Three vectors , , and , each have a magnitude of 52.0 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 29.0 ˚, 191 ˚, and 311 ˚, respectively. What are (a) the magnitude and (b) the angle of the vector (relative to the +x direction in the range of (-180°, 180°)), and (c) the magnitude and (d) the angle of in the range of (-180°, 180°)? What are (e) the magnitude and (f) the angle (in the range of (-180°, 180°)) of a fourth vector such that ?

Answers

Answer 1

Answer:

a)   A´= A  

b)   θ₁´ = 29º,  θ₂´ = - 169º ,  θ₃´ = -49º

Explanation:

In this exercise you are asked to give the magnitudes and angles of the vectors from another system of

reference

a) The magnitudes

The magnitude of a vector, the size of which is a scalar, this does not depend on the reference system, since it is obtained by subtracting the coordinates of the end point minus the coordinate of the origin of the vector

          A = [tex]x_{f}[/tex]- x₀

if the vectors are measured in another reference frame

         x_{f}´ = xx_{f}- U

         x₀´ = x₀ -U

where U is the distance between the two reference frames

         A´ = x_{f}´ - x₀´

we substitute

         A´ = (x_{f} - U) - (x₀-U) = x_{f} - x₀

         A´ = A

 it does not change

b) Angles

The given angles are measured from the positive part of the x axis in a counterclockwise direction, it is asked to give these angles from the x axis

     θ₁ = 29º

     does not change

     θ₁´ = 29º

    θ₂ = 191º

   we measure clockwise

     θ₂´ = θ₂ - 360

     θ₂´ = 191 - 360

     θ₂´ = - 169º

     θ₃ =311º

    we measure clockwise

     θ₃´ = 311 -360

     θ₃´ = -49º


Related Questions

An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of the volume is occupied by liquid water and the rest by vapor. Now an electric resistor placed in the tank is turned on, and the tank is observed to contain saturated water vapor after 20 min. Determine:


(a) the volume of the tank

(b) the final temperature

(c) the electric power rating of the resistor

Answers

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

[tex]$m=\frac{V}{v}$[/tex]

[tex]$m=m_f+m_g$[/tex]

[tex]$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$[/tex]

[tex]$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$[/tex]       (taking the value of [tex]$v_g$[/tex] and [tex]$v_g$[/tex] at 200°C  )

[tex]$V=6.304 \times 10^{-3}$[/tex]

Now quality of vapor

[tex]$x=\frac{m_g}{m}$[/tex]

  [tex]$=3.377 \times 10^{-3}$[/tex]

Internal energy at state 1 can be found out by

[tex]$u_1=u_f+xu_{fg}$[/tex]

    [tex]$=850.65+3.377\times10^{-3}\times 1744.65$[/tex]

    = 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor [tex]$v_2=v_g \text { and }\ x=1$[/tex]

Tank is rigid, so volume of tank is constant.

[tex]$v_g=v_2=\frac{V}{m}$[/tex]

[tex]$v_g=\frac{6.304\times 10^{-3}}{1.4}$[/tex]

[tex]$v_g=4.502 \times 10^{-3} \ m^3 /kg$[/tex]

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

[tex]$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$[/tex]

   = 369.11° C

Internal energy at state 2

[tex]$u_2=2154.9 \ kJ/kg$[/tex]

Now power rating of the resistor

[tex]$P=\frac{m(u_2-u_1)}{t}$[/tex]

[tex]$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$[/tex]

  = 1.51 kW

what do humming birds eat? and how do they get their food?

Answers

bugs insects and stuff like every other bird i think

A student determines the density ρ of steel by taking measurements from a steel wire
Mass- 6.2 +-0.1g
Length- 25.0 +-0.1m
Diameter- 2.00 +-0.01mm
He uses the equation ρ= 4m/πd^2l
What is the percentage uncertainty in his calculated value of density ?

Answers

Answer:

The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].

Explanation:

We can estimate the absolute uncertainty by the definition of total differential. That is:

[tex]\Delta \rho \approx \frac{\partial \rho}{\partial m}\cdot \Delta m + \frac{\partial \rho}{\partial d}\cdot \Delta d + \frac{\partial \rho}{\partial l}\cdot \Delta l[/tex] (1)

Where:

[tex]\frac{\partial \rho}{\partial m}[/tex] - Partial derivative of the density with respect to mass, measured in [tex]\frac{1}{mm^{3}}[/tex].

