First, let's make some convertions:
[tex]268kg*\frac{1 lbm}{0.454kg}= 590.84 lbm[/tex]
[tex]133.7 ft^3*\frac{1m^3}{35.3147ft^3} = 3.78m^3[/tex]
a) weight of the fuel:
Newtons: The weight in newtons is equal to the mass in kilograms times the gravity in m/s^2.
[tex]W = m*g = 268kg*9.81m/s^2=2629.08 N[/tex]
lbf: The weight inlbf is equal to the mass in slugs times the gravity in ft/s^2.
[tex]W= m*g = 590.84 lbm *\frac{1 slug}{32.174lbm} *32.174ft/s^2 = 590.84 lbf[/tex]
b) density:
The density is the mass in kg of the fuel divided by its volume in m^3:
[tex]d = \frac{m}{v} =\frac{268kg}{3.78m^3} =70.9 kg/m^3[/tex]
c) specific volume:
The specific volume is the volume in ft^3 of the fuel divided by its mass in lbm:
[tex]v_{sp} = \frac{v}{m} =\frac{133.7 ft^3}{590.84 lbm} = 0.226 ft^3/lbm[/tex]
Which one of the following situations is NOT Possible? A body has zero velocity and non-zero acceleration. A body travels with a northward velocity and a southward acceleration. A body travels with a northward velocity and a northward acceleration. A body travels with a constant velocity and a constant non-zero acceleration. A body travels with a constant acceleration and a time-varying velocity.
Answer:
Explanation:
A body has zero velocity and non- zero acceleration. It is possible as at the time when a body thrown upwards is at the top of the height. In that case velocity is zero but acceleration is equal to g.
A body travels with northward velocity but southward acceleration . it is also possible as in case when a body is going with northward velocity but when break is applied acceleration becomes southward.
A body travels with northward velocity and northward acceleration . it is also possible as in case when a body is going with northward velocity and when accelerator is applied . Acceleration becomes northward.
A body travels with constant velocity and a constant non zero acceleration . It is not possible . When there is acceleration , there must be a a change in velocity either in terms of magnitude or direction or both.
A body travels with a constant acceleration and a time varying velocity. It is possible. Time varying velocity represents acceleration.
In physics, a constant non-zero acceleration would necessitate a change in velocity over time. Therefore, it's impossible for a body to have a constant velocity and a constant non-zero acceleration at the same time.
Explanation:The situation that is NOT possible is a body traveling with a constant velocity and a constant non-zero acceleration.
Acceleration is the rate at which an object's velocity changes. If the velocity of the body remains constant, it means there is no change in its speed or direction. Therefore, it is not possible for the body to have a non-zero acceleration while traveling with a constant velocity.
On the other hand, scenarios with zero velocity and non-zero acceleration or a body traveling with a northward velocity and a northward acceleration are possible.
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A 20-lb force acts to the west while an 80-lb force acts 45° east of north. The magnitu 80 lb 70 lb 067 lb 100 lb
Answer:
option C
Explanation:
given,
force act on west = 20 lb
force act at 45° east of north = 80 lb
magnitude of force = ?
∑ F y = 80 cos 45⁰
F y = 56.57 lb
magnitude of forces in x- direction
∑ F x = -20 + 80 sin 45⁰
= 36.57 lb
net force
F = [tex]\sqrt{F_x^2+F_y^2}[/tex]
F = [tex]\sqrt{56.57^2+36.57^2}[/tex]
F = 67.36 lb≅ 67 lb
hence, the correct answer is option C
If a proton were traveling the same speed as electrons in a TV picture tube (about 7.1 x 10^7 m/s), what would its de Broglie wavelength be? The mass of a proton is 1.67 x 10^-27 kg. Answer must be in m.
Answer:
The wavelength is [tex]5.591\times10^{-15}\ m[/tex]
Explanation:
Given that,
Speed [tex]v= 7.1\times10^{7}\ m/s[/tex]
Mass of proton [tex]m= 1.67\times10^{-27}\ kg[/tex]
We need to calculate the wavelength
Using formula of de Broglie wavelength
[tex]p=\dfrac{h}{\lambda}[/tex]
[tex]\lambda=\dfrac{h}{p}[/tex]
[tex]\lambda=\dfrac{h}{mv}[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{6.63\times10^{-34}}{1.67\times10^{-27}\times7.1\times10^{7}}[/tex]
[tex]\lambda=5.591\times10^{-15}\ m[/tex]
Hence, The wavelength is [tex]5.591\times10^{-15}\ m[/tex]
Final answer:
The de Broglie wavelength of a proton traveling at the same speed as electrons in a TV picture tube can be calculated by using the equation λ = h / p, where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the proton. The momentum of the proton can be calculated by multiplying its mass by its velocity. Once the momentum is found, the de Broglie wavelength can be calculated. In this case, the de Broglie wavelength of the proton would be 5.61 * 10^-15 m.
Explanation:
The de Broglie wavelength (λ) of a particle is given by the equation:
λ = h / p
where h is the Planck's constant
(6.62607015*10^-34 Js) and p is the momentum of the particle. To find the de Broglie wavelength of a proton traveling at the same speed as electrons in a TV picture tube, we need to first calculate the momentum of the proton. The momentum (p) is given by
p = mass * velocity
Plugging in the values, we get:
p = (1.67 * 10^-27 kg) * (7.1 * 10^7 m/s) = 1.18 * 10^-19 kg m/s
Now we can calculate the de Broglie wavelength:
λ = (6.62607015 * 10^-34 Js) / (1.18 * 10^-19 kg m/s) = 5.61 * 10^-15 m
Therefore, the de Broglie wavelength of the proton would be 5.61 * 10^-15 m.
Helium–neon laser light (λ = 632.8 nm) is sent through a 0.350-mm-wide single slit. What is the width of the central maximum on a screen 2.00 m from the slit?
Answer:
The width of the central bright fringe is 7.24 mm.
