Answer:
11.245 m
Explanation:
The vertical component of the initial velocity v0 is
[tex]v_v = v_0sin58^0 = 0.848v_0[/tex]
This makes the ball reach a maximum height of 8m. If we apply the conservation law of mechanical energy, its kinetic energy is converted to potential energy when it travels to the maximum height
[tex]E_p = E_k[/tex]
[tex]mgh = mv_v^2/2[/tex]
where m is the mass and h = 8 m is the maximum vertical distance traveled, g = 9.81m/s2 is the gravitational acceleration
we can divide both sides by m
[tex]gh = v_v^2/2[/tex]
[tex](0.848v_0)^2 = 2gh = 2*9.81*8 = 156.96[/tex]
[tex]0.848v_0 = \sqrt{156.96} = 12.53[/tex]
[tex]v_0 = 12.53 / 0.848 = 14.77 m/s[/tex]
So if the ball is directed fully upward at v0 speed then we can apply the same equation to find the new H
[tex]E_p = E_k[/tex]
[tex]mgH = mv_0^2/2[/tex]
[tex]H = \frac{v_0^2}{2g} = \frac{14.77^2}{2*9.81} = \frac{218.3}{19.62} = 11.245m[/tex]
the ball would go approximately 0.454 times as high if it were thrown straight upward at speed v0 compared to when it was thrown at an angle of 58.0 degrees above the horizontal.
When a projectile is launched at an angle, its vertical motion is influenced by gravity, while its horizontal motion is independent of gravity. In this case, the ball is initially launched at an angle of 58.0 degrees above the horizontal and reaches a maximum height of 8.0 m.
If the ball were thrown straight upward at the same initial speed (v0), its vertical motion would still be influenced by gravity. In both cases, the initial vertical speed (in the upward direction) is v0. The time it takes to reach the maximum height in both scenarios would also be the same because only the vertical component of velocity affects the vertical motion.
To find how high the ball would go if thrown straight upward, you can use the following kinematic equation:
h = (v0² * sin²(θ)) / (2 * g)
Where:
h is the maximum height (which we want to find).
v0 is the initial speed (given as v0).
θ is the launch angle (58.0 degrees).
g is the acceleration due to gravity (approximately 9.81 m/s²).
First, convert the angle from degrees to radians:
θ (in radians) = 58.0 degrees * (π radians / 180 degrees) ≈ 1.01 radians
Now, plug in the values and solve for h:
h = (v0² * sin²(1.01)) / (2 * 9.81 m/s²)
h ≈ (v0² * 0.454) / 19.62 m/s²
Now, if we consider that the initial speed v0 remains the same and throw the ball straight upward (θ = 90 degrees), the equation becomes:
h_straight_up = (v0² * sin²(π/2)) / (2 * 9.81 m/s²)
h_straight_up ≈ (v0² * 1) / 19.62 m/s²
Now, compare the two expressions for h and h_straight_up:
h / h_straight_up ≈ ((v0² * 0.454) / 19.62 m/s²) / ((v0² * 1) / 19.62 m/s²)
The "v0²" terms cancel out, and you get:
h / h_straight_up ≈ 0.454 / 1
h / h_straight_up ≈ 0.454
So, the ball would go approximately 0.454 times as high if it were thrown straight upward at speed v0 compared to when it was thrown at an angle of 58.0 degrees above the horizontal.
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An electrically neutral model airplane is flying in a horizontal circle on a 2.0-m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is 50 J. Reconsider the same situation, except that now there is a point charge of q on the plane and a point charge of -q at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is 53.5 J. Find the magnitude of the charges.
Answer:
q=3.5*10^-4
Explanation:
concept:
The force acting on both charges is given by the coulomb law:
F=kq1q2/r^2
the centripetal force is given by:
Fc=mv^2/r
The kinetic energy is given by:
KE=1/2mv^2
The tension force:
when the plane is uncharged
T=mv^2/r
T=2(K.E)/r
T=2(50 J)/r
T=100/r
when the plane is charged
T+k*|q|^2/r^2=2(K.E)charged/r
100/r+k*|q|^2/r^2=2(53.5 J)/r
q=√(2r[53.5 J-50 J]/k) √= square root on whole
q=√2(2)(53.5 J-50 J)/8.99*10^9
q=3.5*10^-4
A metal, M, forms an oxide having the formula MO2 containing 59.93% metal by mass. Determine the atomic weight in g/mole of the metal (M). Please provide your answer in 2 decimal places.
Answer: The molar mass of metal (M) is 47.86 g/mol
Explanation:
Let the atomic mass of metal (M) be x
Atomic mass of [tex](MO_2)=(x+32)g/mol[/tex]
To calculate the mass of metal, we use the equation:
[tex]\text{Mass percent of metal}=\frac{\text{Mass of metal}}{\text{Mass of metal oxide}}\times 100[/tex]
Mass percent of metal = 59.93 %
Mass of metal = x g/mol
Mass of metal oxide = (x + 32) g/mol
Putting values in above equation, we get:
[tex]59.93=\frac{x}{(x+32)}\times 100\\\\59.93(x+32)=100x\\\\x=47.86g/mol[/tex]
Hence, the molar mass of metal (M) is 47.86 g/mol
Your electricity bill says 834-kWh of consumption. With this amount of energy, how many 59-W light bulbs can be powered for an average of 7 hours?
Answer:
2019 light bulbs
Explanation:
So the electrical energy consumed by each 59-W light bulb within 7 hours is the product of the power and time duration
E = Pt = 59 * 7 = 413 Wh or 0.413 kWh
If each light bulb consumes 0.413 kWh, then the total number of light bulbs needed to consume 834 kWh would be
834 / 0.413 = 2019 light bulbs
A small boat is moving at a velocity of 3.35m/s when it is accelerated by a river current perpendicular to the initial direction of motion. The current accelerates the boat at 0.750m/s^2. what will the new velocity (magnitude and direction) of the boat be after 5 s?
Final answer:
To calculate the new velocity of the boat after being accelerated by the current for 5 seconds, one must use the Pythagorean theorem and arctangent function to find the magnitude and direction of the resultant velocity, resulting in approximately 5.04 m/s at 48.1 degrees from the original motion.
Explanation:
The student's question is about calculating the resultant velocity of a boat being accelerated by a river current perpendicular to its initial direction of motion. With an initial velocity of 3.35 m/s and a river current accelerating the boat at 0.750 m/s2, we need to find the new velocity after 5 seconds.
