The final velocity of the boy after the collision with snowball is 4.84 m/s at 18.4⁰ north-east
The given parameters;
Mass of the boy, m₁ = 25 kgSpeed of the boy, u₁ = 5 m/sMass of the snowball, m₂ = 1.0 kgSpeed of the snow ball, u₂ = 15 m/sThe initial momentum of the boy is calculated as follows;
[tex]P_y = m_1 u_1\\\\P_y = 25 \times 5\\\\P_y = 125 \ kgm/s \ \ north[/tex]
The initial momentum of the snowball is calculated as follows;
[tex]P_x = m_2 u_2\\\\P_x = 1 \times 15 \ \\\\P_x = 15 \ kgm/s \ \ west[/tex]
The resultant momentum of the boy and the snowball after collision is calculate as follows;
[tex]P_f = \sqrt{P_y^2 + P_x^2} \\\\P_f = \sqrt{125^2 + 15^2} \\\\P_f = 125 .9 \ kgm/s[/tex]
The final velocity of the system boy-snowball system is calculated as;
[tex]v_f(m_1 + m_2)= P_f\\\\v_f = \frac{P_f}{m_1 + m_2} \\\\v_f = \frac{125.9}{25 + 1} \\\\v_f = 4.84 \ m/s[/tex]
The direction of the boy after the collision is calculated as follows;
[tex]\theta = tan^{-1}(\frac{v_y}{v_x} )\\\\\theta = tan^{-1} (\frac{5}{15} )\\\\\theta = 18.4 \ ^0[/tex]
Thus, the final velocity of the boy after the collision with snowball is 4.84 m/s at 18.4⁰ north-east
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Would it be possible to direct the speeds that a coaster will reach before its ever placed on a track?How?
Yes, it is possible to determine the final speed of the rollercoaster if the initial speed and the height to be reached are known.
According the principle of conservation of energy, the total kinetic energy is equal to the total potential energy.
[tex]P.E _i + K.E_i = P.E_f + K.E_f \\\\mgh_i + \frac{1}{2} mv_i^2 = mgh_f + \frac{1}{2} mv_f^2\\\\gh_i + \frac{1}{2} v_i^2 = gh_f + \frac{1}{2} v_f^2\\\\g(0) + \frac{1}{2} v_i^2 = gh_f + \frac{1}{2} v_f^2\\\\\frac{1}{2} v_i^2 = gh_f + \frac{1}{2} v_f^2\\\\v_i^2 = 2gh_f + v_f^2\\\\v_i^2 - 2gh_f = v_f^2\\\\v_f = \sqrt{v_i^2 - 2gh_f}[/tex]
where;
[tex]v_i[/tex] is the initial velocity of the roller coaster[tex]v_f[/tex] is the final velocity of the roller coaster[tex]h_f[/tex] is maximum height reached by the roller coasterThus, it is possible to determine the final speed of the rollercoaster if the initial speed and the height to be reached are known.
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14. Earthworms are crucial for forming soil. As they search for food by digging tunnels,
they expose rocks and minerals to the effects of weathering. Over time, this process
creates new soil. Worms are not the only living things that help to create soil. Plants
also play a part in the weathering process. As the roots of plants grow and seek out
water and nutrients, they help to break large rock fragments into smaller ones. Have
you ever seen a plant growing in a sidewalk? As the plants grows, its roots spread into
tiny cracks in the sidewalk. These roots apply pressure to the cracks, and, over time,
the cracks become larger, ice wedging can occur more readily. As the cracks expand,
more water can flow into them. When the water freezes, it expands and presses against
the walls of the cracks, which makes the cracks larger. Over time, the weathering
caused by water, plants, and worms help to form soil. QUESTION: Ice wedging, as
described in the passage, is an example of which of the following?
A mechanical weathering
B. Oxidation
C. chemical weathering
D. Hydrolysis
Ice wedging, as described in the passage, is an example of mechanical
weathering.
Mechanical weathering is also known as physical weathering and it
involves the breaking of rock into smaller particles without causing changes
in the chemical properties.Mechanical weathering is usually carried through
physical processes such as freezing and thawing etc.
In this scenario, we were told that water freezes, expands and presses
against the walls of the crack thereby breaking into smaller parts which is a
physical process hence mechanical weathering being present.
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Three people are trying to move a box. Which set of forces will result in a net
force on the box of 20 N to the left?