Explanation:
Given that,
Wavelength = 632.8 nm
Width d= 0.350 mm
Distance between screen and slit D= 2.00 m
We need to calculate the distance
Using formula of distance
[tex]y_{m}=\dfrac{\lambda D}{d}[/tex]
Put the value into the formula
[tex]y_{m}=\dfrac{632.8\times10^{-9}\times2.00}{0.350\times10^{-3}}[/tex]
[tex]y_{m}=3.62\ mm[/tex]
We need to calculate the width of the central bright fringe
Using formula of width
[tex]width = 2\times|y_{m}|[/tex]
Put the value into the formula
[tex]width=2\times3.62[/tex]
[tex]width = 7.24\ mm[/tex]
Hence, The width of the central bright fringe is 7.24 mm.
The width of the central maximum on the screen, achieved by sending a Helium-neon laser light through a 0.350-mm-wide single slit and projected on a screen 2.00 m from the slit, would be approximately 3.61 mm.
Explanation:To respond to this question, we need to apply the formula for a single-slit diffraction pattern, which is y = Lλ /w, where y represents the distance from the central maximum to the first minimum (i.e., the width of the central maximum), L is the distance from the slit to the screen, λ is the wavelength, and w is the width of the slit.
The values provided in the problem are L = 2.00 m, λ = 632.8 nm = 632.8 x 10-9 m, and w = 0.350 mm = 0.350 x 10-3 m. Substituting these into our formula gives us:
y = (2.00 m)(632.8 x 10-9 m) / (0.350 x 10-3 m) ≈ 0.00361m or 3.61mm
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Which of the following physical quantities is not a vector? A) velocity B) acceleration C) weight D) mass
Answer: Option (D) is correct.
Explanation:
From the given options, we can state that mass is not a vector quantity. Mass is considered as a scalar quantity. In other words, mass can be referred to as from how much matter a subject is made of. Mass has magnitude but it does not tend to give any indications about the direction of the subject in frame.
A solid conducting sphere of radius 2.00 cm has a charge of 8.84 μC. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a charge of −2.02 μC. Find the electric field at the following radii from the center of this charge configuration.
Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5
Explanation: In order to solve this problem we have to use the gaussian law in the mentioned regions.
Region 1; 0<r<2
∫E.ds=Qinside the gaussian surface/ε0
inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.
Region 2; 2<r<4;
E.4*π*r^2=8,84/ε0
E=8,84/(4*π*ε0*r^2)
Region 3; 4<r<5
E=0 because is inside the conductor.
Finally
Region 4; r>5
E.4*π*r^2=(8,84-2.02)/ε0
Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.6 cm . Two of the particles have a negative charge: q 1 = -7.7 nC and q 2 = -15.4 nC . The remaining particle has a positive charge, q 3 = 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?
Answer:
216.97 X 10⁻⁵ N
Explanation:
Charge q₁ and q₂ will attract q₃ with force F₁ and F₂ .F₁ and F₂ will be calculated as follows
F₁ = [tex]\frac{9\times10^9\times8\times7.7\times10^{-18}}{(2.6\times10^{-2})^2}[/tex]
F₁ = 82.01 X 10⁻⁵ N
F₂= [tex]\frac{9\times10^9\times8\times15.4\times10^{-18}}{(2.6\times10^{-2})^2}[/tex]
F₂ = 164.02 X 10⁻⁵ N
F₁ and F₂ will act at 60 degree so their resultant will be calculated as follows
R² = (82.01 X 10⁻⁵)² +( 164.02 X 10⁻⁵ )² + 2 X 82.01 X 164.02 X 10⁻¹⁰ Cos 60
R² = 47079.48 X 10⁻¹⁰
R = 216.97 X 10⁻⁵ N
A system of 1610 particles, each of which is either an electron or a proton, has a net charge of −5.376×10^−17 C. How many electrons are in this system? What is the mass of this system?
Answer:
Number of electrons in the system = 973.
Total mass of the system = [tex]\rm 1.064\times 10^{-24}\ kg.[/tex]
Explanation:
Assumptions:
[tex]\rm n_e[/tex] = number of electrons in the system.[tex]\rm n_p[/tex] = number of protons in the system.[tex]\rm q_e[/tex] = charge on an electron = [tex]\rm -1.6\times 10^{-19}\ C.[/tex][tex]\rm q_p[/tex] = charge on a proton = [tex]\rm +1.6\times 10^{-19}\ C.[/tex][tex]\rm m_e[/tex] = mass of an electron = [tex]\rm 9.11\times 10^{-31}\ kg.[/tex][tex]\rm m_p[/tex] = mass of a proton = [tex]\rm 1.67\times 10^{-27}\ kg.[/tex]Given:
Total number of particles in the system, N = 1610.Net charge on the system, q = [tex]\rm -5.376\times 10^{-17}\ C.[/tex]Since, the system is comprised of electrons and protons only, therefore,
[tex]\rm N = n_e+n_p\\n_p=N-n_e\ \ \ \ \ \ ................\ (1).[/tex]
The net charge on the system can be written in terms of charges on electrons and protons as
[tex]\rm q=n_eq_e+n_pq_p\ \ \ \ \ ...................\ (2).[/tex]
Putting the value of (2) in (1), we get,
[tex]\rm q=n_eq_e+(N-n_e)q_p\\q=n_eq_e+Nq_p-n_eq_p\\q=n_e(q_e-q_p)+Nq_p\\n_e(q_e-q_p)=q-Nq_p\\n_e=\dfrac{q-Nq_p}{q_e-q_p}\\=\dfrac{-5.3756\times 10^{-17}-1610\times 1.6\times 10^{-19}}{-1.6\times 10^{-19}-1.6\times 10^{-19}}=972.98\\\Rightarrow n_e\approx 973\ electrons.[/tex]
It is the number of electrons in the system.