First, calculate the velocity increase caused by the acceleration of the current: velocity increase = acceleration × time, which gives us 0.750 m/s2 × 5 s = 3.75 m/s. This increase is perpendicular to the initial velocity.
Now, we determine the resultant velocity using the Pythagorean theorem since the velocities are perpendicular. The magnitude of the resultant velocity, Vtotal, is given by √(3.35 m/s)2 + (3.75 m/s)2, which equals to approximately 5.04 m/s. To find the direction, we use the arctangent function: θ = tan-1(3.75 m/s / 3.35 m/s), which results in a direction of approximately 48.1 degrees relative to the original direction of motion. Therefore, the boat will be moving with a velocity of approximately 5.04 m/s at a direction of approximately 48.1 degrees from its original direction, due to the current.
Consider a horizontal semi-transparent plate at a temperature 360 K that is irradiated from above and below. Air at [infinity] 310 K flows over both sides of the plate with a convective coefficient, 45 W/m2·K. The absorptivity of the plate is 0.4. If the radiosity is 4500 W/m2, find the irradiation, in W/m2, and emissivity of the plate. Is the plate diffuse-gray?
Answer:
[tex]G=6750\ W.m^2[/tex] is the irradiation
[tex]\epsilon=0.4725[/tex]
No, it is not a grey body because we don't have its transmissivity is not zero.
Explanation:
Given that:
Temperature of air, [tex]T_{\infty}=310\ K[/tex]
temperature of plate, [tex]T_s=360\ K[/tex]
convective heat transfer coefficient, [tex]h=45\ W.m^{-2}.K^{-1}[/tex]
absorptivity of plate, [tex]\alpha=0.4[/tex]
radiosity of the plate, [tex]J=4500\ W.m^{-2}[/tex]
From the energy balance eq. :
[tex]E_{in}=E_{out}[/tex]
since two surfaces are involved
[tex]2G=2J+2q_{conv}[/tex]
[tex]G=J+q_{conv}[/tex]
where
[tex]q_{conv}=[/tex] heat transfer per unit area due to convection
[tex]G=[/tex] irradiation per unit area
[tex]G=J+h.\Delta T[/tex]
[tex]G=4500+45\times (360-310)[/tex]
[tex]G=6750\ W.m^2[/tex] is the irradiation
Since radiosity includes transmissivity, reflectivity and emissivity.
[tex]J=G.\tau+G.\rho+E[/tex]
[tex]J=G.\tau+G.\rho+\epsilon\times \sigma\times T^{4}[/tex]
where:
[tex]\tau=[/tex] transmissivity
[tex]\rho=[/tex] reflectivity
[tex]\epsilon=[/tex] emissivity
Since the absorptivity of the plate is 0.4 and the plate is semitransparent so the transmissivity and reflectivity = 0.3 each.
[tex]4500=0.3\times 6750+0.3\times 6750+\epsilon\times 5.67\times 10^{-8}\times 360^4[/tex]
[tex]\epsilon=0.4725[/tex]
No, it is not a grey body because we don't have its transmissivity is not zero.
The Stefan-Boltzmann law of radiation can be used to find the irradiation and emissivity of a plate. The irradiation can be found using the net rate of heat transfer equation, and the emissivity can be found by rearranging the equation. The plate is diffuse-gray if its emissivity is between 0 and 1.
Explanation:The irradiation and emissivity of the plate can be found using the Stefan-Boltzmann law of radiation. The rate of heat transfer by emitted radiation is given by P = 6 AeT^4, where o = 5.67 × 10^-8 J/s. m². K^4 is the Stefan-Boltzmann constant, A is the surface area of the object, and T is its temperature in kelvins.
To find the irradiation, we can use the equation Qnet = σe A (T₂^4 – T₁^4), where Qnet is the net rate of heat transfer, σ is the Stefan-Boltzmann constant, e is the emissivity of the body, A is the surface area of the object, T₂ is the temperature of the surrounding air, and T₁ is the temperature of the plate.
To find the emissivity of the plate, we can rearrange the equation Qnet = σe A (T₂^4 – T₁^4) to solve for e. The plate is diffuse-gray if its emissivity is between 0 and 1. If the emissivity is 0, the plate is a perfect reflector, and if the emissivity is 1, the plate is a perfect absorber and emitter of radiation.
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A 0.500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. Calculate the maximum value of its (a) speed and (b) acceleration, (c) the speed and (d) the acceleration when the object is 6.00 cm from the equilibrium position, and (e) the time interval required for the object to move from x = 0 to x = 8.00 cm.
a. 40cm/s,
b. 160 cm/s2,
c. 32cm/s,
d. -96cm/s2,
e. 0.232s
Answer:
a) [tex]v_{max}=0.4\ m.s^{-1}[/tex]
b) [tex]a_{max}=1.6\ m.s^{-2}[/tex]
c) [tex]v_x=0.32\ m.s^{-1}[/tex]
d) [tex]a_x=0.96\ m.s^{-1}[/tex]
e) [tex]\Delta t=0.232\ s[/tex]
Explanation:
Given:
mass of the object attached to the spring, [tex]m=0.5\ kg[/tex]
spring constant of the given spring, [tex]k=8\ N.m^{-1}[/tex]
amplitude of vibration, [tex]A=0.1\ m[/tex]
a)
Now, maximum velocity is obtained at the maximum Kinetic energy and the maximum kinetic energy is obtained when the whole spring potential energy is transformed.
