Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
A subatomic particle created in an experiment exists in a certain state for a time of before decaying into other particles. Apply both the Heisenberg uncertainty principle and the equivalence of energy and mass to determine the minimum uncertainty involved in measuring the mass of this short-lived particle.
Answer:
Δm Δt> h ’/ 2c²
Explanation:
Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions
ΔE Δt> h ’/ 2
h’= h / 2π
to relate this to the masses let's use Einstein's relationship
E = m c²
let's replace
Δ (mc²) Δt> h '/ 2
the speed of light is a constant that we can condense exact, so
Δm Δt> h ’/ 2c²
A 1100 kg car pushes a 1800 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 N.A) What is the magnitude of the force of the car on the truck?B) What is the magnitude of the force of the truck on the car?
Answer:The answer is 3000 N.
Force (F) is the multiplication of mass (m) and acceleration (a).
F = m · a
It is given:
mc = 1000 kg
mt = 2000 kg
total force: F = 4500 N
total mass: m = mc + mt
Let's calculate acceleration which is common:
a = F/m = F/(mc + mt) = 4500/(1000 + 2000) = 4500/3000 = 1.5 m/s²
Now, when we know acceleration, let's calculate force on the truck:
Ft = mt · a = 2000 · 1.5 = 3000 N
Explanation:
An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. The direction of the electric field vector of this wave is:___________
Answer:
either +z direction or -z direction.
Explanation:
The direction of the electric field, in an electromagnetic wave always is perpendicular to the direction of the magnetic field and the direction of propagation of the wave.
You assume a system of coordinates with the negative x axis as the west direction, and the y axis as the up direction
In this case, the wave is propagating toward the west (- x direction), and the magnetic field vector points up (+ y direction), then, it is mandatory that the electric field vector points either +z direction or -z direction.
A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simple pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 6.01 s. What is the height of the tower
Answer:
The height of the tower is 8.96 m.
Explanation:
We have, a visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simple pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 6.01 s.
It is required to find the height of the tower. Let it is l. The time period of a simple pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of pendulum, or height of tower
[tex]l=\dfrac{T^2g}{4\pi^2}\\\\l=\dfrac{(6.01)^2\times 9.8}{4\pi^2}\\\\l=8.96\ m[/tex]
So, the height of the tower is 8.96 m.
A space ship traveling east flies directly over the head of an inertial observer who is at rest on the earth's surface. The speed of the space ship can be found from this relationship: . The navigator's on-board instruments indicate that the length of the space ship is 20 m. If the length of the ship is measured by the inertial earth-bound observer, what value will be obtained
Answer:
10 metres
Explanation:
So, we are given the following data or parameters or information which is going to assist us in solving this particular problem or Question efficiently.
=> The speed of the space ship can be found from this relationship: ✓(1 - [v^2/c^2] ) = 1/2.
=> The length of the space ship = 20 m.
=> Assumption = '' If the length of the ship is measured by the inertial earth-bound observer".
Thus, from the speed of the space ship can be found from this relationship we can determine the value;
✓(1 - [v^2/c^2] ) = 1/2.
V = 20 × 1/2 = 10 metres.
Note that we use the contraction formula to solve for V.
Suppose that 7.4 moles of a monatomic ideal gas (atomic mass = 1.39 × 10-26 kg) are heated from 300 K to 500 K at a constant volume of 0.74 m3. It may help you to recall that CV = 12.47 J/K/mole and CP = 20.79 J/K/mole for a monatomic ideal gas, and that the number of gas molecules is equal to Avagadros number (6.022 × 1023) times the number of moles of the gas.
1) How much energy is transferred by heating during this process?2) How much work is done by the gas during this process?3) What is the pressure of the gas once the final temperature has been reached?4) What is the average speed of a gas molecule after the final temperature has been reached?5) The same gas is now returned to its original temperature using a process that maintains a constant pressure. How much energy is transferred by heating during the constant-pressure process?6) How much work was done on or by the gas during the constant-pressure process?
Answer:
Explanation:
1 ) Since it is a isochoric process , heat energy passed into gas
= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .
= 7.4 x 12.47 x ( 500 - 300 )
= 18455.6 J.
2 ) Since there is no change in volume , work done by the gas is constant.
3 ) from , gas law equation
PV = nRT
P = nRT / V
= 7.4 x 8.3 x 500 / .74
= .415 x 10⁵ Pa.