Therefore, the number of protons is given by
[tex]\rm n_p = N-n_e=1610-973=637.[/tex]
The total mass of the system is given by
[tex]\rm M=n_em_e+n_pm_p\\=(973\times 9.11\times 10^{-31})+(637\times 1.67\times 10^{-27})\\=1.064\times 10^{-24}\ kg.[/tex]
Final answer:
The system consists of 336 electrons and 1274 protons, given the net charge. To calculate the total mass, multiply the number of each particle by its respective mass and then sum the results., which gives 5192.362×10−31 kg as answer.
Explanation:
To find the number of electrons in a system with a net charge, we can use the formula:
Total number of electrons = (Total net charge) / (Charge per electron).Given that the net charge of the system is -5.376×10−17 C and the charge of an electron is approximately -1.6022×10−19 C, we can calculate the total number of electrons using the formula:
Number of electrons = (-5.376×10−17 C) / (-1.6022×10−19 C/electron)
Number of electrons = 3.3555×102
Since we can’t have a fraction of an electron, we round to the nearest whole number, which is 336 electrons.
To determine the mass of the system, first, we need to find the number of protons, which would be 1610 - 336 electrons = 1274 protons. Now we multiply the number of protons by the mass of a proton and the number of electrons by the mass of an electron to get the total mass:
Mass of electrons = 336 × 9.11×10−31 kg
Mass of protons = 1274 × 1.673×10−27 kg
Adding these two gives the total mass of the system, which is 5192.362×10−31 kg.
You have a rod with a length of 146.4 cm. You prop up one end on a brick which is 3.8 cm thick. Your uncertainty in measuring these distances is ±0.05 cm. What is the angle that the rod makes with the table?
_______ degrees
What is the uncertainty in that angle?
________ degrees
Answer:
[tex]\partial \theta = 0.003[/tex]
Explanation:
we know that
[tex]sin\theta = \frac{3.8}{146.4}[/tex]
[tex]\theta = sin^{-1} \frac{3.8}{146.4}[/tex]
[tex]\theta = 1.484°[/tex]
[tex]\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians[/tex]
as we see that [tex]sin\theta = \theta[/tex]
relative error[tex] \frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}[/tex]
Where X_1 IS HEIGHT OF ROCK
[tex]X_2[/tex] IS THE HEIGHT OF ROAD
[tex]\partial X[/tex] = uncertainity in measuring distance
[tex]\partial X = 0.05[/tex]
Putting all value to get uncertainity in angle
[tex]\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}[/tex]
solving for [tex]\partial \theta[/tex] we get
[tex]\partial \theta = 0.003[/tex]
An uncharged metal sphere hangs from a nylon thread. When a positively charged glass rod is brought close to the metalsphere, the sphere is drawntoward the rod. But if the sphere touches the rod, it suddenly flies away from the rod. Explain why the sphere is first attracted and then repelled.
When a charged rod is brought near a neutral metal sphere, the opposite charges are attracted towards the rod, causing an initial attraction. However, when the sphere touches the rod, the charges redistribute, leading to like charges repelling each other with a stronger force than the attraction between opposite charges, resulting in the sphere being repelled.
Explanation:When a charged rod is brought near a neutral substance, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction.
Thus, when a positively charged glass rod is brought close to the neutral metal sphere, the opposite charges in the metal sphere are attracted towards the rod. However, when the sphere touches the rod, it suddenly flies away from the rod. This is because the charges in the metal sphere redistribute, and now the like charges repel each other with a stronger force compared to the attraction between the opposite charges.
This repulsion causes the metal sphere to be repelled away from the rod, creating a net repulsive force.
The metal sphere is first attracted to the positively charged glass rod due to induction, and then repelled after touching the rod due to the repulsion between like charges.
When the positively charged glass rod is brought close to the uncharged metal sphere, the sphere experiences electrostatic induction. The electrons in the metal sphere are repelled by the positive charges on the rod and move to the far side of the sphere, leaving the side closest to the rod with a net positive charge. This positive side of the sphere is attracted to the negative charges in the rod (or the electrons in the rod are attracted to the positive side of the sphere), causing the sphere to be drawn toward the rod.
As the sphere gets closer to the rod, the attraction increases until they make contact. When the sphere touches the rod, electrons from the rod flow onto the sphere, as the rod has a surplus of electrons due to its negative induction on the side opposite the sphere. This transfer of electrons results in the sphere acquiring a net negative charge, as it gains more electrons.
Once the sphere has the same negative charge as the side of the rod it touched, the electrostatic force of repulsion between the like charges takes over. The negatively charged sphere is now repelled by the negatively charged side of the glass rod. This repulsion is strong enough to overcome the gravitational force and the nylon thread's tension, causing the sphere to suddenly fly away from the rod.
An electron with speed v0 = 5.00 x 10^6 m/s is traveling parallel to an electric field of magnitude E = 9900 N/C . 1)How far will the electron travel before it stops?
Express your answer to three significant figures and include the appropriate units.
2)How much time will elapse before it returns to its starting point?
Express your answer to three significant figures and include the appropriate units
Answer:
Explanation:
in the electric field electron will face a force which will create an acceleration
( here - ve ) as follows
Force on electron
= charge on electron x electric field = Q X E
acceleration= Force / mass = Q E / m
mass of electron = 9.1 x 10⁻³¹
acceleration a = [tex]\frac{1.6\times10^{-19}\times 9900}{1.67\times10^{-27}}[/tex]
= 17.4 x 10¹⁴ ms⁻² .
Now initial velocity u = 5 x 10⁶ m/s
Final velocity v = 0
acceleration a = 17.4 x 10¹⁴ ms⁻²
distance of travel = s
v² = u² - 2as
0 = (5 x 10⁶)² - 2 x 17.4 x 10¹⁴ s
s = 7.18 mm
2 ) v = u - at
0 = 5 x 10⁶ - 17.4 x 10¹⁴ t
t = .287 x 10⁻⁸ s
Total time elapsed = 2 x .287 x 10⁻⁸
= .57 x 10⁻⁸ s .