Max. spring potential energy:
[tex]PE_s=\frac{1}{2} .k.A^2[/tex]
[tex]PE_s=0.5\times 8\times 0.1^2[/tex]
[tex]PE_s=0.04\ J[/tex]
When this whole spring potential is converted into kinetic energy:
[tex]KE_{max}=0.04\ J[/tex]
[tex]\frac{1}{2}.m.v_{max}^2=0.04[/tex]
[tex]0.5\times 0.5\times v_{max}^2=0.04[/tex]
[tex]v_{max}=0.4\ m.s^{-1}[/tex]
b)
Max. Force of spring on the mass:
[tex]F_{max}=k.A[/tex]
[tex]F_{max}=8\times 0.1[/tex]
[tex]F_{max}=0.8\ N[/tex]
Now acceleration:
[tex]a_{max}=\frac{F_{max}}{m}[/tex]
[tex]a_{max}=\frac{0.8}{0.5}[/tex]
[tex]a_{max}=1.6\ m.s^{-2}[/tex]
c)
Kinetic energy when the displacement is, [tex]\Delta x=0.06\ m[/tex]:
[tex]KE_x=PE_s-PE_x[/tex]
[tex]\frac{1}{2} .m.v_x^2=PE_s-\frac{1}{2} .k.\Delta x^2[/tex]
[tex]\frac{1}{2}\times 0.5\times v_x^2=0.04-\frac{1}{2} \times 8\times 0.06^2[/tex]
[tex]v_x=0.32\ m.s^{-1}[/tex]
d)
Spring force on the mass at the given position, [tex]\Delta x=0.06\ m[/tex]:
[tex]F=k.\Delta x[/tex]
[tex]F=8\times 0.06[/tex]
[tex]F=0.48\ N[/tex]
therefore acceleration:
[tex]a_x=\frac{F}{m}[/tex]
[tex]a_x=\frac{0.48}{0.5}[/tex]
[tex]a_x=0.96\ m.s^{-1}[/tex]
e)
Frequency of oscillation:
[tex]\omega=\sqrt{\frac{k}{m} }[/tex]
[tex]\omega=\sqrt{\frac{8}{0.5} }[/tex]
[tex]\omega=4\ rad.s^{-1}[/tex]
So the wave equation is:
[tex]x=A.\sin\ (\omega.t)[/tex]
where x = position of the oscillating mass
put x=0
[tex]0=0.1\times \sin\ (4t)[/tex]
[tex]t=0\ s[/tex]
Now put x=0.08
[tex]0.08=0.1\times \sin\ (4t)[/tex]
[tex]t=0.232\ s[/tex]
So, the time taken in going from point x = 0 cm to x = 8 cm is:
[tex]\Delta t=0.232\ s[/tex]
Final answer:
The problem involves calculating the maximum speed, maximum acceleration, speed and acceleration at a certain distance from equilibrium, and the time interval for an object in simple harmonic motion. Using formulas for SHM including maximum speed (v_max = ωA), maximum acceleration (a_max = ω^2A), and expressions for speed and acceleration at a given position, one can determine these values for the object on the spring.
Explanation:
The student has asked us to calculate various properties of an object undergoing simple harmonic motion (SHM) when attached to a spring with a known force constant and amplitude. To answer this question, one needs to use equations that describe SHM.
Maximum Speed (v_max) Calculation:
The maximum speed (v_max) of an object in SHM occurs when it passes through the equilibrium point and can be calculated using the formula v_max = ωA, where ω is the angular frequency (ω = sqrt(k/m)) and A is the amplitude of the motion.
Maximum Acceleration (a_max) Calculation:
The maximum acceleration (a_max) occurs at the maximum displacement and is given by a_max = ω^2A.
Speed at 6 cm from Equilibrium:
To find the speed at a certain position x, we use the formula v = ω sqrt(A^2 - x^2).
Acceleration at 6 cm from Equilibrium:
Acceleration at any position x is a = -ω^2x.
Time Interval to Move from 0 to 8 cm:
The time interval to move from x = 0 to a certain position x can be found using the formula for time in SHM as a function of position.
Your cell phone works as both a radio transmitter and receiver. Say you receive a call at a frequency of 880.65 MHz. What is the wavelength in meters?
Answer:
0.34m
Explanation:
Final answer:
The wavelength of a radio signal received at a frequency of 880.65 MHz is approximately 0.34 meters, using the formula λ = c / f, where c is the speed of light.
Explanation:
If you receive a call at a frequency of 880.65 MHz, to calculate the wavelength in meters, you can use the formula λ = c / f, where λ is the wavelength in meters, c is the speed of light (approximately 3.00 × 108 m/s), and f is the frequency in hertz (Hz). Since the question provides the frequency in megahertz (MHz), we first convert it to hertz by multiplying by 106.
The frequency is 880.65 MHz, which is equal to 880.65 × 106 Hz. Thus, the wavelength λ = 3.00 × 108 m/s / (880.65 × 106 Hz).
Calculating this gives a wavelength of approximately 0.34 meters.
Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a series impedance of (0.4+????2.7) Ω per phase. One of the loads absorbs 560 kVA at 0.707 power factor lagging, and the other 132 kW at unity power factor. The line-to-line voltage at the load end of the line is 2200√3 V. Compute (a) the line-to-line voltage magnitude (in RMS value) at the source end of the line, (b) the total real and reactive power losses in the three-phase line, and (c) the total real and reactive power supplied at the sending end of the line.
Answer:
find answers below
Explanation:
a)
S1 560 e^(j acos 0.707 ) ⋅kVA S1 = ( ) 395.92 396.04j kW+ ⋅
S2 := 132 kW⋅ S2 = 132 kW⋅
Sd S1 +S2 :=
Sd = 527.92 396.04j kW
Sd = 660 kVA ⋅
arg Sd = 36.877 deg ⋅
Vd 2.2 e^(− j⋅0⋅deg)⋅kV
current in the line =(Sd/3 Vd) ⋅
⎯
:=
I Line = 79.988 60.006j A
Line = 99.994 A
arg I( ) Line = −36.877⋅deg
r Line := 0.4⋅Ω resistance in the line xLine := 2.7⋅Ω
Vsan Vd+ (r Line+ j xLine) ILine
Vsan = ( ) kV 2.394 0.192j + ⋅ Vsan = 2.402 kV⋅ arg V( ) san = 4.584 deg ⋅
Vsab 3 e
j 30 ⋅ ⋅deg ⋅ Vsan := ⋅
Vsab = ( ) kV 3.425 2.361j + ⋅ Vsab = 4.16 kV⋅ arg V( ) sab = 34.584 deg
b)
PLine 3 I ( ) Line
2 ⋅ r
Line := ⋅
PLine = 12 kW⋅
QLine 3 I ( ) Line
2 ⋅ xLine := ⋅ QLine = 80.99 kVAR ⋅
c)
Ss 3 Vsan ⋅ I
Line := ⋅
⎯ Ss = ( ) kVA 539.919 477.03j + ⋅
Ss = 720.5 kVA ⋅ arg S( )s = 41.461 deg ⋅
Ps Re S( )s := Ps = 540 kW⋅
Qs Im S( )s := Qs = 477 kVAR ⋅
S1 S2 + PLine + j QLine + ⋅ = ( ) kVA 539.919 477.03j + ⋅ Check
In the compound MgS, the sulfide ion has 1. lost one electrons. 2. lost two electrons. 3. gained one electron. 4. gained two electrons.