4 ) Average kinetic energy of gas molecules after attainment of final temperature
= 3/2 x R/ N x T
= 1.5 x 1.38 x 10⁻²³ x 500
= 1.035 x 10⁻²⁰ J
1/2 m v² = 1.035 x 10⁻²⁰
v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶
= 1.49 x 10⁶
v = 1.22 x 10³ m /s
5 ) In this process , pressure remains constant
gas is cooled from 500 to 300 K
heat will be withdrawn .
heat withdrawn
= n Cp dT
= 7.4 x 20.79 x 200
= 30769.2 J .
6 )
gas will have reduced volume due to cooling
reduced volume = .74 x 300 / 500
= .444 m³
change in volume
= .74 - .444
= .296 m³
work done on the gas
= P x dV
pressure x change in volume
= .415 x 10⁵ x .296
= 12284 J.
Complete the first and second sentences, choosing the correct answer from the given ones.
1. The water temperature in the dish depends on the A / B / C / D.
A. average kinetic energy of water molecules
B. total kinetic energy of water molecules
C. water mass. D. potential energy of the container with water
2. The internal energy of the water in the vessel is E / F / G.
E. potential energy of the vessel with water
F. average kinetic energy of water molecules
G. sum of kinetic energy and potential water molecules
Answer:
Hope this helps :)
Explanation:
1. A
2. G (because the basic definition of internal energy is, the sum of kinetic and potential energies of water molecules)
write the answer:
physics ... i need help
Answer:
6 gallons
Explanation:
At 30 mph, the fuel mileage is 25 mpg.
After 5 hours, the distance traveled is:
30 mi/hr × 5 hr = 150 mi
The amount of gas used is:
150 mi × (1 gal / 25 mi) = 6 gal
a) Write the names of the materials used in the ohm law according to the Figure 1?
b) If the voltage of a circuit is 12 V and the resistance is 40 , What is the generated power?
Answer:
a. i. conducting wire
ii high-pass and low-pass filters
iii. Cobra-4 Xpert-link
iii. voltage source
b. Power generated is 3.6 W.
Explanation:
Ohm's law state that the current passing through a metallic conductor, e.g wire is directly proportional to the potential difference across its ends, provided temperature is constant.
i.e V = IR
i. conducting wire
ii high-pass and low-pass filters
iii. Cobra-4 Xpert-link
iii. voltage source
b. Given that; V = 12 V and R = 40 Ohm's.
P = IV
From Ohm's law, I = [tex]\frac{V}{R}[/tex]
So that;
P = [tex]\frac{V^{2} }{R}[/tex]
= [tex]\frac{12^{2} }{40}[/tex]
= [tex]\frac{144}{40}[/tex]
= 3.6 W
The power is 3.6 W.
Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph . Normally Richard drives at the speed limit, but today he is running late and decides to take his chances by driving at 80 mph. How many minutes does he save?
Answer:
t = 25.5 min
Explanation:
To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.
[tex]t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h[/tex]
Next, you calculate the difference between both times t1 and t2:
[tex]\Delta t=t_1-t_2=2.30h-1.875h=0.425h[/tex]
This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:
[tex]0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s[/tex]
hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph
50 points!! please help :((
Answer:
Loudness: decreases
Amplitude: decreases
Pitch: stays the same
Frequency: stays the same
Explanation:
1.
An oscilloscope measures how much the microphone is vibrating, or how much electricity it is sending. This means that a louder noise will register higher on the oscilloscope. Since the size of the waves at Y is lower than at X, the loudness of the sound has decreased.
2.
Similarly to loudness, amplitude measures how far the crests of the waves are from the nodes. Since Y is closer to the center line than X, it has a lower amplitude.
3 and 4.
The pitch and frequency, for our purposes, are essentially the same thing here. They are dependent on how close together the waves on the oscilloscope are, or how quickly the microphone is vibrated. Since this stays the same throughout the entire sound, they both stay the same.
Hope this helps!
Convert from standard form to scientific notation:
0.00000013
A)1.3 x 10-7
B)13 x 108
C)1.3 x 107
D)13 x 10-8
Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an upstairs bathtub 3.70m above the heater is 414-kPa. What's the flow speed in this pipe?