A bird, accelerating from rest at a constant
rate,experiences a displacement of 28 m in 11s. what is the
averagevelocity?
Answer:
Average velocity of the bird is 2.54 m/s.
Explanation:
Given that,
Initial speed of the bird, u = 0
It experiences a displacement of, d = 28 m
Time taken, t = 11 s
We need to find the average velocity of the bird. Let v is the average velocity. Mathematically, it is given by :
[tex]v=\dfrac{d}{t}[/tex]
[tex]v=\dfrac{28\ m}{11\ s}[/tex]
v = 2.54 m/s
So, the average velocity of the bird is 2.54 m/s. Hence, this is the required solution.
Final answer:
The average velocity of the bird is calculated by dividing the total displacement of 28 m by the total time of 11 s, resulting in an average velocity of 2.545 m/s.
Explanation:
To find the average velocity of a bird that accelerates from rest and experiences a displacement of 28 m in 11 s, we use the formula for average velocity, which is total displacement divided by total time. Since the bird starts from rest and moves in one direction, its average velocity will be the same as its average speed.
Average velocity = Total displacement / Total time = 28 m / 11 s = 2.545 m/s.
Therefore, the average velocity of the bird is 2.545 m/s.
Without effective assessment methods, teachers have no documented proof of childrens specific ____ and _____.
A. Wants, needs
B. Culture, heritage
C. Likes, dislikes
D. strengths, weaknesses
Answer:
D. strengths, weaknesses
Explanation:
Each individual has characteristics that makes them different from one another. Their innate ability or cannot be determined without testing them in various situations.
The various tests which determine a person's aptitude and intelligence. Another aspect to consider is the emotional intelligence of a person i.e., the capability a person has to recognize others' emotions as well as their own emotions.
Testing children in different types of tests which measures their ability in each of them is the only way to know their strength and weaknesses.
Two forces act on a 6.00- kg object. One of the forces is
10.0N. If the object accelerates at 2.00 m/s 2, what is the
greatestpossible magnitue of the other force?
Answer:
Fx = 22N
Explanation:
There are 2 possible scenarios for this problem:
1.- The 10N force is in the same direction of the acceleration. In this case the other force would be:
[tex]F1 - Fx = m*a[/tex] where F1 = 10N, m=6kg, a = 2m/s2
[tex]Fx = F1 - m*a = -2N[/tex] The negative result tells us that this is not possible.
2.- The 10N force is in the opposite direction of the acceleration. In this case the other force would be:
[tex]Fx - F1 = m*a[/tex] [tex]Fx = m*a + F1 = 22N[/tex]
The greatest magnitude of the other force, by which the object accelerates under the application of two forces is 22 N.
What is Newton’s second law of motion?Newton’s second law of motion shows the relation between the force mass and acceleration of a body. It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.
It can be given as,
[tex]\sum F=ma[/tex]
Here, (m) is the mass of the body and (a) is the acceleration.
One of the forces is 10.0N. . Let suppose the magnitude of the other force is [tex]F_x[/tex],
The maximum force we get when both the forces acting on the opposite direction. For this system the summation of force will be,
[tex]\sum F=F_x+(-10)\\\sum F=F_x-10[/tex]
The mass of the object is 6 kg. As the body is accelerating in at 2 m/s. Thus, plug in the values in the above formula as,
[tex]F_x-10=2\times6\\F_x=22\rm N[/tex]
Thus the greatest magnitude of the other force, by which the object accelerates under the application of two forces is 22 N.
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A skater is standing still on a frictionless rink. Herfriend
throws a Frisbee straight at her. In which of the followingcases is
the largest momentum transferred to the skater?
a) the skater catches the frisbee and hols onto it.
b) the sjater catches the frisbee momentarily, but then
dropsit vertically downward.
c) the skater catches the frisbee, holds it momentarily,
andthrows it back to her friend.
Answer:
option C
Explanation:
the correct answer is option C
Skater is standing on the frictionless surface when her friend through Frisbee at her he starts moving in the backward direction as there is no friction acting to stop him.
but when again the skater through Frisbee back to his friend there is backward force which will increase the velocity of the skater.
so, momentum is directly proportional to velocity so, velocity increases momentum also increases.
Final answer:
The largest momentum transfer to the skater happens when she catches and then throws the Frisbee back to her friend, utilizing the conservation of momentum.
Explanation:
The largest momentum transferred to the skater occurs in the scenario where the skater catches the Frisbee, holds it momentarily, and then throws it back to her friend. In this case, not only is the momentum of the incoming Frisbee transferred to the skater when caught, but additional momentum is transferred when she throws it back due to the conservation of momentum.
The driver of a car traveling at a speed of 25.5 m/s slams on the brakes and comes to a stop in 3.4 s. If we assume that the car's speed changed at a constant rate (constant net force), find the following. What was the car's average speed during this 3.4 s interval? ______ m/s
How far did the car go in this 3.4 s interval? ____ m
Answer:
(a). The average speed of the car is 12.75 m/s
(b). The distance is 43.35 m.
Explanation:
Given that,
Initial speed = -25.5 m/s
Final speed = 0
Time = 3.4 s
(a). We need to calculate the average speed
Using formula of average speed
[tex]v_{avg}=\dfrac{v_{f}-v_{i}}{2}[/tex]
Put the value into the formula
[tex]v_{avg}=\dfrac{0-(-25.5)}{2}[/tex]
[tex]v_{avg}=12.75\ m/s[/tex]
(b). We need to calculate the acceleration
Using equation of motion
[tex]v_{f}=v_{i}+at[/tex]
[tex]a =\dfrac{v_{f}-v_{i}}{t}[/tex]
[tex]a=\dfrac{-25.5-0}{3.4}[/tex]
[tex]a=-7.5\ m/s^2[/tex]
We need to calculate the distance
Using equation of motion
[tex]s = ut+\dfrac{1}{2}at^2[/tex]
[tex]s=25.5\times3.4+\dfrac{1}{2}\times(-7.5)\times(3.4)^2[/tex]
[tex]s=43.35\ m[/tex]
Hence, (a). The average speed of the car is 12.75 m/s
(b). The distance is 43.35 m.