[tex]\frac{\partial \rho}{\partial d}[/tex] - Partial derivative of the density with respect to diameter, measured in grams per cubic milimeter.

[tex]\frac{\partial \rho}{\partial l}[/tex] - Partial derivative of the density with respect to length, measured in grams per cubic milimeter.

[tex]\Delta m[/tex] - Mass uncertainty, measured in grams.

[tex]\Delta d[/tex] - Diameter uncertainty, measured in milimeters.

[tex]\Delta l[/tex] - Length uncertainty, measured in milimeters.

[tex]\Delta \rho[/tex] - Density uncertainty, measured in grams per cubic milimeters.

Partial derivatives are, respectively:

[tex]\frac{\partial \rho}{\partial m} = \frac{4}{\pi\cdot d^{2}\cdot l}[/tex] (2)

[tex]\frac{\partial \rho}{\partial d} = -\frac{8\cdot m}{\pi\cdot d^{3}\cdot l}[/tex] (3)

[tex]\frac{\partial \rho}{\partial l} = - \frac{4\cdot m}{\pi\cdot d^{2}\cdot l^{2}}[/tex] (4)

And we expand (1) as follows:

[tex]\Delta \rho \approx \frac{4\cdot \Delta m}{\pi\cdot d^{2}\cdot l} - \frac{8\cdot m\cdot \Delta d}{\pi\cdot d^{3}\cdot l}-\frac{4\cdot m\cdot \Delta l}{\pi\cdot d^{2}\cdot l^{2}}[/tex]

[tex]\Delta \rho \approx \left(\frac{4}{\pi\cdot d^{2}\cdot l}\right)\cdot \left(\Delta m -\frac{m\cdot \Delta d}{d}-\frac{m \cdot \Delta l}{l} \right)[/tex] (5)

If we know that [tex]d = 2\,mm[/tex], [tex]l = 25\,mm[/tex], [tex]m = 6.2\,g[/tex], [tex]\Delta m = \pm 0.1\,g[/tex], [tex]\Delta d = \pm 0.01\,mm[/tex] and [tex]\Delta l = \pm 0.1\,mm[/tex], then the absolute uncertainty is:

[tex]\Delta \rho \approx \pm\left[\frac{4}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)} \right]\cdot \left[(0.1\,g)-\frac{(6.2\,g)\cdot (0.01\,mm)}{2\,mm} -\frac{(6.2\,g)\cdot (0.1\,mm)}{25\,mm} \right][/tex]

[tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex]

And the expected density is:

[tex]\rho = \frac{4\cdot m}{\pi\cdot d^{2}\cdot l}[/tex] (6)

[tex]\rho = \frac{4\cdot (6.2\,g)}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)}[/tex]

[tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex]

The percentage uncertainty in his calculated value of density is:

[tex]\%e = \frac{\Delta \rho}{\rho}\times 100\,\%[/tex] (7)

If we know that [tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex] and [tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex], then the percentage uncertainty is:

[tex]\%e = \frac{\pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} }{78.941\times 10^{-3}\,\frac{g}{mm^{3}} }\times 100\,\%[/tex]

[tex]\%e = \pm 0.713\,\%[/tex]

The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].

Gravitational potential energy is potential energy based on what?
a. temperature
b.height
c. length
d. volume

Answers

B. height
The higher you are, the more gravitational energy you possess

Answer:

Height

Explanation:

the more height the more gravitational energy

Silver has a mass of 10.5 grams and a volume of 19.3 cm3. What is its density?

Answers

Answer:

The answer is 0.54 g/cm³

Explanation:

The density of a substance can be found by using the formula

[tex]density = \frac{mass}{volume} \\ [/tex]

From the question we have

[tex]density = \frac{10.5}{19.3} \\ = 0.54404145...[/tex]

We have the final answer as

0.54 g/cm³

Hope this helps you

help pls!!! this is worth 30 points, can someone pls help even if u can only give a few answers it would be helpful.

