There are two unitless vectors:

F1 = 8.92 i + 17.37 j
F2 = 8.31 i - 10.97 j

A third vector is added to them such that they add up to to the null vector:

F1 + F2 + F3 = 0

What is the angle of the third vector with respect to the +x-axis?

Answer:They have made community members shareholders so they get a share of the profits, which they use for schools and healthcare clinics

Explanation: Edmentum

Answer:

shareholders of the community get profits and that is used for schools and healthcare clinics. cs. 

Explanation:

Answer:

4m/s^2 ( A)

Explanation:

The solution is in the attached file

Answer:

6. 3.9 km SE

7. 17 km NE

8. 3496 seconds

9. 36 m

10. 13.27 km/m

Explanation:

6. and 7. added both numbers

8. multiplied both numbers

9. 60 x 60 = 3600

3600 x 0.01 = 36

10. 60 x 3 = 180

180 + 10 =190

190 / 14.32 = 13.27

Explanation:

[tex]kinetic \: energy \: is \: the \: energy \: of \: a \: body \: in \: motion. \\ that \: is \: energy \: of \: a \: body \: that \: is \: moving.\\ while \\ potential \: energy \: is \: the \: energy \: of \: a \: body \: by \: the \: virtue \: of \: its \: position \: \\ that \: is \: energy \: of \: a \: body \: that \: is \: not \: moving.[/tex]

♨Rage♨

Answer:

m₁ = 70 lb

Explanation:

Here we will use the law of conservation of momentum:

m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃

where,

m₁ = mass of first suitcase = ?

m₂ = mass of second suitcase = 30 lb

m₃ = mass of baggage carrier = 50 lb

u₁ = initial speed of first suitcase = 7.2 ft/s

u₂ = initial speed of second suitcase = 7.2 ft/s

u₃ = initial speed of baggage carrier = 0 ft/s

v₁ = Final speed of first suitcase = 4.8 ft/s

v₂ = Final speed of second suitcase = 4.8 ft/s

v₃ = Final speed of baggage carrier = 4.8 ft/s

because after collision all three will have same speed

Therefore,

(m₁)(7.2 ft/s) + (30 lb)(7.2 ft/s) + (50 lb)(0 ft/s) = (m₁)(4.8 ft/s) + (30 lb)(4.8 ft/s) + (50 lb)(4.8 ft/s)

(m₁)(7.2 ft/s) + (216 lb ft/s) + (0 lb ft/s) = (m₁)(4.8 ft/s) + (144 lb ft/s) + (240 lb ft/s)

(m₁)(7.2 ft/s) - (m₁)(4.8 ft/s) = 168 lb ft/s

m₁ = (168 lb ft/s)/(2.4 ft/s)

m₁ = 70 lb

Answer:

The force of forearm is 239.9 N

The force of elbow is 215.89 N

Explanation:

Suppose, When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.50 kg. If the biceps is connected to the forearm a distance [tex]d_{b}[/tex] 2.50 cm from the elbow, how much force [tex]F_{b}[/tex] must the biceps exert to hold a 950 g ball at the end of the forearm at distance dball J 36.0 cm from the elbow, with the forearm parallel to the floor? How much force [tex]F_{l}[/tex] must the elbow exert,

Given that,

Mass of forearm = 1.50 kg

Distance of forearm = 2.50 cm

Mass of ball = 950 g

Distance of ball = 36.0 cm

We need to calculate the force of forearm

Using balancing torque about elbow

[tex]F_{b}\times d_{b}=w_{f}\times\dfrac{d_{f}}{2}+w_{ball}\times d_{ball}[/tex]

Put the value into the formula

[tex]F_{b}\times 0.025= 1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36[/tex]

[tex]F_{b}=\dfrac{1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36}{0.025}[/tex]

[tex]F_{b}=239.9\ N[/tex]

We need to calculate the force of elbow

Using balancing force

[tex]F_{b}=F_{l}+w_{f}+w_{b}[/tex]

[tex]F_{l}=F_{b}-w_{f}-w_{b}[/tex]

Put the value into the formula

[tex]F_{l}=239.9-(1.50\times9.8)-(0.95\times9.8)[/tex]

[tex]F_{l}=215.89\ N[/tex]

Hence, The force of forearm is 239.9 N

The force of elbow is 215.89 N

Answer:

2.0 m

Explanation:

Energy is conserved.