A student is running at her top speed of 5.3 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 40.7 m from the bus, it starts to pull away, moving with a constant acceleration of 0.168 m/s^2 . a) For how much time does the student have to run at 5.3 m/s before she overtakes the bus?
b) For what distance does the student have to run at 5.3 m/s before she overtakes the bus?
c) When she reaches the bus, how fast is the bus traveling?
d) If the student's top speed is 2.00 m/s , will she catch the bus?
e) What is the minimum speed the student must have to just catch up with the bus?
f) For what time does she have to run in that case?
g) For what distance does she have to run in that case?
Answer:
a) The student has to run for 8.9 s.
b) The student has to run for 47 m.
c) The bus is traveling at 1.5 m/s when the student reaches the bus.
d) No, she will not catch the bus running at 2.00 m/s.
e) The minimum speed the student must have to catch the bus is 3.7 m/s.
f) The student has to run for 21.3 s if she runs at 3.7 m/s to catch the bus.
g) She has to run 79 m to catch the bus running at 3.7 m/s
Explanation:
The equations for a straight movement are:
With constant acceleration:
x = x0 + v0 * t + 1/2 * a * t²
v = v0 + a *t
With constant velocity (a = 0):
x = x0 + v * t
Where
x = position at time t
x0 = initial position
v0 = initial velocity
v = velocity
a = acceleration
t = time
a) When the student catches the bus, the position of the bus and the student are the same:
x student = x bus
The student moves with constant speed while the bus has a constant acceleration. If the origin of the reference system is located where the student starts running, then, x0 student = 0 and x0 bus = 40.7 m. Since the bus starts from rest, v0 = 0.
x student = v * t
x bus = x0 + v0 * t + 1/2 * a * t² = x0 + 1/2 * a * t²
x student = x bus
v * t = x0 + 1/2 * a * t²
Replacing with data:
5.3 m/s * t = 40.7 m + 1/2 (0.168 m/s²) * t²
0 = 40. 7 m - 5.3 m/s * t + 0.084 m/s²¨* t²
Solving the quadratic equation:
t = 8.9 s and t = 54.1 s
We discard the higher value because if the student catches the bus at 8.9 s, she will not catch it again at 54.1 s.
The student has to run for 8.9 s.
b) Using the equation for position of the student:
x = v * t = 5.3 m/s * 8.9 s =47 m
The student has to run for 47 m
c) Using the equation for velocity of the bus:
v = v0 + a * t = 0 m/s + 0.168 m/s² * 8.9 s
v = 1.5 m/s
The bus is traveling at 1.5 m/s when the student reaches the bus
d) The quadratic equation after equalizing the position of the student and the position of the bus would be:
0 = 40. 7 m - 2 m/s * t + 0.084 m/s²¨* t²
If we solve this using the formula to obtain the roots of the parabola we will obtain:
[tex]\frac{-b+\sqrt{b^{2}-4ac} }{2a} = \frac{2+\sqrt{4- 4*40.7*0.084}}{2*0.084}[/tex]
Since the term [tex]\sqrt{4-4*40.7*0.084} = \sqrt{-9,7}[/tex] is not defined in the real numbers, there is no "t" such as the equation of the parabola equals 0. The parabola has no roots. Then, the student will not catch the bus if she runs at 2.00 m/s.
e) The term inside the square root in [tex]\frac{-b+\sqrt{b^{2}-4ac} }{2a}[/tex]
has to be positive or 0, then:
b² - 4* a* c ≥ 0
Notice that "b" is the speed at which the student runs, "a" is 0.084 and "c" is 40. 7 ( see the equation of the parabola obtained in a)). Then:
b² ≥ 4 * a * c
b ≥ √ 4 * a * c
b ≥ √ 4 * 0.084 * 40.7
b ≥ 3.7 m/s
The minimum speed the student must have to catch the bus is 3.7 m/s.
f) Now we have to solve the quadratic equation obtained in a), but using -3.7 as value of "b". Solving the quadratic equation, we will obtain the values of t = 21.3 s and 22.7 s. Again, we discard the higher value.
The student has to run for 21.3 s if she runs at 3.7 m/s to catch the bus.
g) The distance is given by the equation for the position of the student:
x = v * t = 3.7 m/s * 21.3 s = 79 m.
She has to run 79 m to catch the bus.
A woman is reported to have fallen 144 ft from the 17th floor of a building, landing on a metal ventilator box, which she crushed to a depth of 18.0 in. She suffered only minor injuries. Neglecting air resistance, calculate (a) the speed of the woman just before she collided with the ventilator, (b) her average acceleration while in contact with the box, and (c) the time it took to crush the box.
Answer:
(a) 96 ft/s
(b) - 3072 ft/s^2
(c) 0.03125 s
Explanation:
h = 144 ft
u = 0 ft/s
g = 32 ft/s^2
(a) let she strikes the box with velocity v.
Use third equation of motion
[tex]v^{2}=u^{2}+2gh[/tex]
[tex]v^{2}=0^{2}+2\times 32 \times 144[/tex]
v = 96 ft/s
(b) Let the average acceleration is a.
initial velocity, u = 96 ft/s
final velocity, v = 0
h = 18 in = 1.5 ft
Use third equation of motion
[tex]v^{2}=u^{2}+2ah[/tex]
[tex]0^{2}=96^{2}+2\times a \times 1.5[/tex]
a = - 3072 ft/s^2
(c) Let the time taken is t.
Use first equation of motion
v = u + at
0 = 96 - 3072 x t
t = 0.03125 second
Using basic equations of motion, we calculated the speed of the woman just before impact with the ventilator as 29.3 m/s, her average acceleration during the impact as 930 m/s², and the time it took to crush the box as approximately 0.0315 seconds.
