The word _____ in contrast,refers to the accumulation of such a borrowing, year after year

Answers

Answer 1

Answer:

Budget

Explanation:

Have good day!!!


Related Questions

A train is traveling down a straight track at 26 m/s when the engineer applies the brakes, resulting in an acceleration of −1.0 m/s2 as long as the train is in motion. How far does the train move during a 52-s time interval starting at the instant the brakes are applied?______ m

Answers

In order to calculate the distance the train will move, we can use the formula below:

[tex]\Delta S=V_0t+\frac{at^2}{2}[/tex]

Where V0 is the initial speed, t is the time and a is the acceleration.

Since the initial speed is 26 m/s and the train acceleration is -1 m/s², the train will completely stop after 26 seconds. In the remaining 26 seconds to complete the total of 52, the train will be stopped already, so there is no displacement.

Because of that, we will use a time t = 26 seconds.

So, using V0 = 26 m/s and a = -1 m/s², we have:

[tex]\begin{gathered} \Delta S=26\cdot26+\frac{(-1)26^2}{2}\\ \\ \Delta S=676-338\\ \\ \Delta S=338\text{ m} \end{gathered}[/tex]

Therefore the train will move 338 meters.

Moving a charge from point A, where the potential is 369.53 V, to point B, where the potential is 217.85 V, takes 0.55 milliJ of work. What is the value of the charge, in micro-Coulombs?

Answers

Given:

The potential of point A is

[tex]V_A=369.53\text{ V}[/tex]

The potential of point B is

[tex]V_B=\text{ 217.85 V}[/tex]

The work done is W = 0.55 milli Joule.

To find the value of charge in micro Coulomb.

Explanation:

The charge can be calculated as

[tex]\begin{gathered} Q=\frac{W}{V_B-V_A} \\ =\frac{0.55\times10^3\text{ J}}{217.85-369.53} \\ =-3.63\text{ C} \\ =\text{ -3.63 C}\times\frac{10^6\text{ }\mu C}{1\text{ C}} \\ =-3.63\times10^6\text{ }\mu C\text{ } \end{gathered}[/tex]

A heart defibrillator passes 10.3 A through a patient's torso for 5.00 ms in an attempt to restore normal beating.(a) How much charge passed?(b) What voltage was applied if 492 J of energy was dissipated?KV(c) What was the path's resistance?ΚΩ(d) Find the temperature increase caused in the 8.00 kg of affected tissue. The specific heat of tissue is 3500 J/(kg. °C).°C

Answers

a) The formula for calculating the quantity of charge is expressed as

Q = IT

where

Q is the quantity of charge

I is the current

T is the time

From the information given,

I = 10.3

T = 5 ms = 5 x 10^-3 s

Q = 10.3 x 5 x 10^-3

Q = 51.5 x 10^- 5 C

The quantity of charge passed is 51.5 x 10^- 5 C

b) The formula for calculating the energy is expressed as

E = I^2RT

where

R is the resistance

E is the energy

From the information given,

E = 492 J

Thus,

492 = 10.3^2 x R x 5 x 10^-3

R = 492/(5 x 10^-3 x 10.3^2)

R = 927.514 ohms

Voltage, V = IR

Voltage = 10.3 x 927.514

Voltage = 9553.398 V

We would divide by 1000. It becomes

Voltage = 9.553 KV

c) From the calculations,

Resistance = 927.514 ohms

We would divide by 1000. It becomes

Resistance = 0.93 ΚΩ

d) Let the temperature increase be t

mass of tissue, m = 8 kg

Specific heat of tissue = 3500 J/(kg. °C).

°C

The formula for calculating the quantity of heat is

H = mcθ

where

H is the quantity of heat

From the informtaion given,

H = 492

θ = t

Thus,

492 = 8 x 3500 x t

t = 492/(8 x 3500)

t = 0.018

The temperature increase is 0.018 degrees

What is the resistance in a circuit that has a current of 2.5A and a voltage of 40v

Answers

16 ohms

Explanation

Ohm's law relates the strength of a direct current is directly proportional to the potential difference and inversely proportional to the resistance of the circuit,it is given by the expresssion

[tex]\begin{gathered} V=IR \\ if\text{ we isolate R} \\ R=\frac{V}{I} \end{gathered}[/tex]

then

Step 1

a) Let

[tex]\begin{gathered} I=2.5\text{ A} \\ V=40\text{ V} \end{gathered}[/tex]

b)replace in the formula

[tex]\begin{gathered} R=\frac{V}{I} \\ R=\frac{40V}{2.5A} \\ R=16\text{ ohms} \end{gathered}[/tex]

therefore, the answer is 16 ohm

I hope this helps you


Imagine that you are on board a ship that was struck by a rogue wave.
Tell your story, from the calm before the wave hit to its aftermath.