Answers

1- potential 2-kinetic

Answer:

1. potential 2. kinetic 3. potential 4. kinetic 5. kinetic 6. kinetic 7. kinetic 8. kinetic. potential

what is an isoelectronic series

Answers

Answer:

A comparison of the dimensions of atom an ions that have the same number of electron but different nuclear charges is called isoelectronic series

Answer:

A comparison of the dimensions of atoms or ions that have the same number of electrons but different nuclear charges, called an isoelectronic series, shows a clear correlation between increasing nuclear charge and decreasing size

Please help I will give brainliest

Answers

Answer:

ask questions that the findings bring up

Explanation:

hope this helps :)

Answer:

ask questions that the findings bring up

Which two staments explain how a cell's parts help it get nutrients

Answers

Answer:

I think it's A and D.

Hand washing is important especially in this time of pandemic

Answers

so you can kill the bacteria

What is another name for an alpha particle? Include the atomic number and mass number

Answers

Answer:

Alpha particles are Helium nuclei consisting of 2 protons and two neutrons (mass number 4, atomic number 2, no electrons and so carry a 2+ charge).

Explanation:

A stone is thrown at an angle of 34.0° above the horizontal from the top edge of a cliff with an initial speed of 18.3 m/s. A stopwatch measures the stone's trajectory time from the top of the cliff to the bottom at 2.6 s. What is the height of the cliff?

Answers

Answer:

Explanation:

Using the formula for calculating the maximum height in projectile

H = u²sin²theta/2g

u is the initial velocity = 18.3m/s

g is the acceleration due to gravity = 9.8m/s²

theta = 34°

Substitute

H = u²sin²theta/2g

H = 18.3²(sin34)²/2(9.8)

H = 334.89(0.5592)²/19.62

H = 5.337m

Hence the height of the cliff is 5.337m

A dart is thrown horizontally at a target's center that is 5.00\,\text m5.00m5, point, 00, start text, m, end text away. The dart hits the target 0.150\,\text m0.150m0, point, 150, start text, m, end text below the target's center.What was the initial horizontal velocity of the dart?

Answers

Answer:

Explanation:

Given

the range = 5.00m (distance moved in horizontal direction)

Height = 0.150m

Required

Initial velocity of the dart

Using the formula for calculating range of a projectile;

R = U√2H/g

5 = U√2(0.15)/9.8

5 = U√0.0306

5 = 0.1749U

U = 5/0.1749

U = 28.59m/s

Hence the initial horizontal velocity of the dart is 28.59m/s

Answer:

28.59

Explanation:

khan academy


5,000 joules of thermal energy were applied to 1-kg aluminum bar. What was the temperature increase?

Answers

Answer:

ΔT = 4.9°C

Explanation:

The thermal energy of the bar can be given as follows:

Thermal Energy = mCΔT

where,

m = mass of bar = 1 kg

C = specific heat capacity of aluminum = 1020 J/kg.°C

ΔT = Change in Temperature = ?

Therefore,

5000 J = (1 kg)(1020 J/kg.°C)ΔT

ΔT = (5000 J)/(1020 J/°C)

ΔT = 4.9°C

41. A statue weighs 1,000N and exerts a pressure of 20,000 Pa. How big is
the base of the statue in square meters?
please help

Answers

Answer:

The answer is 0.05 m²

Explanation:

The area of the base of the statue can be found by using the formula

[tex]a = \frac{f}{p} \\ [/tex]

f is the force

p is the pressure

From the question we have

[tex]a = \frac{1000}{20000} = \frac{1}{20} \\ [/tex]

We have the final answer as

0.05 m²

Hope this helps you

!!ill give brainliest!
Explain why !

Answers

False because there can be a straight line

a car starts from rest and accelerates to a speed of 30 m/s in a time of 3 seconds. the average acceleration of the car is

Answers

Answer:

10m/s²

Explanation:

v=u+at

30=0+a(3)

∴a=10m/s²

Answer:

10m/s^2. Just curious, what kind of car is it? Is it a lamborghini

Explanation:

how we will solve this question?
36.45cL=______=μL

Answers

Answer:

0.364

I believe... Good luck!

40 POINTS!
A 2,200 kg SUV is traveling at 25 m/s. What is the magnitude of its momentum?
A. 55,000 kg·m/s
B. 550 kg·m/s
C. 2,200 kg·m/s
D. 88 kg·m/s

Answers

Answer:55,000 kg•m/s

Explanation:

Newton's first law of motion states than an object's motion will not change
unless?