Answer:
4.has gained two electrons
Explanation:
There exist electrovalent bonding the compound MgS . In electrovalent bonding, there is a transfer of electrons from the metal to non-metal.
Magnesium atom has an atomic number 12 and its electron configuration is 2,8,2
Sulfur atom , a non-metal has atomic number of 16 and its electron configuration = 2,8,6
This means that magnesium as a metal needs to loose two electrons from its valence shell to attain its stable structure.Also sulfur requires two more electron to achieve its octet structure.
Hence a transfer of electrons will take place from magnesium atom to sulfur atom, sulfur gaining two electrons.
Final answer:
In MgS, the sulfide ion has gained two electrons, resulting in a charge of -2, represented as S²-. Magnesium loses two electrons to form Mg²+, balancing the electron transfer to create a stable ionic compound.
Explanation:
In the compound MgS, the sulfide ion has gained two electrons to achieve a stable electron configuration. When sulfur (S), which has an atomic number of 16, gains two electrons, it results in an ion with 18 electrons and 16 protons. This gives the ion a charge of -2, since there are two more negative electrons than positive protons. Therefore, the sulfide ion is represented as S²-. The magnesium atom loses its two valence electrons to become a Mg²+ cation, as magnesium is in Group 2A of the periodic table and tends to lose two electrons to achieve a noble gas electron configuration. This electron transfer process is balanced, meaning the number of electrons lost by magnesium is equal to the number of electrons gained by sulfur, forming a stable ionic compound.
A steady electric current flows through a wire. If 9.0 C of charge passes a particular spot in the wire in a time period of 2.0 s, what is the current in the wire? 4.5 A 18 A 0.22 A 9.0 A If the current is a constant 3.0 A, how long will it take for 14.0 C of charge to move past a particular spot in the wire? 0.21 s 4.7 s 14.0 s 42 s
1) Current: 4.5 A
2) Time taken: 4.7 s
Explanation:
1)
The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:
[tex]I=\frac{q}{t}[/tex]
where
I is the current
q is the amount of charge passing a given point in a time t
For the wire in this problem, we have
q = 9.0 C is the amount of charge
t = 2.0 s is the time interval
Solving for I, we find the current:
[tex]I=\frac{9.0}{2.0}=4.5 A[/tex]
2)
To solve this problem, we can use again the same formula
[tex]I=\frac{q}{t}[/tex]
where
I is the current
q is the amount of charge passing a given point in a time t
In this problem, we have:
I = 3.0 A (current)
q = 14.0 C (charge)
Therefore, the time taken for the charge to move past a particular spot in the wire is
[tex]t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s[/tex]
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The electric current in the wire in the first scenario is 4.5 A, and it will take 4.7 seconds for 14.0 C of charge to move past a particular spot on the wire with a constant current of 3.0 A in the second scenario.
Explanation:The amount of electric current in a circuit is defined by the amount of electric charge that passes through it in a given amount of time. This is represented by the formula I = Q / t, where I is the current, Q is the charge and t is the time.
In the first part of your question, we are given that Q = 9.0 C and t = 2.0 s. We can substitute these values into the formula to find the current: I = 9.0 C / 2.0 s = 4.5 A.
In the second part, we are given that I = 3.0 A and Q = 14.0 C. This time, we need to rearrange the formula to find t: t = Q / I = 14.0 C / 3.0 A = 4.7 s.
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The value of q is and the value of r is 75.0 cm. Note that, in this question, you are only asked to find the magnitude of the net force, but you should also think about the direction of the net force. What is the magnitude of the net force on the ball of mass m that is located on the positive y-axis, because of the other four balls
Answer:
F_net = - 0.365 N (Down-ward direction)
Explanation:
Given:
- Value of r = 0.75 m
- Charges on x axis are -2*q
- Charge +q on origin
- Charge on - y axis is -2q
- Charge on + y axis is +q
- q = 5.00 * 10^-6 C
Find:
-What is the magnitude of the net force on the ball of mass m that is located on the positive y-axis, because of the other four balls?
Solution:
- Force due to each of the two charges on x axis:
F_x = k*(-2*q)*(+q) / r*^2
r* = sqrt(2)*r
F_x = -k*q^2 / r^2 (Down-wards)
- Force due to +q charge on origin:
F_y = k*(+q)*(+q) / r^2
F_y = + k*q^2 / r^2 (Up-wards)
- Force due to -2*q charge on y-axis:
F_-2y = k*(-2*q)*(+q) / 4*r^2
F_-2y = - k*q^2 / 2*r^2 (Downwards-wards)
- Total net Force on charge +q on + y-axis:
2*F_x*sin(45) + F_y + F_-2y = F_net
-sqrt(2)*k*q^2 / r^2 + k*q^2 / r^2 - k*q^2 / 2*r^2 = F_net
(0.5-sqrt(2))*k*q^2 / r^2 = F_net
F_net = (0.5-sqrt(2))*(8.99*10^9)*(5*10^-6)^2 / 0.75^2
F_net = - 0.365 N
Whose view was that the Cosmos was based on the belief that every occurrence in the physical universe had logos behind it and that is where life originated?
Answer:
Aristotle
Explanation:
Aristotelian theory of the Universe
For two millennia, the philosophical tradition considered that the universe was eternal and did not change. The wise Aristotle said so, with total clarity and his ideas dominated Western thought for more than two thousand years.
This distinguished philosopher believed that the stars are made of an imperishable matter and that the landscapes of the sky are immutable.
From the time of Aristotle until the beginning of the twentieth century, the idea that the universe was static, that the cosmos had been eternally equal, was admitted.
In those years, the origin of the universe was not really considered in a scientific way, since it was based on the basis that the gods had created everything that exists, at the time they wanted it, according to their omnimous power.
So that all the efforts of the wise men of the time focused on discovering the existing organization in the universe created by the gods.
According to Aristotle and the thinkers of the fourth century B.C. what is below the Moon is a changing world, what is beyond the Moon is an immutable world.