Answers

The possibility of a wave toppling cruise liner is extremely low. They are made wide and have enough ballast on the lower decks to be heavy enough rogue waves. On the side, the crew's neglect also be necessary.

What are rogue waves?

A wave that is double the region's major wave height is typically considered a rogue wave. The highest one-third of waves on average over a period of time make up the noteworthy wave height. Even the biggest ships and oil rigs can be rendered useless and sunk by rogue waves.

Have rogue waves ever struck a cruise ship?

Rogue waves have occasionally hit cruise ships, although it is not frequently. Four cruise ships have collided with rogue waves since rogue wave records were first kept in 1995. All sustained damage, and some people reported injuries, but there have been no confirmed fatalities on cruise ships due to rogue waves.

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A NASA probe is moving horizontally above the surface of the moon at a constant speed to the right, as depicted in the diagram below. It releases an instrument package when it is directly above Point P. As seen from the lunar surface, which path would the package likely follow after the release and why?
B, because the gravity of the moon will pull the instrument to the ground with constant acceleration and the lack of an atmosphere allows the package to fall straight down
D, because the gravity of the moon will pull the instrument to the ground with constant acceleration so that its vertical velocity increases as it falls, while the horizontal component of the velocity remains constant
C, because the reduced gravity of the moon pulls the package down vertically at a constant speed while the package travels horizontally at a constant speed resulting in a straight-line trajectory to the lunar surface.
A, because the gravity of the moon pulls the package down with constant acceleration, while the atmosphere of the moon creates horizontal drag on the package which reduces the horizontal component of the package’s velocity causing the package to be pulled backward as it falls

Answers

As seen from the lunar surface, the path followed by the package after the release is D, because the gravity of the moon will pull the instrument to the ground with constant a, so that its vertical velocity increases.

A is incorrect because there is no wind pushing the package backwards. B is incorrect because the package has an initial velocity. C is incorrect because vertical velocity is not constant due to the presence of gravity. E and F are incorrect because gravity acts  immediately after the package is dropped.

D is correct because the horizontal component remains constant because there is no horizontal force acting on the package. This is because in outer space there is no atmosphere, so there will be no air resistance. The vertical component increases with respect to time because of constant acceleration due to gravitational pull on the package.

Therefore, as seen from the lunar surface, the path followed by the package after the release is D, because the gravity of the moon will pull the instrument to the ground with constant acceleration so that its vertical velocity increases as it falls, while the horizontal component of the velocity remains constant

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Tritium has a half-life of 12.3 years. How many years will have elapsed when the radioactivity of a tritium sample has decreased to 10 percent of its original value?

Answers

Given

Half life of tritium is

[tex]t_{\frac{1}{2}}=12.3\text{ years}[/tex]

The sample is reduced to 10% of its original value.

To find

How many years will have elapsed when the radioactivity of a tritium sample has decreased to 10 percent of its original value?

Explanation

The activity is given by

[tex]\begin{gathered} A=A_oe^{-\lambda t} \\ \Rightarrow0.1A_o=A_oe^{-\frac{0.693}{t_{\frac{1}{2}}}t} \\ \Rightarrow ln(0.1)=-\frac{0.693}{1.23}t \\ \Rightarrow-2.302=-\frac{0.693}{1.23}t \\ \Rightarrow t=4.08 \end{gathered}[/tex]

Conclusion

The time taken is 4.08 years

Convert 500BTU/ft.s.F to g.cal/cm^2.s.°C

Answers

The requirement of the prompt is for 500 British Thermal Units /Calories (Cal) to be converted to Gram Calories. Hence, 500 BTU (cal) = 126,082cm²

What is a British Thermal Unit?

A British thermal unit is a heat unit defined as the amount of heat needed to raise the temperature of one pound of water by one degree Fahrenheit. It is also one of the customary units used in the United States.

To covert 500 BTU, one must note that according to established scientific metrics.:

1 BTU = 252.164 Gram (Cal).

Hence,

500 BTU = 252.164  x 500

= 126,082 cm²

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a 2403 kg racecar has a total momentum of 9.912*10^4kgm/s at one point in the race. calculate the speed of the racecar at that point

Answers

In order to calculate the speed, we can use the formula for the momentum:

[tex]p=m\cdot v[/tex]

Where p is the momentum (in kg m/s), m is the mass (in kg) and v is the speed (in m/s).