Answer: push force gravity tension and reverse force
Explanation: sense they are pushing the box there is a push force gravity because this is likely on earth tension because it is the reverse of gravity and the reverse force because you have to have the reverse of push
a spring with a spring constant of 65 n/m is displaced -94cm. what is the magnitude of the force exerted by the spring
Answer:
Explanation:
F = kx = 65 N/m(0.94 m) = 61.1 = 61 N
Look at the circuit given below. It consists of a cell, a bulb with two terminals X, Y and wires. P, Q, R and S are positions marked. What is the direction of the flow of current? a) PQXYRS b) SRYXQP c) SPQXYR d) PSRYXQ
The conventions for the current allow to find the result for which is the direction of the current the is circuit is:
It leaves the positive pole, goes to the blue cable, to the light bulb, from there to the red cable and reaches the negative pole.
Current is the movement of electrons in a circuit per unit time.
In an electrical circuit the electrons that have a negative charge leave the negative electrode towards the positive electrode, by convention the current goes in the opposite direction from the positive electrode towards the negative electrode.
When analyzing the given circuit, the current leaves the positive pole at the top of the battery and moves through the blue wire, reaches the light bulb and moves the red wire and reaches the negative pole of the battery.
In conclusion using the conventions for the current we can find the result for which is the direction of the current the is circuit is:
It leaves the positive pole, goes to the blue cable, to the light bulb, from there to the red cable and reaches the negative pole.
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HELP ME the mean free path λ and the mean collision time τ of the molecules of a diatomic gas of molecular mass 6.00 × 10⁻²⁵ kg and radius r = 1.0 x 10⁻¹⁰ m are measured. From these microscopic data can we obtain macroscopic properties such as temperature T and pressure P? If so, consider λ = 4.32 x 10⁻⁸ m and τ = 3.00 x 10⁻¹⁰ s and calculate T and P.
The temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.
The given parameters;
Mass of the gas molecules, m = 6 x 10⁻²⁵ kgRadius of the gas, r = 1 x 10⁻¹⁰ mMean free path, [tex]\lambda_{rms}[/tex] = 4.32 x 10⁻⁸ mMean collision time, [tex]\tau = 3 \times 10^{-10} \ s[/tex]The mean velocity of the gas molecules is calculated as follows;
[tex]\tau = \frac{\lambda _{rms}}{V_{rms}} \\\\V_{rms} = \frac{\lambda _{rms}}{\tau} \\\\V_{rms} = \frac{4.32 \times 10^{-8} }{3 \times 10^{-10}} \\\\V_{rms} = 144 \ m/s[/tex]
The temperature of the gas molecules is calculated as follows;
[tex]V_{rms} = \sqrt{\frac{3kT}{M} } \\\\V_{rms}^2 = \frac{3kT}{M} \\\\T = \frac{V_{rms} ^2 M}{3k}[/tex]
where;
k is Boltzmann constant[tex]T = \frac{V_{rms} ^2 M}{3k} \\\\T = \frac{(144)^2 \times (6.0 \times 10^{-25})}{3 \times 1.38 \times 10^{-23}} \\\\T = 300.5 \ K[/tex]
The number of gas molecules per unit volume is calculated as follows;
[tex]\lambda = \frac{1}{4\pi \sqrt{2} \ r^2 n} \\\\n = \frac{1}{\lambda 4\pi \sqrt{2} \ r^2} \\\\n = \frac{1}{(4.32 \times 10^{-8}) \times 4 \pi \times \sqrt{2} \ \times (1\times 10^{-10})^2} \\\\n = 1.303 \times 10^{26} \ molcules/m^3[/tex]
The pressure of the gas molecule is calculated as follows;
[tex]n = \frac{P}{kT} \\\\P = nkT\\\\P = (1.303 \times 10^{26} ) \times (1.38 \times 10^{-23}) \times (300.5)\\\\P = 540,341.07 \ Pa\\\\P = 5.33 \ atm[/tex]
Thus, the temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.
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A 2.8 kg rectangular air mattress is 2.00 m long, 0.500 m wide, and
0.100 m thick. What mass can it support in water before sinking?
160 m long When a truck ulls
Mass the air mattress support before sinking will be M = 97.2 kg
What will be the mass mattress will support?Given that
mass of mattress m=2.8 kg
Length = 2.00 m
width =0.5 m
The mattress can support a total weight that is equal to the weight of water that the air mattress displaces.