A spaceship moves radially away from Earth with acceleration 29.4 m/s 2 (about 3g). How much time does it take for sodium streetlamps (λ = 589 nm) on Earth to be invisible to the astronauts who look with a powerful telescope upon the city streets of Earth?
Answer:
doppler shift's formula for source and receiver moving away from each other:
λ'=λ°√(1+β/1-β)
Explanation:
acceleration of spaceship=α=29.4m/s²
wavelength of sodium lamp=λ°=589nm
as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm
using doppler shift's formula:
λ'=λ°√(1+β/1-β)
putting the values:
700nm=589nm√(1+β/1-β)
after simplifying:
β=0.17
by this we can say that speed at that time is: v=0.17c
to calculate velocity at an acceleration of a=29.4m/s²
we suppose that spaceship started from rest so,
v=v₀+at
where v₀=0
so v=at
as we want to calculate t so:-
t=v/a v=0.17c ,c=3x10⁸ ,a=29.4m/s²
putting values:
=0.17(3x10⁸m/s)/29.4m/s²
t=1.73x10⁶
Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 6.00 m/s. The velocity of the ball relative to Mia is 5.00 m/s in a direction 30.0 east of south. What are the magnitude and direction of the velocity of the ball relative to the ground?
Answer:
v_b = 10.628 m /s (13.605 degrees east of south)
Explanation:
Given:
- Velocity of mia v_m = + 6 j m/s
- Velocity of ball wrt mia v_bm = 5.0 m/s 30 degree due east of south
Find:
What are the magnitude and direction of the velocity of the ball relative to the ground? v_b
Solution:
- The relation of velocity in two different frame is given:
v_b - v_m = v_bm
- Components along the direction of v_b,m:
v_b*cos(Q) - v_m*cos(30) = 5
v_b*cos(Q) = 5 + 6 sqrt(3) / 2
v_b*cos(Q) = 5 + 3sqrt(3)
- Components orthogonal the direction of v_b,m:
-v_m*sin(30) = v_b*sin(Q)
-6*0.5 = v_b*sin(Q)
-3 = v_b*sin(Q)
- Divide two equations:
tan(Q) = - 3 / 5 + 3sqrt(3)
Q = arctan(- 3 / 5 + 3sqrt(3)
Q = -16.395 degrees
v_b = -3 / sin(-16.395)
v_b = 10.63 m/s
The magnitude of the soccer ball's velocity relative to the ground is 2.89 m/s, and its direction is approximately 56.31° north of east. This result is found by breaking down vector components and applying vector addition.
Explanation:You're asking about the velocity of the ball relative to the ground when soccer players Mia and Alice are running, and Alice passes the ball to Mia. To solve this, we'll use vector addition. Mia is running due north at 6.00 m/s, and the velocity of the ball relative to Mia is 5.00 m/s at 30° east of south. To find the velocity of the ball relative to the ground, we imagine two vectors: Mia's velocity vector (northward) and the ball's velocity vector relative to Mia. The latter will have an east and a south component due to the 30° angle.
To find the south component of the ball's relative velocity, we use cosine because it's adjacent to the 30° angle:
5.00 m/s * cos(30°) = 4.33 m/s. The east component is found using sine:
5.00 m/s * sin(30°) = 2.50 m/s.
Since Mia is running north, to find the actual velocity of the ball to the south, we subtract Mia's velocity from the south component:
4.33 m/s - 6.00 m/s = -1.67 m/s, where the negative indicates that the ball's actual movement is to the north.
Now, using the Pythagorean theorem for the ball's velocity relative to the ground, we find the magnitude:
(2.50 m/s)² + (-1.67 m/s)² = √(6.25 + 2.7889)
√8.3389 m²/s² = 2.89 m/s.
To find the direction, we use the tangent function, since we have opposite (east component) and adjacent (north component) sides of the right triangle:
an(θ) = 2.50 / 1.67; θ = tan(⁻¹)(2.50 / 1.67);
θ ≈ 56.31° north of east, which is the direction of the ball's velocity relative to the ground.
Two balls of different radii, 53 cm and 26 cm move directly toward each other with the same speed. If they are originally at a center-to-center distance 223 cm and it takes them 18.9 s to collide, how fast were the balls moving?
Answer:
[tex]v = 3.81\ m/s[/tex]
Explanation:
given,
Radius of the ball, r₁ = 53 cm
Radius of another ball, r₂ = 26 cm
center to center distance between the balls = 223 cm
time, t = 18.9 s
surface to surface distance between them
S = 223 - (53+26)
S = 144 cm
Speed of the ball = ?
[tex]relative\ speed = \dfrac{distance}{time}[/tex]
[tex]2 v = \dfrac{144}{18.9}[/tex]
both the balls are moving towards each other so, speed doubles.
[tex]v= \dfrac{7.62}{2}[/tex]
[tex]v = 3.81\ m/s[/tex]
Speed of the balls is equal to 3.81 m/s
Two small plastic spheres each have a mass of 1.2 g and a charge of -56.0 nC . They are placed 3.0 cm apart (center to center). Part A What is the magnitude of the electric force on each sphere? Express your answer to two significant figures and include the appropriate units.
Answer:
F_e = +/- 0.013133 N
Explanation:
Given:
- charge on each sphere q = -56 nC
- separation of spheres r = 3.0 cm
- charge constant k = 8.99*10^9
Find:
- Magnitude of Electric Force F_e on each sphere.
Solution:
The magnitude of electric Force F_e of one sphere on other is:
F_e = k*q^2 / r^2
- Plugging the given values :
F_e = (8.99*10^9) * (56*10^-19)^2 / (0.03)^2
F_e = 0.013133 N
- An equal and opposite force is experienced on the other sphere. Hence, F_e = + / - 0.013133 N
Two identical loudspeakers separated by distance dd emit 200 Hz sound waves along the x-axis. As you walk along the axis, away from the speakers, you don't hear anything even though both speakers are on.
What are the three lowest possible values of d? Assume a sound speed of 340 m/s.