So, using p = 99120 kg m/s and m = 2403 kg, we have:

[tex]\begin{gathered} 99120=2403\cdot v\\ \\ v=\frac{99120}{2403}\\ \\ v=41.25\text{ m/s} \end{gathered}[/tex]

person pulls a toboggan for a distance of 35.0 m along the snow with a rope directed 32.0o above the snow. The tension in the rope is 105.0 N. How much work is done on the toboggan by the tension force?

Answers

The amount of work done on the toboggan by the tension force of 105 N with a rope directed 32° above the snow is 3.12 KJ

W = F d cos θ

W = Work done

F = Force

d = Distance

θ = Angle between force and displacement vector

d = 35 m

F = 105 N

θ = 32°

W = 105 * 35 * cos 35°

W = 105 * 35 * 0.85

W = 3123.75 N m

W = 3.12 KJ

Work done is energy transferred to make an object move to a distance. Its unit is Joules which is denoted as J. It is the amount of work done by a force of 1 Newton to move a distance of 1 meter.

Therefore, the amount of work done on the toboggan by the tension force is 3.12 KJ

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You accidentally throw your car keys horizontally at 5.0 m/s from a cliff 45 m high. How far from the base of the cliff should you look for your keys?

Answers

The distance from the base of the cliff you should look for your key is 15 m

How to determine the distance from the cliff

To answer the question given above, we shall determine the time taken for the keyl to land. This can be obtained as follow:

Height (h) = 45 mAcceleration due to gravity (g) = 9.8 m/s²Time (t) = ?

h = ½gt²

45 = ½ × 9.8 × t²

45 = 4.9 × t²

Divide both side by 4.9

t² = 45 / 4.9

Take the square root of both side

t = √(45 / 4.9)

t = 3 s

Finally, we shall determin the distance from the base of the cliff the key will land. This can be obatined as follow:

Horizontal velocity (u) = 5 m/sTime (t) = 3 sDistance (s) = ?

s = ut

s = 5 × 3

s = 15 m

Thus, you should look for your key at a distance of 15 from the base of cliff

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A duck is paddling due East at 1.1 m/s across the river while it flows due South at 0.55 m/s What is her resultant velocity?

Answers

The diagram representing this scenario is shown below

A right angle triangle is formed. R represents the resultant velocity. To find R, we would apply the pythagorean theorem which is expressed as

hypotenuse^2 = one leg^2 + other leg^2

From the diagram,

hypotenuse = R

one leg = 1.1

other leg = 0.55

Thus,

R^2 = 1.1^2 + 0.55^2 = 1.5125

Taking square root of both sides,

R = square root of 1.5125

R = 1.23

The resultant velocity is 1.23 m/s

A kid is on a stationary sled, onsnowy ground with fls 0.105.It takes 71.2 N of force to setthe sled moving. How muchnormal force is acting?(Unit = N)

Answers

ANSWER

[tex]678.10N[/tex]

EXPLANATION

Parameters given:

Coefficient of static friction, μs = 0.105

Force, F = 71.2 N

The minimum force required to set the sled moving must be equal to the frictional force acting on the sled.

The frictional force is given mathematically as:

[tex]F_f=\mu_sN[/tex]

where N = normal force

Since the force required is equal to the frictional force, we have that:

[tex]F=F_f[/tex]

Therefore:

[tex]F=\mu_sN[/tex]

Solve for N:

[tex]\begin{gathered} N=\frac{F}{\mu_s} \\ N=\frac{71.2}{0.105} \\ N=678.10N \end{gathered}[/tex]

That is the normal force.

Whats the percent of 10 of 20

Answers

In order to determine the associated percent of 10 related to 20, proceed as follow:

If x is the percentage, then, you can write:

[tex]\frac{x}{100}\cdot20=10[/tex]

which means that x percentage of 20 is equal to 10. By solving for x, you get:

[tex]\begin{gathered} x=\frac{10}{20}\cdot100 \\ x=50 \end{gathered}[/tex]

Hence, 10 is the 50% of 20

Answer:2

Explanation:multiply 0.20 times 10 you get it

3. An object of mass 8 kg is sliding down a friction lessinclined plane of length 11 m that makes an angleof 70 deg with the horizontal. Calculate thework done by gravitational force as the objectslides from the top of the inclined plane to theground. (1 point)A. 099.045 JB. O1135.872 JC.810.391 JD.499.917

Answers

Given

Mass of the object, m=8 kg

Length of the inclined plane, l=11 m

Angle of inclination,

[tex]\theta=70^o[/tex]

To find

Calculate the work done by gravitational force as the object slides from the top of the inclined plane to the

ground.