Initial KE = Final PE + Work done by friction

½ mv² = mgh + Fd

½ mv² = mgh + mgμd

½ v² = gh + gμd

½ v² − gh = gμd

d = (½ v² − gh) / (gμ)

d = (½ (7.5 m/s)² − (10 m/s²) (2.1 m)) / ((10 m/s²) (0.35))

d = 2.0 m

Answer:

False

Explanation:

The cryosphere is all the part of the earth where water is in solid form. Where wind interacts with rocks, it is an example of atmosphere - geosphere interaction.

Answer:

The rowboat will move away from sally more quickly because the rowboat because the sailboat is larger in mass

Explanation:

Gnerally the row boat will move away from her quicker than the sailboat this is because the mass of the sail boat is larger than the row boat , hence the frictional force that opposes motion will be greater in the sailboat than in the row boat.

Answer:

The correct answer is A. A government that wants to increase its GDP would most likely increase the money supply to make it easier to borrow money.

Explanation:

If the government wanted to increase its GDP, the most appropriate way to do so would be to increase the money supply both through issuance and through a reduction in bank reserve requirements, thereby increasing the circulating money in the hands of society.

This, in turn, would make citizens reinvest that money, increasing economic production and, therefore, the national GDP.

Answer: A. Increase the money supply to make it easier to borrow money

Explanation: I just took the test on Ap ex

Answer:

significant

Explanation:

The digits in a measurement that are considered significant are all of those digits that represent marked calibrations on the measuring device plus one additional digit to represent the estimated digit (tenths of the smallest calibration).

Answer:

The energy is   [tex]U =  18.98 \  J [/tex]

Explanation:

From the question we are told that

   The inductor is  [tex]L  =  4.74 \ H[/tex]

    The resistance of the resistor is [tex]R =  9.33 \  \Omega[/tex]

    The voltage of the battery is [tex]V =  26.4 \  V[/tex]

Generally the current flowing in the circuit is mathematically represented as

      [tex]I =  \frac{V}{R}[/tex]

=>   [tex]I =  \frac{26.4}{9.33 }[/tex]

=>   [tex]I =  2.83 \ A[/tex]

Generally the corresponding energy stored in the circuit is  

       [tex]U =  \frac{1}{2} * L  *  I^2[/tex]

        [tex]U =  \frac{1}{2} *  4.74  *  2.83 ^2[/tex]

       [tex]U =  18.98 \  J [/tex]

Answer:

t = 5.77 s

Explanation:

This exercise will use Newton's second law for rotational motion

            τ = I α

             α = τ / I

             α = 66/175

             α = 0.3771   rad/s²

now we can use the rotational kinematics relations, remember that all angles must be in radians

            θ = 1 rev = 2π radians

            θ = w₀ t + ½ α t²

as the wheel starts from rest w₀ = 0

             t = √ (2θ/α)

let's calculate

             t = √ (2 2π / 0.3771)

             t = 5.77 s

I think it is D. Force

Answer:

Explanation:

Do you mean tire tracks?

If you do then the tracks show that there must have been a lot of heat involved. When the tires skid, the friction produced must be awfully hot to stop and 800 kg vehical (which is pretty small). The tire tracks that you see are bits of melted rubber taken from the tire.

Force is involved, but heat is the better answer.

gravity is necessary for friction to take place. If you were out in space, you could not get enough friction to leave tire tracks. (Force is awfully small).

Lubrication is there to give you a 4th choice. The exact opposite is what takes place. Lubrication reduces friction.

Answer:

61 m/s

Explanation:

        [tex]v_{f}^{2} - v_{o}^ {2} = 2* g* h (1)[/tex]

Answer:

25.  Approximately 8.1 meters

26. North 1.31 km, and East 2.81 km

Explanation:

25.