Answer:
The velocity is [tex]v_2= 0.45 \ m/s[/tex]
Explanation:
From the question we are told that
The initial speed of the hot water is [tex]v_1 = 0.85 \ m/s[/tex]
The pressure from the heater [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]
The height of the hot water before flowing is [tex]h_1 = 0 \ m[/tex]
The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]
The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]
The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]
Apply Bernoulli equation
[tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]
Substituting values
[tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]
=> [tex]v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}[/tex]
=> [tex]v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}[/tex]
=> [tex]v_2= 0.45 \ m/s[/tex]
Two vectors having magnitudes of 5.00 and 9.00 respectively. If the value of their dot product is 12.0, find the angle between the two vectors.
Answer:
C = 74.53°
Explanation:
Let the magnitudes of 5.00 and 9.00 be vectors A and B respectively, hence the dot product of this vector is defined as
A.B = |A||B|cosC; let C be the angle between the vectors
12 = 5×9 cos C
Hence cos C = 12/45
C = cos^-1(12/45)
C = 74.53°
A standing wave on a string that is fixed at both ends has frequency 80.0 Hz. The distance between adjacent antinodes of the standing wave is 12.0 cm. What is the speed of the waves on the string, in m/s
Answer:
v = 19.2 m/s
Explanation:
In order to find the speed of the string you use the following formula:
[tex]f=\frac{v}{2L}[/tex] (1)
f: frequency of the string = 80.0Hz
v: speed of the wave = ?
L: length of the string = 12.0cm = 0.12m
The length of the string coincides with the wavelength of the wave for the fundamental mode.
Then, you solve for v in the equation (1), and replace the values of the other parameters:
[tex]v=2Lf=2(0.12m)(80.0Hz)=19.2\frac{m}{s}[/tex]
The speed of the wave is 19.2 m/s
Inside a stereo speaker, you will find two permanent magnets: one on the cone and one near the cone. True of false?
Answer:
false
Explanation:
g it as been suggested that solar powered space ships could get a boost from a laser either on earth or in orbit around earth. the laser would have to be very powerful to give any measurable benefit to the ship. if the laser produces a 0.18-m diameter beam of 490-nm light, what is the minimum angular spread of the beam?
Answer:
The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians (or 3.32 μrad).
Explanation:
The minimum angular spread of a wave is the ratio of its narrowest diameter to its wavelength.
From Rayleigh's formula,
Angular spread = 1.22 (wavelength ÷ diameter)
= 1.22 (λ ÷ D)
Given that:
diameter, D = 0.18 m and wavelength, λ = 490 nm, then;
Angular spread of the laser beam = 1.22 (λ ÷ D)
= 1.22[tex](\frac{490*10^{-9} }{0.18})[/tex]
= 1.22× 2.7222 × [tex]10^{-6}[/tex]
= 3.3211 × [tex]10^{-6}[/tex] rad
The minimum angular spread of the laser beam is 3.32 × [tex]10^{-6}[/tex] radians.
A potential difference of 71 mV is developed across the ends of a 12.0-cm-long wire as it moves through a 0.27 T uniform magnetic field at a speed of 6.0 m/s. The magnetic field is perpendicular to the axis of the wire.
Required:
What is the angle between the magnetic field and the wire's velocity?
Answer:
Explanation: please see attached file I attached the answer to your question.
The angle between the magnetic field and the wire's velocity is 33.2 degrees.
Calculation of the angle:Since the potential difference = 71mv = 71 *10 ^-3 V
The length is 12 cm = 0.12m
The magnetic field i.e. B = 0.27T
The speed or v = 4 m/s
here we assume [tex]\theta[/tex] be the angle
So,
e = Bvl sin[tex]\theta[/tex]
So,
[tex]Sin\theta[/tex] = e/bvl
= 71*10^-3 / 0.27 *4*0.12
= 0.5478
= 33.2 degrees
Therefore, the angle should be 33.2 degrees
Learn more about an angle here: https://brainly.com/question/14661707
The Gulf Stream off the east coast of the United States can flow at a rapid 3.8 m/s to the north. A ship in this current has a cruising speed of 8.0 m/s . The captain would like to reach land at a point due west from the current position.
At this heading, what is the ship's speed with respect to land?
Answer:
61.6° west of South
Explanation:
The ship goes to the south at an equal rate just like water flows to the north. Thus, the velocities would balance making the ship move towards the west.