Answers

Answer:

it is hit by an external force

Explanation:

4. Describe the forces acting on a softball after it leaves the pitcher's hand. Ignore the effects of
air resistance.

Answers

Answer:

Drag, which is a force preventing the ball from going faster than it was thrown, this increases as velocity increases. Gravity, which Is pulling down on the ball.

Explanation:

Grade 8 Science Admin. May 2018 Released
37 A physics teacher performed a demonstration for a science class by pulling a crate across the
floor and measuring the force with a spring scale. While she pulled, a student measured the
acceleration of the crate with a handheld electronic device. The results of three trials are
shown below
Motion
F = 900 N
Actual Acceleration
of Crate
Trial
1
140 kg
Acceleration
(m/s)
0.36
0.34
0.38
0.36
2
3
Average
The teacher asked the class to calculate the acceleration of the crate based on the crate's
mass and the force she applied. What conclusion can be made about the difference between
the calculated acceleration and the actual acceleration that occurred in the trials?
A Another force in the direction of the motion produced a lower acceleration than calculated.
B An opposing force caused by friction produced a lower acceleration than calculated.
C Another force in the direction of the motion produced a higher acceleration than
calculated.
D An opposing force caused by friction produced a higher acceleration than calculated,

Answers

Answer:

B. An opposing force caused by friction produced a lower acceleration than calculated

Explanation:

Answer:

An opposing force caused by friction produced a lower acceleration than calculated.

Explanation:

A box sitting still on the ground by itself has a Normal Force of 700N, what is the mass? (gravity’s acceleration is 9.80 m/s²)

Answers

Answer:

Mass, m = 71.42 kg

Explanation:

Given that,

Normal force acting on a box, F = 700 N

We need to find the mass of a box. Let it is m. Normal force acting on an object is balanced by its weight such that,

F = mg

Where m is the mass of the box

[tex]m=\dfrac{F}{g}\\\\m=\dfrac{700\ N}{9.8\ m/s^2}\\\\m=71.42\ kg[/tex]

So, the mass of the box is 71.42 kg.

PLEASE HELP ME QUICKLY!!!!!!!!!!!!!!​

Answers

Answer:

I think thats the correct answer

Explanation:

A system has both thermal energy and chemical potential energy. According
to the law of conservation of energy, which statement describes the total
energy of this system?
O A. It will decrease over time as chemical potential energy is
transformed into thermal energy.
B. It will decrease as thermal energy is transferred out of the system.
C. It will remain constant because the total amount of thermal energy
plus chemical potential energy stays the same.
OD. It will increase as thermal energy is created from chemical
potential energy

Answers

Answer:

C. It will remain constant because the total amount of thermal energy  plus chemical potential energy stays the same.

Explanation:

According to the law of conservation of energy, the total energy of this system will remain constant because the total amount of thermal energy plus chemical potential energy stays the same.

The law of conservation of energy states that "energy is neither created nor destroyed in system but transformed from one form to another". Total energy in a system will always be conserved. It will remain constant.

What two factors are a part of thermohaline circulation

Answers

Answer:

These deep-ocean currents are driven by differences in the water's density, which is controlled by temperature (thermo) and salinity (haline). This process is known as thermohaline circulation.

Explanation:

Which reaction is an acid-base neutralization reaction?

A. Mg + 2HCl → MgCl2 + H2

B. HCl + NaOH → NaCl2 + H2O

C. SiCl4 → Si2 + 2Cl2

D. N2 + 3H2 → 2NH3

Answers

Answer:

option B is correct

Explanation:

neutralization is the reaction between acid and base to produce water and salt therefore option B is correct

I WILL GIVE BRAINLIEST!!!
The Earth's diameter at the equators is ____ its diameter at the poles.

equal to

less than

greater than

Answers

i believe the answer greater than

can a chemical change be undone?​

Answers

Answer:

no

Explanation:

because it will bind together in the molecular level

What must follow the division of the nucleus to complete the process of binary fission

Answers

Answer:

Replication of the DNA must occur. Segregation of the "original" and its "replica" follow. Cytokinesis ends the cell division process. Whether the cell was eukaryotic or prokaryotic, these basic events must occur.

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