Answer:
[tex]200.38^0[/tex]
Explanation:
Given the forces:
F1 = 8.92 i + 17.37 j
F2 = 8.31 i - 10.97 j
If a third vector is added to them such that they add up to to the null vector as F1 + F2 + F3 = 0
then to get F3:
F3 = -F2-F1
F3 = -(8.31 i - 10.97 j)-(8.92 i + 17.37 j )
F3 = -8.31 i + 10.97 j-8.92 i - 17.37 j
F3 = -8.31i-8.92i+10.97j-17.37j
F3 = -17.23i-6.4j
from the vector:
x = -17.23 and y = -6.4
angle of the third vector with respect to the +x-axis is expressed as:
[tex]\theta = tan^{-1}\frac{y}{x}\\ \theta = tan^{-1}\frac{-6.4}{-17.23}\\\theta = tan^{-1}0.3715\\\theta = 20.38^0[/tex]
Hence the angle the vector makes with the x axis will be [tex]\theta = 180+20.38 = 200.38^0[/tex]
Determine the acceleration that results when a 12 N net force is applied to a 3 kg object.
a. 4 m/s 2
b. 6 m/s 2
c. 12 m/s 2
d. 36 m/s 2
Answer:
4m/s^2 ( A)
Explanation:
The solution is in the attached file
In the figure below, a block of 1.67 slides on a track with different levels, which has friction only at the highest point where the kinetic coefficient of friction is uk = 0.35. If the block has an initial speed V0 = 7.5m/s and the highest point of the track is at ℎ = 2.1 above the initial position of the block, calculate the distance where the friction force for the block is.
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
What is the magnitude of the force exerted by the biceps FbicepsFbicepsF_biceps? What is the magnitude of the force exerted by the elbow FelbowFelbowF_elbow
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
A man is passing barrels of water to another person from a height of 7 meters. The barrel is attached to a roof that is 10 meters above the ground. The rope holding the barrel is 8 meters long and will break if the tension exceeds 638 N. How many liters of water can the man put in the barrel without the rope breaking (assuming that the barrel is massless)? (The density of water = 1000 kg/m3 and 1 m3 = 1000 L. Density = mass/volume.)
***ECONOMICS***
A government that wants to increase its GDP would most likely take which
action?
A. Increase the money supply to make it easier to borrow money
Ο Ο
B. Decrease the money supply to slow the growth of inflation
C. Increase taxes on businesses that operate outside the country
O O
D. Decrease taxes on citizens who are poor or unemployed
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
Wind eroding the rocks on a moutain is an example of the atmosphere interacting with the cryosphere
true or false
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.
Rocks are part of the geosphere The geosphere is the part of the earth made up of solid rocks. Wind erosion occurs when wind wears down part of the geosphere.Sally is on a large sailboat that comes to a stop a small distance from the dock. Since it is such a small distance, Sally decides to jump to the dock. She makes the jump, but the large sailboat moves away from her as she jumps. Since Sally is interested to see what happens on other boats, she makes the same jump from a rowboat that is much smaller than the large sailboat. Which boat will move away from Sally more slowly
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.
4. Tires on the road show how friction produces
a. lubrication
b. heat
c. gravity
d. force
help :(
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.
In an RL series circuit, an inductor of 4.74 H and a resistor of 9.33 Ω are connected to a 26.4 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Answer in units of J.
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]
The name for this type of energy is
A.potential energy
B.motion
C.position
D.kinetic energy
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♨
Suppose the angle of incidence of a light ray is 42°.What is the angle of reflection?
Answer:
angle of reflection will be also 42°Explanation:
we know that ------------- angle of incidence=angle of reflectionSuppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m57.0 m . If the track is completely flat and the race car is traveling at a constant 24.5 m/s24.5 m/s (about 55 mph55 mph ) around the turn, what is the race car's centripetal (radial) acceleration
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
PLS ANSWER WILL MARK BRANLIEST!!!!!!!!!!!!
Describe the life cycle of a star before it collapses into a black hole.
Describe the life cycle of a star before it becomes a black dwarf.
What is the likely outcome of our sun? *
The sun will supernova and become a black hole.
The sun will swell, encompassing the inner planets and collapses into a dwarf star.
The sun will become a pulsar.
How Do You Know?
