engine is used:

[tex]Work\;done = Force \times distance[/tex]

[tex]Work\;done = 1111 \times 22[/tex]

Work done = 24,442 Nm.

e. The reason why a bigger engine is used is, so that the engine can easily do more work in the same amount of distance.

f. Since we know that work requires a change in energy, the bigger engine would use more gas to go 22 meters.

g. An empty semi truck uses much more gas than a car because it has more weight (mass) and as such requires more energy, in accordance with the law of inertia.

Read more: https://brainly.com/question/22599382

Answer:

The value is  [tex]|\vec B| = 1.267 *0^{-8} \ T[/tex]

Explanation:

From the question we are told that  

   The magnitude of the electric fields is  [tex]E = 3.80 V/m[/tex]

Generally speed of light is mathematically represented as

        [tex]c = \frac{|\vec E|}{ |\vec B|}[/tex]

Here c is the speed of light with value  [tex]c = 3.0*10^{8} \ m/s[/tex]

        [tex]|\vec B |[/tex]  is the  magnitude of the magnetic field so  

=>         [tex]|\vec B| = \frac{|\vec E|}{c}[/tex]

=>         [tex]|\vec B| = \frac{ 3.80 }{3.0*10^{8}}[/tex]

=>         [tex]|\vec B| = 1.267 *0^{-8} \ T[/tex]

Given :

A 12 kg coyote runs towards a rabbit in the Ortega Mountains  with a velocity of 8 m/s.

To Find :

The momentum of the  coyote.

Solution :

The given in the question is mass of coyote which is 12 kg and velocity which is 8 m/s .

Momentum is given by :

M = mass × velocity

M = 12 × 8 kg m/s

M = 96 kg m/s

Hence, this is the required solution.

Explaining the concept:

To explain this, we first need to know what density actually is:

Density is the mass of a given substance per unit volume, which means, that is the amount of mass present in unit volume (1 cm³ , 1m³)

Impure water is water, but with impurities added to it

Why is the density of Impure water is greater:

When we compare the mass of equal volumes of Water and Impure water, we will find that the mass of Impure water will be more, that is because the extra mass is of the impurities

Since impure water has impurities uniformly spread in it, we will find more matter in unit volume of Impure water than in Pure water

That is why the density of Impure water is greater

Answer:

a). a = F/m

Explanation:

Formula is F=ma

The car accelerates with magnitude a such that

31.3 m/s = 25.6 m/s + a (5.4 s)

→   a = (31.3 m/s - 25.6 m/s) / (5.4 s) ≈ 1.056 m/s²

Then the applied force has a magnitude F of

F = (1430 kg) a ≈ 1500 N

Answer:

F = 2389.603 N

Explanation:

Given:

Mass m = 1,369.4 kg

Initial velocity u = 28.9 m/s

Final velocity v = 20 m/s

Time t = 5.1 s

Find:

Net force

Computation:

a = (v - u)/t

a = (20 - 28.9)/5.1

a = -1.745 m/s²

F = ma

F = (1369.4)(1.745)

F = 2389.603 N

Answer:

the last time i had my tempature taken was at disney prings about a week ago -_- they used one of those gun things that dont touch u and the SHOVED me forward so i guess i was fine

Explanation:

Answer:

thermometer

Explanation:

Answer:

4 joules

Explanation:

Answer:

force acting on the parent = 25 N .

Explanation:

According to third law of Newton , there is equal and opposite reaction to every action . Here force by the parent on child is action and the force by child on parent is reaction . The former is given as 25 N so force by child on parent will also be 25 N .

Answer is 25 N .

Answer:

c

Explanation: beacuse newtons law of phisics says ''what goes up must come down''

The force of gravity between any two ordinary objects on the earth is stronger if the objects are closer to each other.

The force of gravity between two object on the earth surface is given Newton's law of universal gravitation;

[tex]F= \frac{Gm_1m_2}{r^2}[/tex]

where;

The distance between the object is inversely proportional to the force of gravity between the objects. The smaller the distance between the two objects, the greater the force of gravity and vice versa.

Thus, we can conclude that the force of gravity between any two ordinary objects on the earth is stronger if the objects are closer to each other.

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Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 30 × 10 × 15

We have the final answer as

Hope this helps you

Answer:

 F = 1.68 N

Explanation:

Let's solve this exercise in parts.

Let's use the concept of conservation of the mechanical nerve

initial

    Em₀ = 500 J

The energy is totally kinetic

     Em₀ = K = ½ m v₀²

     v₀ = [tex]\sqrt{\frac{2 Em_{o} }{m} }[/tex]

     v₀ = √ (2 500/32)

     v₀ = 5.59 m / s

now with kinematics we can find a space

      v² = v₀² - 2 a x

the negative sign is because the body is stopping

       a =[tex]( \frac{v_{o}^{2} - v^{2} }{2x} )[/tex]  

let's calculate

       a = (5.59² - 5.1²) / 2 50  

       a = 0.0524 m / s²

Finally let's use Newton's second law

     F = ma

     F = 32 0.0524

     F = 1.68 N