Notice that the displacements: 6 meters east and 5.4 south create the legs of a right angle triangle. The hypotenuse of that triangle will be the distance (d) needed to cover in order to get the ball in the hole in one putt. That is:

[tex]d=\sqrt{6^2+5.4^2} =\sqrt{65.16} \approx 8.072\,\,\,m[/tex]

which can be rounded to 8.1 m.

26.

Notice that the 3.1 km at an angle of 25 degrees north of east, is the hypotenuse of a right angle triangle that has for legs the east and north components of that distance.

We can find the leg corresponding to the east displacement using the cosine function (that relates adjacent side with hypotenuse):

[tex]east\,\, comp=3.1 * cos(25^o)\approx 2.809\,\,km[/tex]

and we can calculate the north component using the sine function that relates the opposite side to the angle with the hypotenuse.

[tex]north\,\,component = 3.1 * sin(25^o) \approx 1.31 \,\,km[/tex]

Answer:

b

Explanation:

Answer:

Explanation:

Given data

mass m= 1200kg

radius r=92m

coefficient of friction μ=0.65

We know that

F=mv2/r

F=μmg  

mv2/r = μsmg  

v^2/r = μsg  

vmax = √(rμg)

Substituting our data we have

vmax=√(92*0.65*9.81)

vmax=√586.638

vmax=24.22m/s

ans: 4m/s²

and 2m/s²

step:

f=m.a

in first condition

f=12N , M=3kg

so,

12=3.a

a=12/3

a=4m/s².

similarly

in second condition

f=12N ,M=6Kg

so,

12=a.6

a=12/6

a=2m/s²

Answer:

10) The distance between the archer and the tree is 50.074 meters.

11) The speed of the banana when it hits the water is approximately 13.554 meters per second.

Explanation:

10) The arrow experiments a parabolic motion, which is the combination of horizontal motion at constant velocity and vertical uniform accelerated motion. In this case we need to find the horizontal distance between the archer and the tree, calculated by the following kinematic equation:

[tex]x = x_{o} +v_{o}\cdot t \cdot \cos \theta[/tex] (Eq. 1)

Where:

[tex]x_{o}[/tex] - Initial position of the arrow, measured in meters.

[tex]x[/tex] - Final position of the arrow, measured in meters.

[tex]v_{o}[/tex] - Initial speed of the arrow, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]\theta[/tex] - Launch angle, measured in sexagesimal degrees.

If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 65\,\frac{m}{s}[/tex], [tex]t = 0.85\,s[/tex] and [tex]\theta = 25^{\circ}[/tex], the horizontal distance between the archer and the tree is:

[tex]x = 0\,m + \left(65\,\frac{m}{s}\right)\cdot (0.85\,s)\cdot \cos 25^{\circ}[/tex]

[tex]x = 50.074\,m[/tex]

The distance between the archer and the tree is 50.074 meters.

11) The final speed of the banana ([tex]v[/tex]), measured in meters per second, just before hitting the water is determined by the Pythagorean Theorem:

[tex]v = \sqrt{v_{x}^{2}+v_{y}^{2}}[/tex] (Eq. 2)

Where:

[tex]v_{x}[/tex] - Horizontal speed of the banana, measured in meters per second.

[tex]v_{y}[/tex] - Vertical speed of the banana, measured in meters per second.

Each component of the speed are obtained by using these kinematic equations:

[tex]v_{x} = v_{o}\cdot \cos \theta[/tex] (Eq. 3)

[tex]v_{y} = v_{o}\cdot \sin \theta +g\cdot t[/tex] (Eq. 4)

Where [tex]g[/tex] is the gravitational acceleration, measured in meters per square second.