Since we're dealing with water, the ship goes 3.8 m / s to the South, but a lot still remains to the west. Finding this would require us drawing a triangle. 3.8 m/s point down side and the hypotenuse is 8
cos(θ) = [adjacent/hypotenuse]
Cos θ = 3.8/8
Cos θ = 0.475
θ = cos^-1 (0.475)
θ = 61.6°
Therefore the angle is 61.6° west of South.
a 15-nC point charge is at the center of a thin spherical shell of radius 10cm, carrying -22nC of charge distributed uniformly over its surface. find the magnitude and direction of the electric field (a) 2.2cm,(b)5.6cm,and (c)14 cm from the point charge.
Answer:
A) E = 278925.62 N/C with direction; radially out.
B) E = 43048.47 N/C with direction radially out.
C) E = -3214.29 N/C with direction radially in.
Explanation:
From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;
E = kQ/r²
where;
Q is the net charge within the distance r.
We are given the charge Q = 15-nC and
spherical shell of radius 10cm
A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C
While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²
E = 278925.62 N/C
This will be radially out ,since the net charge is positive.
B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C
While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²
E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²
E = 43048.47 N/C
This will be radially out ,since the net charge is positive.
C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;
Q = 15 nC - 22 nC
Q = -7 nC = -7 x 10^(-9) C
and;
E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²
E = -3214.29 N/C
This will be radially in, since the net charge is negative. You can indicate this with a negative answer.
A) When The distance r is = 2.2 cm = 0.022 m is between the surface and also the point charge, also that so only the point charge lies within this distance and also Q = 15 NC = 15 x 10^(-9) C
Then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²Then E = 278925.62 N/CThen This will be radially out since the net charge is positive.
B) When The distance r = 5.6 cm = 0.056 m is between the surface and also the point charge, so only the point charge lies within this distance and also Q = 15 nC = 15 x 10^(-9) C
then While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²When E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²Then E = 43048.47 N/CAfter that This will be radially out since the net charge is positive.
C) Then when The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;
Then Q = 15 nC - 22 nCAfter that Q = -7 nC = -7 x 10^(-9) CWhen E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²Then E = -3214.29 N/C Thus, This will be radially in, since the net charge is negative.Find out more information about magnitude here:
https://brainly.com/question/13502329
An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object
Answer:
v = 25.45 m/s
Explanation:
In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:
[tex]h_{max}=\frac{v_o^2}{g}[/tex] (1)
vo: initial speed of the object = 18 m/s
g: gravitational acceleration = 9.8 m/s²
Furthermore you use the following formula for the final speed of the object:
[tex]v^2=v_o^2-2gh[/tex] (2)
h: height
You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:
[tex]v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2[/tex]
Then, you solve the previous result for vo:
[tex]v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}[/tex]
The initial speed of the object was 25.45 m/s
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven-the driving force is transferred to the object, which oscillates instead of the entire building X 50%
Part (a) What effective force constant, in N/m, should the springs have to make them oscillate with a period of 1.2 s? k = 9.5 * 106 9500000 X Attempts Remain 50%
Part (b) What energy, in joules, is stored in the springs for a 1.6 m displacement from equilibrium?
Answer:
The force constant is [tex]k =1.316 *10^{7} \ N/m[/tex]
The energy stored in the spring is [tex]E = 1.68 *10^{7} \ J[/tex]
Explanation:
From the question we are told that
The mass of the object is [tex]M = 4.8*10^{5} \ kg[/tex]
The period is [tex]T = 1.2 \ s[/tex]
The period of the spring oscillation is mathematically represented as
[tex]T =2 \pi \sqrt{ \frac{M}{k}}[/tex]
where k is the force constant
So making k the subject
[tex]k = \frac{4 \pi ^2 M }{T^2}[/tex]
substituting values
[tex]k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}[/tex]
[tex]k =1.316 *10^{7} \ N/m[/tex]
The energy stored in the spring is mathematically represented as
[tex]E = \frac{1}{2} k x^2[/tex]
Where x is the spring displacement which is given as
[tex]x = 1.6 \ m[/tex]
substituting values
[tex]E = \frac{1}{2} (1.316 *10^{7}) (1.6)^2[/tex]
[tex]E = 1.68 *10^{7} \ J[/tex]
A bicycle wheel has an initial angular velocity of 1.10 rad/s . Part A If its angular acceleration is constant and equal to 0.200 rad/s2 , what is its angular velocity at t = 2.50 s ? (Assume the acceleration and velocity have the same direction) Express your answer in radians per second. ω = nothing rads Request Answer Part B Through what angle has the wheel turned between t = 0 and t = 2.50 s ? Express your answer in radians. Δθ = nothing rad Request Answer Provide Feedback
Let [tex]\theta[/tex], [tex]\omega[/tex], and [tex]\alpha[/tex] denote the angular displacement, velocity, and acceleration of the wheel, respectively.