Explanation:The subject of this question is physics, specifically dealing with concepts of kinematics and mechanics. The falling woman problem requires application of the laws of free fall in physics along with some basic algebraic manipulation, and is quite suitable for a high school student.
(a) The speed of the woman just before she collided with the ventilator can be calculated using the formula v=√(2g h) where g is the gravity (9.8 m/s²) and h is the height (144 ft, which is approximately 43.9 m). Plugging in these values gives us v=√(2*9.8*43.9) ≈ 29.3 m/s.
(b) Knowing the depth the box was crushed (converted to meters) we can use the third equation of motion v² = u² + 2a s (s is displacement) we get a = (v² - u²) / 2s = (29.3 m/s)² / 2*0.46m = 930 m/s² as the average acceleration.
(c) Finally, using v = u + at, as initial speed is 0, we get time t = v/a = 29.3m/s / 930m/s² ≈ 0.0315 seconds.
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In a vacuum, two particles have charges of q1 and q2, where q1 = +3.2 µC. They are separated by a distance of 0.28 m, and particle 1 experiences an attractive force of 2.9 N. What is q2 (magnitude and sign)?
Answer:
q2 = - 8 × [tex]10^{-6}[/tex] C
negative sign because attract together
Explanation:
given data
q1 = +3.2 µC = 3.2 × [tex]10^{-6}[/tex] C
distance r = 0.28 m
force F = 2.9 N
to find out
q2 (magnitude and sign)
solution
we know that here if 2 charge is unlike charge
than there will be electrostatic force of attraction , between them
now we apply coulomb law that is
F = [tex]\frac{q1q2}{4\pi \epsilon *r^{2}}[/tex] ............1
here we know [tex]\frac{1}{4\pi \epsilon}}[/tex] = 9 × [tex]10^{9}[/tex] Nm²/C²
so from equation 1
2.9 = 9 × [tex]10^{9}[/tex] × [tex]\frac{3.2*10^{-6}*q2}{0.28^{2}}[/tex]
q2 = [tex]\frac{2.9*0.28^2}{9*10^9*3.2*10^{-6}}[/tex]
q2 = - 8 × [tex]10^{-6}[/tex] C
Final answer:
Using Coulomb's Law, it's found that the magnitude of charge q2 is 3.2 µC. Since the force experienced by particle 1 is attractive, q2 must have an opposite sign to q1. Therefore, q2 is -3.2 µC.
Explanation:
To determine the magnitude and sign of charge q2, we can use Coulomb's Law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation for Coulomb's Law is F = k * |q1 * q2| / r², where F is the force between the charges, k is Coulomb's constant (8.988 × 10⁹ N·m²/C²), q1 and q2 are the charges, and r is the distance between the charges.
Given that q1 = +3.2 µC (+3.2 × 10⁻⁶ C), r = 0.28 m, and F = 2.9 N, we can rearrange the equation to solve for q2:
|q2| = (F * r²) / (k * |q1|).
Substituting the given values into the equation, we get:
|q2| = (2.9 N * (0.28 m)²) / (8.988 × 10⁹ N·m²/C² * 3.2 × 10⁻⁶ C),
|q2| = 3.2 µC or -3.2 µC. Because particle 1 experiences an attractive force, it implies that q2 must have an opposite sign to q1. Thus, q2 = -3.2 µC.
Consider an electron that is 100 m from an alpha particle ( = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electric field (in N/C) due to the alpha particle at the location of the electron? N/C (b) What is the electric field (in N/C) due to the electron at the location of the alpha particle? N/C (c) What is the electric force (in N) on the alpha particle? On the electron? electric force on alpha particle electric force on electron
Answer:
a)[tex]E=2.88*10^{-13}N/C[/tex]
b)[tex]E=1.44*10^{-13}N/C[/tex]
c)[tex]F=4.61*10^{-32}N[/tex]
Explanation:
The definition of a electric field produced by a point charge is:
[tex]E=k*q/r^2[/tex]
a)Electric Field due to the alpha particle:
[tex]E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(100)^2=2.88*10^{-13}N/C[/tex]
b)Electric Field due to the electron:
[tex]E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(100})^2=1.44*10^{-13}N/C[/tex]
c)Electric Force on the alpha particle, on the electron:
The alpha particle and electron feel the same force magnitude but with opposite direction:
[tex]F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(100)^2=4.61*10^{-32}N[/tex]
The electric output of a power plant is 716 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows at a rate of 1.35 x 10^8 L/Hr. The water enters the plant at 25.4°C and exits at 30.7°C. (a) What is the power plant's total thermal power? (MWT (Megawatt thermal)
(b) What is the efficiency of the power plant?
Answer:
(a) 83475 MW
(b) 85.8 %
Explanation:
Output power = 716 MW = 716 x 10^6 W
Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3
mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000
= 1.35 x 10^8 kg
Time, t = 1 hr = 3600 second
T1 = 25.4° C, T2 = 30.7° C
Specific heat of water, c = 4200 J/kg°C
(a) Total energy, Q = m x c x ΔT
Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J
Power = Energy / time
Power input = [tex]P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W[/tex]
Power input = 83475 MW
(b) The efficiency of the plant is defined as the ratio of output power to the input power.
[tex]\eta =\frac{Power output}{Power input}[/tex]
[tex]\eta =\frac{716}{83475}=0.858[/tex]
Thus, the efficiency is 85.8 %.
A force in the +x-direction has magnitude F = b/xn, where b and n are constants. For n > 1, calculate the work done on a particle by this force when the particle moves along the x-axis from x = x0 to infinity.
The work done on a particle by the force in the +x-direction from x = x0 to infinity can be calculated using the formula Work = ∫(b/x^n) dx. The limits of integration for the integral are from x0 to infinity.