If we know that [tex]v_{o} = 7.6\,\frac{m}{s}[/tex], [tex]\theta = -40^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]t = 0.75\,s[/tex], the components of final speed are, respectively:

[tex]v_{x} = \left(7.6\,\frac{m}{s} \right)\cdot \cos (-40^{\circ})[/tex]

[tex]v_{x} = 5.822\,\frac{m}{s}[/tex]

[tex]v_{y} = \left(7.6\,\frac{m}{s}\right)\cdot \sin (-40^{\circ})+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.75\,s)[/tex]

[tex]v_{y} = -12.240\,\frac{m}{s}[/tex]

And the speed of the banana right before hitting the water is:

[tex]v = \sqrt{\left(5.822\,\frac{m}{s} \right)^{2}+\left(-12.240\,\frac{m}{s} \right)^{2}}[/tex]

[tex]v \approx 13.554\,\frac{m}{s}[/tex]

The speed of the banana when it hits the water is approximately 13.554 meters per second.

Answer:

La aceleración necesaria para detener el avión es - 10.42 m/s².

Explanation:

Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.

Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en  M.U.A se cumple:

Vf² - Vo² = 2*a*d

donde:

En este  caso:

Reemplazando:

(0 m/s)² - (50 m/s)² = 2*a*120 m

Resolviendo:

[tex]a=\frac{(0 m/s)^{2} -(50 m/s)^{2} }{2*120 m}[/tex]

a= - 10.42 m/s²

La aceleración necesaria para detener el avión es - 10.42 m/s².

1) describe the life cycle of a star before it collapses into a black hole.

1) describe the life cycle of a star before it collapses into a black hole.ans: A star's life cycle is determined by its mass. The larger its mass, the shorter its life cycle. A star's mass is determined by the amount of matter that is available in its nebula, the giant cloud of gas and dust from which it was born. Over time, the hydrogen gas in the nebula is pulled together by gravity and it begins to spin. As the gas spins faster, it heats up and becomes as a protostar. Eventually the temperature reaches 15,000,000 degrees and nuclear fusion occurs in the cloud's core. The cloud begins to glow brightly, contracts a little, and becomes stable. It is now a main sequence star and will remain in this stage, shining for millions to billions of years to come. This is the stage our Sun is at right now.

2) describe the life cycle of a star before it becomes a dwarf.

ans: The life cycle of a low mass star (left oval) and a high mass star (right oval). ... As the core collapses, the outer layers of the star are expelled. A planetary nebula is formed by the outer layers. The core remains as a white dwarf and eventually cools to become a black dwarf.

3) what is the likely outcome of our sun?

ans: All stars die, and eventually — in about 5 billion years — our sun will, too. Once its supply of hydrogen is exhausted, the final, dramatic stages of its life will unfold, as our host star expands to become a red giant and then tears its body to pieces to condense into a white dwarf.

Answer:

Explanation:

Answer:

throw them away............

Answer:

Force Exerted by Sedan on Truck = - F

Explanation:

This question can easily be solved by using Newton's Third Law of Motion. Newton's Third Law of Motion clearly states that for every action force there exists an equal in magnitude reaction force. However, the direction of the reaction force is opposite to the direction of the action force. Mathematically,

Action Force = - Reaction Force

Hence, we have here:

Force Exerted by Semi Truck on Sedan = F

So, from Newton's Third Law of motion:

Force Exerted by Sedan on Truck = - Force Exerted by Semi Truck on Sedan

Force Exerted by Sedan on Truck = - F

Answer:

10.53m/s²

Explanation:

Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:

[tex]a = \frac{v^2}{r}[/tex]

v is the velocity of the car = 24.5m/s

r is the radius of the track = 57.0m

Substitute the given values into the formula:

[tex]a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}[/tex]

Hence the centripetal acceleration of the race car is 10.53m/s²

Answer:

Airplane X has more gravitational potential energy than Airplane W

Explanation:

Gravitational potential energy is defined as "the energy acquired by an object due to its positional change in presence of gravitational force."

That being said, gravitational potential energy depends on the height of an object above the ground. It also depends on the mass of the object and even further, the amount of gravitational force that is applied.

And if we take a look at the question again, we'd agree that the two airplanes are flying at different heights, this means their gravitational potential energy will be different. And as such, Airplane X has more gravitational potential energy than Airplane W