(A) The wheel has angular velocity at time [tex]t[/tex] according to
[tex]\omega=\omega_0+\alpha t[/tex]
so that after 2.50 s, the wheel will have attained an angular velocity of
[tex]\omega=1.10\dfrac{\rm rad}{\rm s}+\left(0.200\dfrac{\rm rad}{\mathrm s^2}\right)(2.50\,\mathrm s)=\boxed{1.60\dfrac{\rm rad}{\rm s}}[/tex]
(B) The angular displacement of the wheel is given by
[tex]\theta=\theta_0+\omega_0t+\dfrac\alpha2t^2\implies\Delta\theta=\omega_0t+\dfrac\alpha2t^2[/tex]
After 2.50 s, the wheel will have turned an angle [tex]\Delta\theta[/tex] equal to
[tex]\Delta\theta=\left(1.10\dfrac{\rm rad}{\rm s}\right)(2.50\,\mathrm s)+\dfrac12\left(0.200\dfrac{\rm ram}{\mathrm s^2}\right)(2.50\,\mathrm s)^2=\boxed{3.38\,\mathrm{rad}}[/tex]
A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance . For the next loading, the spring is compressed a distance . How much faster does the second dart leave the gun compared with the first
Complete question is;
A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?
Answer:
The second dart leaves the gun two times faster than the first one.
Explanation:
If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;
Potential energy = kinetic energy
Thus;
½kx² = ½mv²
Making velocity "v" the subject, we have;
v = √(kx²/m)
Since the initial distance is "x", thus initial launching velocity is;
v1 = √(kx²/m)
Since next distance is 2x, thus, second launch velocity is;
v2 = √(k(2x)²/m)
Expanding, we have;
v2 = √(4kx²/m)
v2 = 2√(kx²/m)
Comparing this to the one gotten for v1 earlier, we can see that it is double v1.
So, v2 = 2v1
Hence, The second dart leaves the gun two times faster than the first one.
Some runners train with parachutes that trail behind them to provide a large drag force. These parachutes are designed to have a large drag coefficient. One model expands to a square 1.8 m on a side, with a drag coefficient of 1.4. A runner completes a 200 m run at 5.0 m/s with this chute trailing behind. Part A How much thermal energy is added to the air by the drag force
Answer:
13.9 kJ
Explanation:
Given that
Length of the side, l = 1.8 m
Drag coefficient, C(d) = 1.4
Distance of run, d = 200 m
Velocity of run, v = 5 m/s
Density, ρ = 1.23
Using the Aerodynamics Drag Force formula. We have
F(d) = 1/2.ρ.A.C(d).v²
The Area, A needed is 1.8 * 1.8 = 3.24 m². So that,
F(d) = 1/2 * 1.23 * 3.24 * 1.4 * 5²
F(d) = 139.482/2
F(d) = 69.74
recall that, energy =
W = F * d
W = 69.74 * 200
W= 13948
W = 13.9kJ
Therefore, the thermal energy added to the air by the drag force is 13.9kJ
Consider a weather balloon floating in the air. There are three forces acting on this balloon: the force of gravity is FG, the force from lift towards balloon is FL, and the force from the wind is labeled Fw. The orientation of these forces along with a coordinate system is given below:
Assume that || FG || = 20 N, || FL ||= 25 N, and || Fw ll = 15 N.
Required:
Find the magnitude of the resultant force acting on the weather balloon and round your answer to two decimal places.
When Marcel finds the distance L from the previous part, it turns out to be greater than Lend, the distance from the pivot to the end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is at height h above the pivot. (Figure 3)With what force in the rightward direction, Fx, should Marcel push? If your expression would give a negative result (using actual values) that just means the force should be toward the left.Express your answer in terms of W, Lend, w, L2, L3, and h.