The volume of the air mattress will be equal to
[tex]V= (2\times 0.5\times 0.1)[/tex]
[tex]V= 0.1 m^{3}[/tex]
This volume is equal to the maximum amount of water that the mattress can displace in the water before it begins sinking.
We know the density of water is 1000 kg / m²
We can calculate the weight of the maximum amount of water that mattress will displace
[tex]V\times density of water\times g=0.1\times 1000\times 9.81=981N[/tex]
Let M represent the mass that the mattress is supporting.
The total weight of the mattress and the supported mass is:
[tex](2.8+M)\times9.81[/tex]
Setting this weight equal to the weight of the water displaced gives:
[tex](2.8+M)\times9.81=981[/tex]
[tex]2.8+M=100[/tex]
m = 97.2 kg
Hence Mass the air mattress support before sinking will be M = 97.2 kg
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A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) what is the mass of the block office? (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?
Answer:
(a) 91 kg (2 s.f.) (b) 22 m
Explanation:
Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.
(a)
[tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}[/tex]
Subsequently,
[tex]F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})[/tex]
*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.
(b) To find the final velocity of the ice block at the end of the first 5 seconds,
[tex]v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}[/tex]
According to Newton's First Law which states objects will remain at rest
or in uniform motion (moving at constant velocity) unless acted upon by
an external force. Hence, the block of ice by the end of the first 5
seconds, experiences no acceleration (a = 0) but travels with a constant
velocity of 4.4 [tex]m \ s^{-1}[/tex].
[tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}[/tex]
Therefore, the ice block traveled 22 m in the next 5 seconds after the
worker stops pushing it.
2. Define Lightning conductor. How does it work?
In Building, there are a host of protective devices that are installed to protect lives and properties, one of them is the Lightning Conductor that a metal rod mounted on a structure and intended to protect the structure from a lightning strike. It works on the principle of induction
Principle of Operation of Lightning ConductorThe lightning conductor works on the principle of induction:
When a charged cloud passes by the building hosting the Lightning conductor, it gets a charge opposite to that of the cloud through the process of induction. They are Typically made from copper material.Most Lightening conductors are made from copper materials.
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Which ball moved at the same speed as Ball 3?
Answer:
you forgot to attach the image
Can someone help me solve this problems please? It's a physics problem.
Answer:
i cant see
Explanation:
but im smart
Two rods of mass m, length L are stuck together to form an X shape and spun around the center. What is the rotational inertia
A change in momentum is also called:
a. Impact
b. Imput
c. Impulse
d. Impole
Answer:
c. Impulse
Explanation:
a uniform thin rod of length l and mass m is allowed to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position. a.) What is the speed of the center of gravity when the rod reaches its lowest position? b.) What is the tangential speed of the lowest point of the rod when the rod reaches its lowest position?
Answer:
Explanation:
Potential energy gets converted to rotational kinetic energy
a) ½Iω² = mgh
½(mL²/3)ω² = mgL/2
(L/3)ω² = g
ω = [tex]\sqrt{3g/L}[/tex]
v(CG) = (L/2) [tex]\sqrt{3g/L}[/tex]
Not sure if you wanted angular speed or tangential speed of the CG so I gave both.
b) v = Rω = L [tex]\sqrt{3g/L}[/tex]
The part of the moon that is visible will appear to grow and shrink during the lunar cycle. This occurs in the direction of _____ to _____.
Answer:
left to right
Explanation:
give me brain pls
Answer:
of the moon to .......
Select the best reason for studying the past and its effect on us today based on "The Terror of the Middle Ages." A. to learn what people did on a daily basis B. to enjoy stories about where people used to live O C. to study the causes of diseases and learn to prevent them D. to learn about earlier cultures and lifestyles
it is D for sure im good with history
Answer: D
Explanation: Its just s that guy ca get brainlist :D
In the following free body diagram, what is the net force on the object?
A.10n
B.5 N to the right
C.20 N to the right
D.7 N to the right
Answer:
B
Explanation:
Simply take all forces pointing to the right of the box as positive and all of the forces pointing to the left of the box as negative and add all values.
ΣF = 7 + 18 + (-20) = 5N to the right
Hello I am absolutely stumped on these six physics problems. Please help me on them.