Explanation:To calculate the work done by the force in the +x-direction, we can use the formula:
Work = ∫F dx = ∫(b/x^n) dx
Since the force is in the +x-direction and the particle moves along the x-axis from x = x0 to infinity, the limits of integration for the integral are from x0 to infinity.
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A particle with a charge of -1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s)î + (-3.85 X 104 m/s)j. What is the force exerted on this particle by a magnetic field (a) = B(1.40 T)i and (b) È = (1.40 T)k?
Answer:
(a) F= 6.68*10¹¹⁴ N (-k)
(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
Explanation
To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:
F=q* v X B Formula (1 )
q: charge (C)
v: velocity (m/s)
B: magnetic field (T)
vXB : cross product between the velocity vector and the magnetic field vector
Data
q= -1.24 * 10¹¹⁰ C
v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j
B =(1.40 T)i
B =(1.40 T)k
Problem development
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =
= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)
F= 6.68*10¹¹⁴ N (-k)
a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =
=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T
1T= 1 N/ C*m/s
We apply the formula (1)
F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s
F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N
The force on a charged particle in a magnetic field is calculated using the Lorentz force equation. For the given particle with charge -1.24 x 10⁻⁹ C and velocity vector, the force is determined by the cross product of the velocity and the magnetic field, taking the charge into account.
Explanation:A particle with a charge of -1.24 x 10⁻⁹ C is moving with an instantaneous velocity of (4.19 X 10⁴ m/s)î + (-3.85 X 10⁴ m/s)ᴇ. To find the force exerted on this particle by a magnetic field we use the Lorentz force equation, which is F = q(v x B), where F is the force on the particle, q is the charge, v is the velocity of the particle, and B is the magnetic field.
For part (a) where the magnetic field B is (1.40 T)î, the velocity vector v is perpendicular to B since v has no i-component, thus the force can be found simply by calculating the magnitude as q*v*B since sin(θ) is 1 for θ = 90°. The direction of the force is given by the right hand rule, considering that both v and B are vectors and the charged particle is negatively charged.
For part (b) where the magnetic field B is (1.40 T)ᴅ, the velocity vector v has no k-component, thus v is perpendicular to B and the same principle applies.
The magnitude of the force in both scenarios is computed using the charge, the magnitude of velocity (which is the combination of both the i and j components), and the magnitude of the magnetic field. The direction will differ based on the cross product between v and B.
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A professional diver steps off of a cliff that is 18 m high. Draw a sketch of the cliff, defining your origin and final position. (Careful with negative and positive signs.) Unlike the WB assignment, assume the diver jumps up first and has an initial vertical velocity is 4 m/s. (Ignore air resistance.) (a) How long does it take the diver to hit the water? (b) What's the diver's velocity on impact with the water? (Careful with negative and positive signs.)
Answer:
19.2 m/s
Explanation:
We set a frame of reference with the origin at the cliff top and the positive X axis pointing down.
Then the initial position is:
X0 = 0
The initial speed is:
V0 = -4 m/s
It is negative because it is speed upwards and the frame of reference is positive downwards.
Since the diver is in free fall, he is affected only by the acceleration of gravity, we can consider him moving under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
He will hit the water at X = 18 m, so:
18 = 0 - 4 * t + 1/2 * 9.81 * t^2
4.9 * t^2 - 4 * t - 18 = 0
Solving this equation electronically:
t = 2.37 s
The diver will hit the water 2.37 s after jumping.
The equation for speed under constant acceleration is:
V(t) = V0 + a * t
V(2.37) = -4 + 9.81 * 2.37 = 19.2 m/s
In a historical movie, two knight on horseback start from
rest88.0 m apart and ride directly toward each other to dobattle.
Sir George's acceleration has a magnitude of0.300 m/s2
while Sir Alfred's has a magnitude of0.200 m/s2
Relative to Sir George's starting point, where do the
knightscollide?
Answer:
The knights collide 53.0 m from the starting point of sir George.
Explanation:
The equation for the position in a straight accelerated movement is as follows:
x = x0 + v0 t + 1/2 a t²
where
x = position at time t
x0 = initial position
v0 = initial speed
a = acceleration
t = time
The position of the two knights is the same when they collide. Since they start from rest, v0 = 0:
Sir George´s position:
xGeorge = 0 m + 0 m + 1/2 * 0.300 m/s² * t²
Considering the center of the reference system as Sir George´s initial position, the initial position of sir Alfred will be 88.0 m. The acceleration of sir Afred will be negative because he rides in opposite direction to sir George:
xAlfred = 88.0 m + 0 m - 1/2 * 0.200 m/s² * t²
When the knights collide:
xGeorge = x Alfred
1/2 * 0.300 m/s² * t² = 88.0 m - 1/2 * 0.200 m/s² * t²
0.150 m/s² * t² = 88.0 m - 0.100 m/s² * t²
0.150 m/s² * t² + 0.100 m/s² * t² = 88.0 m
0.250 m/s² * t² = 88.0 m
t² = 88.0 m / 0.250 m/s²
t = 18.8 s
At t = 18.8 s the position of sir George will be
x = 1/2 * 0.300 m/s² * (18.8 s)² = 53.0 m
The question asks for the calculation of the collision point of two knights charging toward each other with different accelerations from a certain distance apart, using principles of kinematics.
Explanation:The question involves kinematics in one dimension, specifically the calculation of the point of collision of two knights with different accelerations.
Sir George has an acceleration of 0.300 m/s2 and Sir Alfred an acceleration of 0.200 m/s2. They start 88.0 m apart.
To determine the collision point, we set up equations based on the formula for distance covered under constant acceleration from rest, which is s = 0.5 * a * t2, where s is the distance, a is acceleration, and t is time.
Since they begin at the same time and collide at the same time, we have:
For Sir George: sG = 0.5 * 0.300 * t2
For Sir Alfred: sA = 0.5 * 0.200 * t2
As they cover the 88.0 m together, sG + sA = 88.0. Substituting the equations for sG and sA and solving for t can give us the individual distances they covered, and hence, the point of collision relative to Sir George's starting point.