Answer:
Fx = - (1/h)( wL2 + wL3 - wLend )
Explanation:
Assuming The twins Gilles and Jean has a weight ( w ) each
The torque that would balance the equation would be = wL2 + wL3 -------- 1
THEREFORE the ccw torques are = wLend + Fh ----------- 2
hence equation 2 equals equation 1
= wLend + Fh = wL2 + wL3 --------- 3
equation 3 can as well be represented as
F = ( 1/h) ( wL2 + wL3 - wLend )---------- 4
From equation 4 it can be seen that F is on the left hand side therefore the value of Fx is negative
therefore equation 4 is represented as
Fx = - (1/h)( wL2 + wL3 - wLend )
A pendulum on a planet, where gravitational acceleration is unknown, oscillates with a time period 5 sec. If the mass is increased six times, what is the time period of the pendulum?
Explanation:
We have, a pendulum on a planet, oscillates with a time period 5 sec. The formula used to find the time period is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of the pendulum
g is acceleration due to gravity on which it is placed
It is clear that, the time period of pendulum is independent of the mass. Hence, if the mass is increased six times, its time period remains the same.
A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from zero to 4.30 rev/s in 3.05 s . Part A What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00 rev/s
Answer:
[tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]
Explanation:
Angular acceleration
[tex]\begin{aligned}
\alpha &=\frac{\left(\omega_{f}-\omega_{i}\right)}{t} \\
\omega_{i} &=0 \\
\omega_{f} &=4.30 \mathrm{rev} / \mathrm{s} \\
&=4.30 \times 2 \pi \mathrm{rad} / \mathrm{s} \\
&=27.02 \mathrm{rad} / \mathrm{s} \\
\alpha &=\frac{(27.02-0)}{3.15} \\
&=8.57 \mathrm{m} / \mathrm{s}^{2}
\end{aligned}[/tex]
a)Tangential acceleration
[tex]\begin{aligned}
a &=r \alpha \\
&=\frac{12}{2} \times 10^{-2} \times 8.57 \\
a &=0.51 \mathrm{m} / \mathrm{s}^{2}
\end{aligned}[/tex]
The tangential acceleration of the disc is [tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]
This question involves the concepts of the equations of motion for angular motion.
The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed reaches 2 rev/s will be "0.532 m/s²".
First, we will use the first equation of motion for the angular motion to find out the angular acceleration:
[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]
where,
[tex]\alpha[/tex] = angular acceleration = ?
[tex]\omega_f[/tex] = final angular speed = (4.3 rev/s)[tex](\frac{2\pi\ rad}{1\ rev})[/tex] = 27.02 rad/s
[tex]\omega_i[/tex] = initial angular speed = 0 rad/s
t = time taken = 3.05 s
Therefore,
[tex]\alpha =\frac{27.02\ rad/s-0\ rad/s}{3.05\ s}\\\\\alpha= 8.86\ rad/s^2[/tex]
Now, the tangential acceleration can be given as follows:
[tex]a=r\alpha\\a=(\frac{diameter}{2})(8.86\ rad/s^2)\\\\a=(\frac{0.12\ m}{2})(8.86\ rad/s^2)\\\\[/tex]
a = 0.532 m/s²
Learn more about the angular motion here:
brainly.com/question/14979994?referrer=searchResults
The attached picture shows the angular equations of motion.
A ball is thrown upward from the ground with an initial speed of 19.2 m/s; at the same instant, another ball is dropped from a building 18 m high. After how long will the balls be at the same height above the ground?
Answer:
0.938 seconds
Explanation:
For the ball thrown upwards, we use the formula below to solve it:
[tex]s = ut - \frac{1}{2}gt^2[/tex]
where s = distance moved
u = initial speed = 19.2 m/s
t = time taken
g = acceleration due to gravity = 9.8 [tex]m/s^2[/tex]
Let x be the height at which both balls are level, this means that:
=> [tex]x = 19.2t - 4.9t^2[/tex]________(1)
For the ball dropped downwards, we use the formula below:
[tex]s = ut + \frac{1}{2}gt^2[/tex]
u = 0 m/s
At the point where both balls are level:
s = 18 - x
=> [tex]18 - x = 0 + 4.9t^2[/tex]
=> [tex]x = 18 - 4.9t^2[/tex]__________(2)
Equating both (1) and (2):
[tex]19.2t - 4.9t^2 = 18 - 4.9t^2\\\\=> 19.2t = 18\\\\t = 18/19.2 = 0.938 secs[/tex]
They will be level after 0.938 seconds