Answer:
The first possible value of d is 0.85 m
The second possible value of d is 2.55 m
The third possible value of d is 4.25 m
Explanation:
Given that,
Distance =d
Frequency of sound wave= 200 Hz
We need to calculate the wavelength
Using formula of wavelength
[tex]\lambda=\dfrac{v}{f}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{340}{200}[/tex]
[tex]\lambda=1.7\ m[/tex]
The separation between the speakers in the destructive interference is
[tex]\Delta x= d[/tex]
The equation for destructive interference
[tex]2\pi\times\dfrac{\Delta x}{\lambda}-\Delta\phi_{0}=(m+\dfrac{1}{2})2\pi[/tex]
The loudspeakers are in phase
So, [tex]\Delta\phi_{0}=0[/tex]
The equation for destructive interference is
[tex]2\pi\times\dfrac{d}{\lambda}=(m+\dfrac{1}{2})2\pi[/tex]....(I)
Here, m = 0,1,2,3.....
We need to calculate the first possible value of d
For, m = 0
Put the value in the equation (I)
[tex]2\pi\times\dfrac{d_{1}}{1.7}=(0+\dfrac{1}{2})2\pi[/tex]
[tex]d_{1}=\dfrac{1.7}{2}[/tex]
[tex]d_{1}=0.85\ m[/tex]
We need to calculate the second possible value of d
For, m = 1
Put the value in the equation (I)
[tex]2\pi\times\dfrac{d_{2}}{1.7}=(1+\dfrac{1}{2})2\pi[/tex]
[tex]d_{2}=\dfrac{1.7\times3}{2}[/tex]
[tex]d_{2}=2.55\ m[/tex]
We need to calculate the third possible value of d
For, m = 1
Put the value in the equation (I)
[tex]2\pi\times\dfrac{d_{3}}{1.7}=(2+\dfrac{1}{2})2\pi[/tex]
[tex]d_{3}=\dfrac{1.7\times5}{2}[/tex]
[tex]d_{3}=4.25\ m[/tex]
Hence, The first possible value of d is 0.85 m
The second possible value of d is 2.55 m
The third possible value of d is 4.25 m
An X-ray telescope located in Antarctica would not work well because of (a) the extreme cold; (b) the ozone hole; (c) continuous daylight; (d) Earth’s atmosphere.
The entrance of certain frequencies through the atmosphere varies depending on the sector. This concept of permittivity by which the rays can enter the earth is called the Radio Window and has a spectrum that ranges from 5MHz to 30GHz. Of the radiation produced by astronomical objects, one part is 'filtered' and the other has the ability to be absorbed by the earth due to its opacity effect.
Ultraviolet rays, X-rays and even gamma rays are completely affected by the ozone layer in Earth's atmosphere.
This makes it difficult to use X-ray telescopes located in Antarctica.
The correct option is D.
Earth's atmosphere limits X-ray observations, making Antarctica unsuitable due to extreme cold.
Earth's atmosphere acts as a barrier for X-ray observations, as it blocks most radiation at wavelengths shorter than visible light. Telescopes located in Antarctica would not work well due to the extreme cold which can affect their performance.
If you are at the equator and driving north at a speed of 90 m/s, what is direction of the magnetic force on your head? 1. north 2. south 3. downward 4. east 5. upward 6. west 7. There is no force.
Answer:
7. The force is zero
Explanation:
The force is zero when your velocity is parallel to the magnetic field
A model of a helicopter rotor has four blades, each 3.40 m long from the central shaft to the blade tip. The model is rotated in a wind tunnel at 550 rev/min. (a) What is the linear speed of the blade tip, in m/s? (b) What is the radial acceleration of the blade tip expressed as a multiple of g?
the radial acceleration of the blade tip is approximately 1210.50 times g .
To find the linear speed of the blade tip and the radial acceleration of the blade tip, we can use the following formulas:
(a) Linear speed of the blade tip:
[tex]\[ \text{Linear speed} = \text{Angular speed} \times \text{Radius} \]\[ v = \omega \times r \][/tex]
(b) Radial acceleration of the blade tip:
[tex]\[ \text{Radial acceleration} = \omega^2 \times r \]\[ a_r = \omega^2 \times r \][/tex]
Given:
- Angular speed [tex](\( \omega \))[/tex] = 550 rev/min
- Radius [tex](\( r \))[/tex] = 3.40 m
- Acceleration due to gravity[tex](\( g \))[/tex] = 9.81 m/s²
First, let's convert the angular speed from rev/min to rad/s:
[tex]\[ \omega = \frac{550 \text{ rev/min} \times 2\pi \text{ rad/rev}}{60 \text{ s/min}} \]\[ \omega = \frac{550 \times 2\pi}{60} \text{ rad/s} \]\[ \omega \approx 57.91 \text{ rad/s} \][/tex]
(a) Linear speed of the blade tip:
[tex]\[ v = \omega \times r \]\[ v = 57.91 \times 3.40 \]\[ v \approx 197.04 \text{ m/s} \][/tex]
So, the linear speed of the blade tip is approximately 197.04 m/s.
(b) Radial acceleration of the blade tip:
[tex]\[ a_r = \omega^2 \times r \]\[ a_r = (57.91)^2 \times 3.40 \]\[ a_r \approx 11854.76 \text{ m/s}^2 \][/tex]
Now, express the radial acceleration as a multiple of ( g ):
[tex]\[ \text{Multiple of } g = \frac{a_r}{g} \]\[ \text{Multiple of } g = \frac{11854.76}{9.81} \]\[ \text{Multiple of } g \approx 1210.50 \][/tex]
So, the radial acceleration of the blade tip is approximately 1210.50 times ( g ).
A 3.0 L cylinder is heated from an initial temperature of 273 K at a pressure of 105 kPa to a final temperature of 381 K. 381 K. Assuming the amount of gas and the volume remain the same, what is the pressure (in kilopascals) of the cylinder after being heated?
Answer:
[tex]{P_2}=146.53\ kPa[/tex]
Explanation:
Volume ,V = 3 L
Initial temperature ,T₁ = 273 K
Initial pressure ,P₁ = 105 kPa
Final temperature ,T₂ = 381 K
Lets take final pressure =P₂
We know that ,If the volume of the gas is constant ,then we can say that
[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}[/tex]
[tex]{P_2}=P_1\times \dfrac{T_2}{T_1}[/tex]
Now by putting the values in the above equation we get
[tex]{P_2}=105\times \dfrac{381}{273}\ kPa[/tex]
[tex]{P_2}=146.53\ kPa[/tex]
Therefore the final pressure will be 146.53 kPa.
The final pressure of the cylinder after being heated is 75.57 kPa.