Answer:
Explanation:
20° from the normal = 110° from parallel
1a) τ = (200sin90)[6] + (75sin110)[3] - (100sin110)[3]
τ = 1,129.52305... = 1100 N•m CCW
1b) τ = 200(6)(sin90) + 75(-3)(sin(360-110)) + 100(3)(sin(270 + 20)
τ = 1,129.52305... = 1100 N•m CCW
1c) My directions agree, both are positive z by right hand rule.
1d) Moment of inertia for a thin rod about an axis perpendicular to its center is
I = (1/12)mL²
τ = Iα
α = τ/I = 1129.523 / ((1/12)(200)(12²)) = 0.4706 rad/s²
θ = ½αt²
θ = ½(0.4706)(2•60)² = 3,388.56916... radians
θ = 3400 radians
at which time is is spinning about 9 revolutions per second
1. What is the frequency of light waves with wavelength of 5 x 10-⁷ m?
Answer:
Speed of light =m/s
wavelength = m
frequency = ?
we have
Speed = frequency × wavelength
[tex]3* 10^8[/tex] = frequency × [tex]5 * 10^{-7}[/tex]
Frequency = [tex]\frac{3*10^8}{5*10^{-7}}=6*10^{14}[/tex]hz
Taking into account the definition of wavelength, frecuency and propagation speed, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.
Definition of wavelengthFirst of all, wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).
Definition of frequencyOn the other side, frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
Definition of propagation speedFinally, the propagation speed is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement.
The propagation speed relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation:
v = f× λ
All electromagnetic waves propagate in a vacuum at a constant speed of 3×10⁸ m/s, the speed of light.
Frequency of light waves with wavelength of 5×10⁻⁷ mIn this case, you know:
v= 3×10⁸ m/sf= ?λ= 5×10⁻⁷ mReplacing in the definition of propagation speed:
3×10⁸ m/s = f× 5×10⁻⁷ m
Solving:
3×10⁸ m/s ÷ 5×10⁻⁷ m= f
f= 6×10¹⁴ Hz
In summary, the frequency of light waves with wavelength of 5×10⁻⁷ m is 6×10¹⁴ Hz.
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How large is the tension in a rope that is being used to accelerate a 100 kg box upward at 2m/s2?
[tex]\\ \sf\Rrightarrow T=F[/tex]
[tex]\\ \sf\Rrightarrow T=ma[/tex]
[tex]\\ \sf\Rrightarrow T=100(2)[/tex]
[tex]\\ \sf\Rrightarrow T=200N[/tex]
How can whales descend quickly and face no problems
Answer:
Whales face an increasing number of threats including: For hundreds of years people hunted whales for their oil to fuel lamps and candles, to lubricate machinery and to make margarine, lipsticks and other products. They also used baleen whales to make tennis racquets and corsets!
Explanation:
A moving car initially has kinetic energy K. The car then moves in the opposite direction with four times its initial speed. What is the kinetic energy now
If initial speed was v, Kinetic energy was K = 1/2mv^2
When speed will be 4v, KE will be:
= 1/2 m (4v)^2
=1/2m 16 v^2
=16K
Now kinetic energy will become 16 times of initial Kinetic Energy
Predict changes in state according to change in particle motion. Know the vocabulary used to describe changes of state.
The change in the state of matter causes change in the motion of the particles of the matter. The gaseous state of matter has the greatest speed while the solid state has the least speed.
The change in state of every matter is accompanied by lost or gained of energy.
Example is water.
The solid state of water is ice. The motion of particles of the water is relatively zero because the molecules are held at a fixed position.
The liquid state of water occurs when the temperature of the ice is increased above zero degree Celsius. The speed of the particles of water in liquid state is greater than solid state.
The gaseous state of water occurs when the temperature of the liquid water is increased beyond 100 degree Celsius. The speed of water in gaseous state is greater than liquid state.
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PLEASE HELP ASP WILL GIVE 50 POINT AND BRAINLIEST!!!!!!!!!!!!!
A. In the Bohr model of the hydrogen atom, the speed of the electron is approximately
2.16 × 10⁶ m/s. Find the central force acting on the electron as it revolves in a circular orbit of radius 5.17 × 10⁻¹¹ m.
Answer in units of N.