Explanation

The height of the ramp

[tex]h=lsin\theta=11sin70^o[/tex]

The work done by gravitational force,

[tex]\begin{gathered} W=mgh \\ \Rightarrow W=8\times9.8\times11sin70^o \\ \Rightarrow W=810.390J \end{gathered}[/tex]

Conclusion

The work done is C.810.391

Atmospheric pressure is about 1.00 × 105 Pa on average.A. What is the downward force of the air on a desktop with surface area 2.59 m2?B. Convert the downward force of the air on a desktop with surface area 2.59 m2 to pounds to help others understand how large it is.

Answers

A)

The formula for calculating pressure is expressed as

pressure = force/area

From the information given,

pressure = 1.0 x 10^5 pa

Recall, 1 pa = 1 N/m^2

This means that

pressure = 1.0 x 10^5 N/m^2

surface area = 2.59 m^2

Force = pressure x area

Force = 1.0 x 10^5 x 2.59 = 259000 N

Recall,

B)

1 newton = 0.224808943 pounds

259000 newtons = 259000 x 0.224808943

= 58226 pounds

A 2.0 m long frictionless pendulum of mass 1.6 kg is released from point A at an angle & of
15 degrees. What is the speed of the pendulum at Point C.

Answers

The speed of pendulum at point C would be 0.47 m/sec when a 2 meter long frictionless pendulum of mass 1.6 kg is released from point A.

What is speed and how it is calculated out to be 0.47 meter/second?Speed is a quantity used to measure the distance travelled per unit time of the given time period.Here is given the length of rope be 2 meter and the mass of pendulum 1.6 kg released from point A.Using L cos theta = L ( 1 - cos theta) , we can write v^2/2g and putting all the numerical values we will get the speed be 0.47 m/s.Putting the angle that is theta be 15 degrees and getting the v the speed be 0.47 m/s.Hence the speed of the pendulum at point C would be 0.47 m/s.

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Consider the graph shown. Which of the motions is consistent with the graph?a) The object has a constant velocity in the negative direction.b) The object is moving in the negative direction with a changing speed.c) The object is moving in the positive direction and slowing down.

Answers

Given:

The graph of velocity vs time of an object

To find:

Which of the motions is consistent with the graph?

Explanation:

We see here that the object's initial velocity is positive, and the final velocity is zero. So, the object is slowing down. as the initial velocity is positive, the object's direction of movement is positive.

Hence, The object is moving in a positive direction and slowing down.

I need help on this science homework I forgot on what to do through number 4-7 including a,b,c,d, and e.

Answers

Answer:

[tex]\begin{gathered} (a)\Rightarrow v=2ms^{-1} \\ (b)\Rightarrow v=1ms^{-1} \\ (c)\Rightarrow v=6.67ms^{-1}_{} \\ (d)\Rightarrow v=1.2ms^{-1} \\ (e)\Rightarrow v=0ms^{-1} \end{gathered}[/tex]

Explanation: We need to calculate the speed on intervals a b c d and e, the speed can be calculated with the following formula:

[tex]v=\frac{\Delta S}{\Delta t}\Rightarrow(1)[/tex]

(a) 0-5 seconds Interval:

[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(10m-0m)}{(5s-0s)}=\frac{10m}{5s}=2ms^{-1} \\ v=2ms^{-1} \end{gathered}[/tex]

(b) 5-15 seconds Interval:

[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(20m-10m)}{(15s-5s)}=\frac{10m}{10s}=1ms^{-1} \\ v=1ms^{-1} \end{gathered}[/tex]

(c) 15-18 seconds Interval:

[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(40m-20m)}{(18s-15s)}=\frac{20m}{3s}=6.67ms^{-1} \\ v=6.67ms^{-1} \end{gathered}[/tex]

(d) 18-23 seconds Interval:

[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(46m-40m)}{(23s-18s)}=\frac{6m}{5s}=1.2ms^{-1} \\ v=1.2ms^{-1} \end{gathered}[/tex]

(e) 23-25 seconds Intervals:

[tex]\begin{gathered} v=\frac{\Delta S}{\Delta t}\Rightarrow v=\frac{(40m-40m)}{(25s-23s)}=\frac{0m}{2s}=0ms^{-1} \\ v=0ms^{-1} \end{gathered}[/tex]

A person is running at a velocity vv. By which factor would they need to increase their speed in order to double their kinetic energy?