Explanation:To solve this problem, we can use the equation for Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature. In this case, the initial volume is 3.0 L and the initial temperature is 273 K. The final temperature is 381 K. Now we can set up a proportion:
(Initial volume) / (Initial temperature) = (Final volume) / (Final temperature)
Plugging in the numbers, we get:
(3.0 L) / (273 K) = (Final volume) / (381 K)
Solving for the final volume gives us:
Final volume = [(3.0 L) / (273 K)] x (381 K) = 4.1732 L
Since the volume remains the same, the pressure is inversely proportional to the new volume. So, if the initial pressure is 105 kPa, the final pressure can be calculated using the following equation:
(Initial pressure) x (Initial volume) = (Final pressure) x (Final volume)
Plugging in the numbers, we get:
(105 kPa) x (3.0 L) = (Final pressure) x (4.1732 L)
Solving for the final pressure gives us:
Final pressure = [(105 kPa) x (3.0 L)] / (4.1732 L) = 75.57 kPa
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Perform the following unit conversions:
a. 1 L to in.3
b. 650 J to Btu
c. 0.135 kW to ft · lbf/s
d. 378 g/s to lb/min
e. 304 kPa to lbf/in.2
f. 55 m3/h to ft3/s
g. 50 km/h to ft/s
h. 8896 N to ton (=2000 lbf)
Explanation:
a. 1 liter (L) is equal to 61.0237 cubic inches ([tex]in^3[/tex]), So:
[tex]1L*\frac{61.0237in^3}{1L}=61.0237in^3[/tex]
b. 1 J is equal to [tex]9.47817*10^{-4}[/tex] BTU. Thus:
[tex]650J*\frac{9.47817*10^{-4}BTU}{1J}=6.16081*10^{-1}BTU[/tex]
c. 1 kW is equal to 737.56 [tex]\frac{ft\cdot lbf}{s}[/tex]. So:
[tex]0.135kW*\frac{737.56\frac{ft\cdot lbf}{s}}{1kW}=99.57\frac{ft\cdot lbf}{s}[/tex]
d. 1 [tex]\frac{g}{s}[/tex] is equal to 0.1323 [tex]\frac{lb}{min}[/tex]. Therefore:
[tex]378\frac{g}{s}*\frac{0.1323\frac{lb}{min}}{1\frac{g}{s}}=50.01\frac{lb}{min}[/tex]
e. 1 kPa is equal to 0.145[tex]\frac{lbf}{in^2}[/tex]. Thus:
[tex]304kPa*\frac{0.145\frac{lbf}{in^2}}{1kPa}=44.08\frac{lbf}{in^2}[/tex]
f. 1 [tex]\frac{m^3}{h}[/tex] is equal to [tex]9.81*10^{-3}\frac{ft^3}{s}[/tex]. So:
[tex]55\frac{m^3}{h}*\frac{9.81*10^{-3}\frac{ft^3}{s}}{1\frac{m^3}{h}}=5.40*10^{-1}\frac{ft^3}{s}[/tex]
g. 1 [tex]\frac{km}{h}[/tex] is equal to 0.911 [tex]\frac{ft}{s}[/tex]. Therefore:
[tex]50\frac{km}{h}*\frac{0.911\frac{ft}{s}}{1\frac{km}{h}}=45.55\frac{ft}{s}[/tex]
h. 1 N is equal to [tex]1.1*10^{-4}[/tex] ton. Thus:
[tex]8896N*\frac{1.1*10^{-4}ton}{1N}=0.98ton[/tex]
2)It is known that the connecting rodS exerts on the crankBCa 2.5-kN force directed down andto the left along the centerline ofAB. Determine the moment of this force about for the two casesshown at below.
Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
Baseball scouts often use a radar gun to measure the speed of a pitch. One particular model of radar gun emits a microwave signal at a frequency of 10.525 GHz. What will be the increase in frequency if these waves are reflected from a 95.0mi/h fastball headed straight toward the gun?
Since the Units presented are not in the International System we will proceed to convert them. We know that,
[tex]1 mi/h = 0.447 m/s[/tex]
So the speed in SI would be
[tex]V=95mi/h(\frac{0.447m/s}{1mi/h})[/tex]
[tex]V=42.465 m/s[/tex]
The change in frequency when the wave is reflected is
[tex]f'=f(1+\frac{V}{c})[/tex]
Or we can rearrange the equation as
[tex]f' = f + f\frac{V}{c}[/tex]
f' = Apparent frequency
f = Original Frequency
c = Speed of light
[tex]f'-f = f\frac{V}{c}[/tex]
[tex]\Delta f = f\frac{V}{c}[/tex]
Replacing,
[tex]\Delta f = (10.525*10^9)(\frac{42.465}{3*10^8})[/tex]
[tex]\Delta f =1489.8 Hz[/tex]
Since the waves are reflected, hence the change in frequency at the gun is equal to twice the change in frequency
[tex]\Delta f_T = 2 \Delta f[/tex]
[tex]\Delta f_T = 2(1489.8Hz)[/tex]
[tex]\Delta f_T = 2979.63Hz[/tex]
Therefore the increase in frequency is 2979.63Hz
The increase in frequency of these waves is equal to 2979.6 Hertz.
Given the following data:
Frequency = 10.525 GHz.Velocity = 95.0 mi/h.Conversion:
1 mi/h = 0.447 m/s.
95.0 mi/h = [tex]95 \times 0.447[/tex] = 42.465 m/s.
To determine the increase in frequency:
How to calculate the increase in frequency.Mathematically, the change in frequency of a wave is given by this formula:
[tex]F' = F(1+\frac{V}{c} )\\\\F' = F+F\frac{V}{c}\\\\F' - F=F\frac{V}{c}\\\\\Delta F = F\frac{V}{c}[/tex]
Where:
F is the observed frequency.[tex]F'[/tex] is the apparent frequency.c is the speed of light.V is the velocity of an object.Substituting the given parameters into the formula, we have;
[tex]\Delta F = 10.525 \times 10^9 \times \frac{42.465}{3 \times 10^8}\\\\\Delta F = \frac{446.944 \times 10^9}{3 \times 10^8}\\\\\Delta F = 1489.8\;Hertz[/tex]
Since the waves were reflected, the increase in frequency toward the gun is double (twice) the change in frequency. Thus, we have;
[tex]F_2 = 2\Delta F\\\\F_2 = 2\times 1489.8\\\\F_2 = 2979.6\;Hertz[/tex]
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Suppose a hydrogen atom in its ground state moves 130 cm through and perpendicular to a vertical magnetic field that has a magnetic field gradient dB/dz = 1.2 × 102 T/m. (a) What is the magnitude of the force exerted by the field gradient on the atom due to the magnetic moment of the atom's electron, which we take to be 1 Bohr magneton? (b) What is the vertical displacement of the atom in the 130 cm of travel if its speed is 2.2 × 105 m/s?