B. Find the centripetal acceleration of the electron
Explanation:
A. The centripetal force experienced by an electron as it goes around a hydrogen nucleus is given by
[tex]F_c = m_e\dfrac{v^2}{r}[/tex]
where [tex]m_e = \text{electron\:mass} = 9.11×10^{-31}\:\text{kg}[/tex]
[tex]r = 5.17×10^{-11}\:\text{m}[/tex] = orbital radius
[tex]v = 2.16×10^6\;\text{m/s}[/tex] = orbital velocity
so the centripetal force is
[tex]F_c = (9.11×10^{-31}\:\text{kg})\dfrac{(2.16×10^6\;\text{m/s})^2}{5.17×10^{-11}\:\text{m}}[/tex]
[tex]\;\;\;=8.22×10^{-8}\:\text{N}[/tex]
B. The electron's centripetal acceleration is given by
[tex]a_c = \dfrac{v^2}{r}[/tex]
Using the values from (A), we get
[tex]a_c = \dfrac{(2.16×10^6\;\text{m/s})^2}{5.17×10^{-11}\:\text{m}}[/tex]
[tex]\;\;\;=9.02×10^{22}\:\text{m/s}^2[/tex]
Which of the following is NOT a function of the lens in the eye?
It can perform minor adjustments for distance.
It flattens when light rays from distant objects are to be focused.
It is a light receptor that generates nerve signals that are sent to the brain.
It maintains its spherical shape to view nearby objects.
Answer:
it is a light receptor that generates nerve signals that are sent to the brain
Explanation:
the lens are like the glasses, this means that is used to see things better. You just put them in your eye and that's all it's not connected to the brain
Answer:
c
Explanation:
Raghu studies in grade 6th. He wants a cricket bat to be made by a carpenter. He tells the
that the length of the bat should be 7 hand spans. The tall carpenter tells Raghu that it
will be ready by tomorrow. When Raghu went to collect the bat the next day, he was very
disappointed. Why? Was the bat longer or shorter than what Raghu expected? Give reason.
carpenter
Based on the information given, it can be noted that the bat was either shorter or longer than what he expected.
From the information given, it was stated that Raghu wants a cricket bat to be made by a carpenter and he tells the carpenter that the length of the bat should be 7 hand spans.
Since he got disappointed when he collected the bat, the reason for this will be because the bat was either shorter or longer than what he requested.
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True or False. Abraham and Sarah were in agreement regarding God's command to sacrifice Isaac.
Answer:
Explanation:
Not in Genesis. The proposed sacrifice of Isaac was all Abraham's doing. Sarah was not present and could not, therefore offer any input. It might have been covered in other Jewish writings, but it is not recorded in Genesis.
Since the book most widely used by Everyone is the Bible, I would answer false.
The graph below shows the variation with distance r from the nucleus of the square of the wave function, Ψ^2, of a hydrogen atom according to Schrödinger theory.
A. is most likely to be near a.
B. is always a.
C. is always less than a.
D. is always greater than a.
The region a represents the distance of the electron from the nucleus.
According to the wave mechanical model of the atom, the probability of finding an electron within a given volume element (representing the atom) is the square of the wave function psi.
Since a is the region in space where there is the greatest probability of finding the electron in the atom, it follows that distance of the electron form the atom is always a.
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As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constants. The angular speed changes from 3.70 rad/s at t = 0 to 2.00 rad/s at t = 8.60 s.
a. Use this information to determine σ and ω0.
σ = _______s−1
ωo = ______rad/s
b. Determine the magnitude of the angular acceleration at t = 3.00 s.
______rad/s2
c. Determine the number of revolutions the wheel makes in the first 2.50 s
_______rev
d. Determine the number of revolutions it makes before coming to rest.
_______rev
Hi there!
a.
We can use the initial conditions to solve for w₀.
It is given that:
[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]
We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:
[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]
Now, we can solve for sigma using the other given condition:
[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]
b.
The angular acceleration is the DERIVATIVE of the angular velocity function, so:
[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]
c.
The angular displacement is the INTEGRAL of the angular velocity function.
[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]
[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]
[tex]\theta = 8.471 rad[/tex]
Convert this to rev:
[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]
d.
We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.
[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]
Evaluate the improper integral:
[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]
Convert to rev:
[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]
A 2.2 kg model rocket is shot straight up in the air from the ground, with an initial velocity of 36.4 m/s. The rocket reaches its maximum height, and falls back to the ground. What is the maximum height of the rocket? Round your answer to 2 decimal places.
Answer:
Explanation:
Ignoring friction, the initial kinetic energy will convert to maximum potential energy at its highest point.
PE = KE
mgh = ½mv²
h = v²/2g
h = 36.4²/ (2(9.81))
h = 67.53109...
h = 67.53 m