Answers

The kinetic energy is given by:

[tex]K=\frac{1}{2}mv^2[/tex]

If we want to double it then we have:

[tex]2(\frac{1}{2}mv^2)=\frac{1}{2}m(\sqrt[]{2}v)^2[/tex]

Therefore the speed has to increas by a factor of square root of 2

which are neutrally charged, are found in thenucleus of the atom.

Answers

Given:

Nucleus of the atom

Required:

Neutra

what do fusion and fission have in common

Answers

Answer:

Explanation:

they both involve nuclear reactions that produce energy, but the application are not the same  

Ahmed wants to measure the weight of an object. What instrument should he use?

Answers

Answer:

a scale

Explanation:

you use it to weigh things

Answer:

its not a scale

Explanation:

state the dimension of energy in physics
hint: Energy = force × distance

Answers

Force = Mass * Acceleration = kg* m/s^2= MLT^-2

Distance = metres= L

Energy = MLT^-2 * L =ML^2T^-2

You place a box weighing 276 N on an inclined plane that makes a 44.5° angle with the horizontal.
Compute the component of the gravitational force acting down the inclined plane.
Answer in units of N.

Answers

Answer:

this is the answer

Explanation:

hope it helps

Can you help me match these with the correct word

Answers

When an object is thrown up into the air the time up is equal to the time down.

When an object is thrown up into the air and reacher the apex, the velocity of the object is zero.

If you throw an object straight up the air the velocity of the object is decreasing on its way up.

An object has 25 J of kinetic energy and 35 J of potential energy. What isthe total energy possessed by the object? *

Answers

The total energy possessed by the object is 60 J

Given:

The kinetic energy of the object, K=25 J

The potential energy of the object, P=35 J

To find:

The total energy possessed by the object.

Explanation:

The kinetic energy of an object is the energy possessed by it due to its motion. The kinetic energy is directly proportional to the square of the velocity of the object.

The potential energy of an object is the energy possessed by it due to its position. The gravitational potential energy of an object is directly proportioinal to the mass ofthe

On a fishing trip Justin rides in a boat 12 km south. The fish aren't biting so they go 4 km west. They then follow a school of fish 1 km north. What is his total displacement?

Answers

ANSWER:

The displacement is 11.7 km

STEP-BY-STEP EXPLANATION:

The displacement is the vector sum from the start point to the end point. Vector addition is applying the formula of the distance between two points.. We will use this formula:

[tex]d=\sqrt[]{(x_2-x^{^{}}_1)^2+(y_2-y_1)^2^{}}[/tex]

To better understand the exercise, we will draw a picture of the situation, it would look like this:

Replacing the points (-4, -11) and (0, 0)

[tex]\begin{gathered} d=\sqrt[]{(-4-0^{}^{}_{})^2+(-11_{}-0_{})^2} \\ d=\sqrt[]{16+121^{}} \\ d=\sqrt[]{137} \\ d=11.7 \end{gathered}[/tex]

Based on the periodic table why are He, Ne, Ar, Kr, and Xe the same
column/group/family?

Answers

The noble gases or inert gases helium (He), neon (Ne), argon (Ar), krypton (Kr), and radon (Rn) are found in group 8A of the periodic table. The name comes from little these elements react with other substances.

What is periodic table ?

The periodic table is a tabular arrangement of chemical elements arranged according to number of protons, starting with hydrogen and going up to the element the with highest atomic number. The quantity of protons found in the nucleus of an element's atom is referred as the element's atomic number.

Why is the periodic table important?

All known elements were grouped together in the periodic table of elements into groups with attributes. It becomes a crucial tool for chemists, nanotechnologists, and other scientists as a result. One can forecast what chemicals will react if you learn to use and understand the periodic table.

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given q= 10^-6 c, v= (3i + 4j )m/(s) , b=1i tesla find magnetic force

Answers

Lorentz magnetic force is the force exerted on a charged particle 'q' which moves with a velocity 'v' through a magnetic field B. The Lorentz forces exerted by moving charges on one another are not equal and opposite.

Given:

Charge, q = [tex]10^{-6}[/tex] C

Velocity, v = (3i + 4j) m/s

Magnetic field, B = 1i T.

To find: Magnetic force, F.

From Lorentz force law, the magnetic force is given by,

F = q × (v × B)

Substituting the given values,

F =  [tex]10^{-6}[/tex] × ((3i + 4j) × 1i)

Since i×i = 0, and j×i = k,

F =  [tex]10^{-6}[/tex] × 4k

∴ F = 4 × [tex]10^{-6}[/tex] k newtons.

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