Answer:a)1.11×10^-21Nm
b) 1.16×10^-3m
Explanation:see attachment
A bullet moving with an initial speed of vo strikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height h. Which of the following statements is true of the collision? o The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h. the bullet immediately after the collision the potential energy of the bullet and block at the instant they reach the maximum O The initial momentum of the bullet before the collision is equal to the momentum of o The kinetic energy of the bullet and block immediately after the collision is equal to height h. energy of the bullet and block when they reach the maximum heighth energy of the bullet and block immediately after the colision O The initial kinetic energy of the bullet before the collision is equal to the potential o The initial kinetic energy of the bullet before the collision is equal to the kinetic
Final answer:
The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h. Option A
Explanation:
The correct statement of the collision in this scenario is A) The initial momentum of the bullet before the collision is equal to the momentum of the bullet and block at the instant they reach the maximum height h.
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, as long as there are no external forces acting on the system. In this case, the bullet and block form a closed system, and therefore, the initial momentum of the bullet is equal to the momentum of the bullet and block at the maximum height.
To understand why this statement is true, we can consider the momentum of the bullet and block system at different points in the motion. Initially, the bullet has momentum in the positive x-direction, and this momentum is transferred to the bullet and block as they rise to the maximum height. Therefore, the statement A is correct.
The energy of a photon is inversely proportional to the wavelength of the radiation. (T/F)
Answer:
true
Explanation:
The statement "the energy of a photon is inversely proportional to the wavelength of the radiation" is definitely true.
What is Wavelength?Wavelength may be characterized as the space or distance between two successive crests of a wave that especially includes the points in a sound wave or electromagnetic wave. The wavelength may be represented by a letter known as lambda (λ).
When the energy of the radiation increases, the waves travel faster which directs the small gap between two successive crests. This means that the wavelength decreases.
On contrary, when the energy of the radiation decreases, the waves travel slower which leads to a huge gap between two successive crests. This means that the wavelength increases.
Therefore, the statement "the energy of a photon is inversely proportional to the wavelength of the radiation" is definitely true.
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Two equal positive point charges are placed at two of the co ?rners of an equilateral triangle of side A. What is the magnitude of the net electric field at the center of the triangle ?
Answer:
Therefore the magnitude of the net electric filed at the center of the triangle is [tex]=\frac{6K}{A^2}[/tex] N/C
Explanation:
Given,A = side of the triangle. q = 1 C
The center of a triangle is the centroid of the triangle.
The distance between centroid to any vertices of a equilateral triangle is
[tex]=\frac{2}{3}[/tex] of the height of the equilateral triangle
[tex]=\frac{2}{3}\times \frac{\sqrt{3} }{2} \times A[/tex]
[tex]=\frac{1}{\sqrt{3} } A[/tex]
Electric field= [tex]\frac{Kq}{d^2}[/tex] K=8.99×10⁹ Nm²/C², q=charge and d = distance
Therefore the magnitude of the net electric filed at the center of the triangle is
=2[tex]\frac{Kq}{d^2}[/tex] [both charge are at same distance from the centroid]
=[tex]\frac{2k}{(\frac{1}{\sqrt{3} }A)^2 }[/tex]
[tex]=\frac{6K}{A^2}[/tex] N /C [K=8.99×10⁹ Nm²/C²]
A bobsledder pushes her sled across horizontal snow to get it going, then jumps in. After she jumps in, the sled gradually slows to a halt. What forces act on the sled just after she's jumped in?
a) Gravity and kinetic friction
b) Gravity and normal force
c) Gravity and the force of the push
d) Gravity, a normal force, and kinetic friction
e) Gravity, a normal force, kinetic friction, and the force of the push
Answer:
d) Gravity, a normal force, and kinetic friction
Explanation:
When the bobsledder pushes her sled across horizontal snow to get it going, after she jumps into the sled there acts a force of gravity on the total mass of the sled including the bobsledder.The sled moves horizontally and not vertically this means that there is a normal force acting in the vertically upward direction opposite to the gravity.While the sled moves on the horizontal surface and comes to the rest there acts a kinetic frictional force on the body in the direction opposite to the direction of motion.Suppose you walk 17.5 m straight west and then 22.0 m straight north. How far are you from your starting point (in m)
Answer:
Explanation:
Given
Man walks 17.5 m straight to west 17.5 m
So position vector is given by
[tex]\vec{r_1}=-17.5\hat{i}[/tex]
Now he walks 22 m North
so position vector is
[tex]r_{21}=22\hat{j}[/tex]
Position of man from initial Position
[tex]\vec{r_{2}}=\vec{r_2}-\vec{r_1}[/tex]
[tex]\vec{r_{2}}=22\hat{j}-(-17.5\hat{i})[/tex]
[tex]\vec{r_{2}}=17.5\hat{i}+22\hat{j}[/tex]
So Magnitude of distance is given by
[tex]|\vec{r_{2}}|=\sqrt{17.5^2+22^2}[/tex]
[tex]|\vec{r_{2}}|=28.11\ m[/tex]
To determine the distance from the starting point after walking 17.5 m west and 22.0 m north, use the Pythagorean theorem. Calculating this gives a distance of approximately 28.1 meters from the starting point.
Calculating Distance Using Pythagorean Theorem
To find how far you are from your starting point after walking 17.5 m west and then 22.0 m north, we can use the Pythagorean theorem. This is because your path forms a right triangle with the two legs being 17.5 m and 22.0 m.
The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the distance from your starting point) is equal to the sum of the squares of the other two sides:
a² + b² = c²
a = 17.5 m (west)b = 22.0 m (north)Substituting these values into the Pythagorean theorem gives:
17.52 + 22.02 = c²
Calculating the squares:
306.25 + 484.00 = c²
Adding them together:
790.25 = c²
Taking the square root of both sides:
c ≈ 28.1 m
Therefore, you are approximately 28.1 meters from your starting point.