A car is driven east for a distance of 50 km, then north for 28 km, and then in a direction 28° east of north for 21 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.
Answer:
A ) displacement=75.69km
B) Angle[tex]= 28.92^o[/tex]
Explanation:
This is a trigonometric problem:
in order to answer A and B , we first need to know the total displacement on east and north.
the tricky part is when the car goes to the direction northeast, but we know that:
[tex]sin(\alpha )=\frac{opposite}{hypotenuse} \\where:\\opposite=north side\\hypotenuse=distance[/tex]
North'=9.86km
and we also know:
[tex]cos(\alpha )=\frac{adjacent}{hypotenuse} \\where:\\adjacent=east side\\hypotenuse=distance[/tex]
East'=18.54km
So know we have to total displacement
North=28km+North'=37.86km
East=50km+East'=68.54km
To calculate the total displacement, we have to find the hypotenuse, that is:
[tex]Td=\sqrt{North^2+East^2} =75.69km[/tex]
we can find the angle with:
[tex]\alpha = arctg (\frac {North} {East}) \\\\ \alpha = arctg (\frac {37.86} {68.54})= 28.92^o[/tex]
A rectangular container measuring 20cm x 30cm x 50cm is filled with water. What is the mass of this volume of water in kilograms? A. 30 kg B. 30,000 kg C. 30 milligrams D. 30,000 lbs
Answer:
A. 30 kg
Explanation:
As we know that,
[tex]1 litre=1000cm^{3}[/tex]
And 1 litre is equivalent to 1kg.
Given that, The volume of the rectangular container is,
[tex]V=20cm\times 30cm\times 50cm\\V=30000cm^{3}[/tex]
And this volume will be equal to [tex]V=30000cm^{3}=30litres[/tex]
And this litres in kg will be equal to,
[tex]V=30litres=30 kg[/tex]
Therefore the mass of this volume of water is 30 kg.
A helicopter blade spins at exactly 110 revolutions per minute. Its tip is 4.50 m from the center of rotation. (a) Calculate the average speed (in m/s) of the blade tip in the helicopter's frame of reference. m/s
(b) What is its average velocity (in m/s) over one revolution? m/s
Answer:
(a). The average speed is 51.83 m/s.
(b). The average velocity over one revolution is zero.
Explanation:
Given that,
Angular velocity = 110 rev/m
Radius = 4.50 m
(a). We need to calculate the average speed
Using formula of average speed
[tex]v=r\omega[/tex]
[tex]v = 4.50\times110\times\dfrac{2\pi}{60}[/tex]
[tex]v=51.83\ m/s[/tex]
(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.
Hence, (a). The average speed is 51.83 m/s.
(b). The average velocity over one revolution is zero.
The average speed of the helicopter blade tip is 51.83 m/s, and its average velocity over one revolution is 0 m/s because it returns to its starting point.
Explanation:To calculate the average speed of the blade tip, we use the formula for the circumference of a circle (C = 2πr) and the given rotational speed.
The radius (r) is 4.50 m, and the helicopter blade spins at 110 revolutions per minute (rpm). First, we need to find the distance one tip of the blade travels in one revolution, which is its circumference:
C = 2πr = 2π(4.50 m) ≈ 28.27 m
Then, to find the distance per minute, we multiply the circumference by the number of revolutions per minute:
Distance per minute = 28.27 m × 110 = 3109.7 m/min
Now, to find the average speed in meters per second, we convert the minute to seconds:
Average speed = 3109.7 m/min × (1 min/60 s) ≈ 51.83 m/s
For part (b), the average velocity over one revolution is 0 m/s since the blade tip returns to its starting point, making the displacement over one revolution zero.
Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibrium, each string makes and angle θ =50 with the vertical. Find the size of the charge on each spere.
Answer:
0.247 μC
Explanation:
As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:
[tex]F_y: T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N[/tex]
[tex]T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N[/tex]
[tex]T_x = T*sin(50) = 0.0234 N[/tex]
The electric force is given by the expression:
[tex]F = k*\frac{q_1*q_2}{r^2}[/tex]
In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):
[tex]r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m[/tex]
And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.
[tex]F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}[/tex]
[tex]q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C[/tex]
O 0.247 μC
An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and sticks 16 m above the ground. You may ignore air resistance. a) What was the initial speed of the arrow? b) At what angle above horizontal was the arrow shot?
Answer:
(a). The initial speed of the arrow is 49.96 m/s.
(b). The angle is 39.90°.
Explanation:
Given that,
Horizontal distance = 230 m
Time t = 6 sec
Vertical distance = 16 m
We need to calculate the horizontal component
Using formula of horizontal component
[tex]R =u\cos\theta t [/tex]
Put the value into the formula
[tex]\dfrac{230}{6} = u\cos\theta[/tex]
[tex]u\cos\theta=38.33[/tex].....(I)
We need to calculate the height
Using vertical component
[tex]H=u\sin\theta t-\dfrac{1}{2}gt^2[/tex]
Put the value in the equation
[tex]16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2[/tex]
[tex]u\sin\theta=\dfrac{16+9.8\times18}{6}[/tex]
[tex]u\sin\theta=32.06[/tex].....(II)
Dividing equation (II) and (I)
[tex]\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}[/tex]
[tex]\tan\theta=0.8364[/tex]
[tex]\theta=\tan^{-1}0.8364[/tex]
[tex]\theta=39.90^{\circ}[/tex]
(a). We need to calculate the initial speed
Using equation (I)
[tex]u\cos\theta\times t=38.33[/tex]
Put the value into the formula
[tex]u =\dfrac{230}{6\times\cos39.90}[/tex]
[tex]u=49.96\ m/s[/tex]
(b). We have already calculate the angle.
Hence, (a). The initial speed of the arrow is 49.96 m/s.
(b). The angle is 39.90°.