P.S. the how do you know is only for the last question
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.
QUESTION 10
An archer fires an arrow towards a tree with initial speed 65 m/s and angle 25 degrees above the horizontal. If the arrow takes 0.85
seconds to hit the tree, calculate the horizontal distance between the archer and the tree.
QUESTION 11
A monkey throws a banana from a tree into a nearby river. The banana has initial speed 7.6 m/s, is angled 40 degrees below the
horizontal, and takes 0.75 seconds to land in the river. Calculate the speed of the banana when it hits the water.
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.
what is a proper way and a safe way to dispose batteries
Answer:
throw them away............
What is the resultant velocity of a plane that is traveling at 245 m/s North and encounters a tailwind of 55 m/s North?
Answer:
b
Explanation:
How have the owners of the game reserve invested in the local community?
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:
The drawing shows two identical airplanes at an air show. The airplanes are flying at the same speed. Airplane W is flying 50 m higher than airplane X. Which statement best describes the energy of the two airplanes?
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
Which of the following is not true about taxes? A. Mandatory sum of money by government so that it can operate B. Due on April 15th C. largely collected to support private businesses D. Collected by the Internal Revenue Service (IRS)
What is the maximum speed with which a 1200-kg car can round a turn of radius 92.0 m on a flat road if the coefficient of static friction between tires and road is 0.65
Answer:
24.22m/sExplanation:
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
un avión aterriza en la superficie de un portaaviones a 50 m/s y se detiene por completo en 120 metros, ¿cuál es la aceleración necesaria para detenerlo?
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:
Vf: Velocidad final Vo: Velocidad inicial a: Aceleración d: Distancia recorridaEn este caso:
Vf: 0 m/s, porque el avión se detieneVo: 50 m/sa: ?d: 120 mReemplazando:
(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.
6. A cat walks 1.5km South and then 2.4km East. What is the total displacement of the cat?
7. A boy scout troop hikes 10.0 km East and then hikes an additional 7km North. What is the total displacement of the boy scout troop?
8. A go-kart is moving 23 m/s on the track. The track is 152m long. How long will take for the go-kart to reach the end of the track?
9. If a snail moves at a pace of .01m/s and it travels for 1 hr. How far does the snail get?
10. It takes 3 hr and 10 min to ride the tram to the top of Pike’s Peak in Colorado. The tram will travel a total distance of 14.32 km. What is the speed of the tram?
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
Question 1 of 15
All digits shown on the measuring device, plus one estimated digit, are
considered
Answer here
SUBMIT
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).
A stone is released from rest from the edge of a building roof 190 m above the ground. Neglecting air resistance, the speed of the stone, just before striking the ground, is:___________.
Answer:
61 m/s
Explanation:
If the stone is realeased from rest, this means that its initial velocity is 0.As tha stone is only influenced by gravity, and the acceleration due to it is constant (near the surface of the Earth), we can apply the following kinematic equation:[tex]v_{f}^{2} - v_{o}^ {2} = 2* g* h (1)[/tex]
Replacing by the values of g=9.8 m/s², and h=190 m, rearranging and solving for vf, we get:vf = √2*g*h =√2*9.8 m/s²*190 m = 61 m/s (assuming that the downward direction is the positive one).A 66-N ⋅ m torque acts on a wheel with a moment of inertia 175 kg ⋅ m2. If the wheel starts from rest, how long will it take the wheel to make one revolution?
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
An airline employee tosses two suitcases in rapid succession with a horizontal velocity of 7.2 ft/s onto a 50-lb baggage carrier which is initially at rest. Problem 14.003.a Conservation of momentum: two colliding suitcases Knowing that the final velocity of the baggage carrier is 4.8 ft/s and that the first suitcase the employee tosses onto the carrier has a weight of 30 lb, determine the weight of the other suitcase. (You must provide an answer before moving on to the next part.) The weight of the other suitcase is lb.
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
Forces are expressed in ________. (newtons or mass)
A large semi-truck, with mass 31x crashes into a small sedan with mass x . If the semi-truck exerts a force F on the sedan, what force will the sedan exert on the semi-truck
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
Determine the accelerations that result when a 12-N net force is applied to a 3-kg object and then to a 6-kg object.
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
Help solve these two problems im having trouble trying to start these problems?
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]