Answer:
0.29[tex]\text{m}\text{s}^{2}[/tex]
Explanation:
Given: Haley is driving down a straight highway at 73 mph. A construction sign warns that the speed limit will drop to 55 mph in 0.50 mi.
To Find: constant acceleration (in [tex]\text{m}\text{s}^{2}[/tex]) that will bring Haley to this lower speed in the distance available
Solution:
we know that,
[tex]1[/tex] mile=[tex]1.609[/tex]km
[tex]1[/tex] km[tex]\setminus[/tex]h = [tex]5\setminus18 m\setminus s[/tex]
Initial Speed of Haley(u)=[tex]73[/tex] mph
[tex]73\times 1.609\times 5\setminus 18[/tex]
[tex]32.63[/tex][tex]\text{m}\setminus\text{s}[/tex]
Final Speed of Haley(v)= [tex]55[/tex] mph
[tex]55\times 1.609\times 5\setminus 18[/tex]
[tex]24.58[/tex][tex]\text{m}\setminus\text{s}[/tex]
The distance to be travelled while lowering the speed(S)=[tex]0.5[/tex] mile
[tex]0.805[/tex] km
[tex]805[/tex] m
according to third equation of motion,
[tex]\text{v}^{2}-\text{u}^{2}=2\text{a}\text{S}[/tex]
as speed is lowering down the acceleration will be in the opposite direction of motion, hence acceleration will be negative, equation will become
[tex]\text{u}^{2} -\text{v}^{2}=2a\text{S}[/tex]
putting values,
[tex]32.63^{2}[/tex]-[tex]24.58^{2}[/tex]=[tex]2a\times805[/tex]
a=[tex]460.55\setminus1610[/tex]
a≅[tex]0.29[/tex][tex]\text{m}\text{s}^{2}[/tex]
the constant acceleration that will drop speed to [tex]55 \text{mph}[/tex] is [tex]0.29[/tex][tex]\text{m}\text{s}^{2}[/tex]
The constant acceleration of the driver is -0.286 m/s²
The given parameters;
initial velocity of the driver, v = 73 mph
final velocity of the driver, v = 55 mph
distance available, d = 0.5 mile
The constant acceleration of the driver is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{55^2 - 73^2}{2(0.5)} \\\\a = -2,304\ mi/h^2[/tex]
The acceleration in m/s²
1 mile = 1609 m
[tex]a = \frac{-2304 \ mi}{hr^2} \times \frac{1609 \ m}{1 \ mile} \times \frac{1 \ hr^2}{(3600)^2 \ s^2} \\\\a = -0.286 \ m/s^2[/tex]
Thus, the constant acceleration of the driver is -0.286 m/s²
Learn more here: https://brainly.com/question/18677887
You are at an amusement park and decide to ride the biggest roller coaster there. The coaster gets to the top of its highest hill and stops there for a dramatic pause. At this point the coaster has what type of energy? A. Electrical B. Nuclear C. Potential D. Kinetic
Answer:
C. Potential energy
Explanation:
At this point the roller coaster is just waiting to be dropped into moving energy, kinetic energy. But at the moment that its at the top, its just potential energy.
At the highest point of the hill, the coaster has potential energy, therefore the correct option is C.
What is mechanical energy?Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total stored energy in the system which is represented by total potential energy.
As total mechanical energy is the sum of all the kinetic as well as potential energy stored in the system.
As the coaster is stopped at the highest point it has zero kinetic energy the only form of energy it contains is potential energy.
Learn more about mechanical energy
brainly.com/question/12319302
#SPJ5
A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. The coefficient of static friction is mu_s, and the acceleration due to gravity is g. Find the total work W done on the car by the external forces.
Answer:
Work done by external force is given as
[tex]Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
Explanation:
As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car
so we will have
[tex]Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
now we have
[tex]W_{gravity} = -mg(Lsin\theta)[/tex]
[tex]W_{friction} = -\mu mgcos(\theta) L[/tex]
so from above equation
[tex]Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
so from above equation work done by external force is given as
[tex]Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/tex]
When electric current is flowing in a metal, the electrons are movinga. at nearly the speed of light.b. at the speed of sound in air.c. at the speed of sound in the metal.d. at none of the above speeds.e. at the speed of light.
Answer:
Option (d)
Explanation:
The electrons in a conductor moves with the drift velocity when the electric current is flowing through the conductor.
The drift velocity is due to the applied electric field across the conductor.
While on a camping trip, a man would like to carry water from the lake to his campsite. He fills two, non‑identical buckets with water and attaches them to a 1.43 m long rod. Since the buckets are not identical, he finds that the rod balances about a point located at a 5.59 cm from its midpoint. Which bucket holds more water?
The bucket that holds more water is the bucket that is 65.9 cm from the balance point.
Principle of momentThe bucket that exerts the greatest force will be determined by applying the pinciple of moment as shown below;
Let the weight of the first bucket = W1Let the weight of the second bucket = W2Let the weight of the rod = WMid point of the rod = 0.715 m|___________1.43 m_________________|
---------------------------------------------------------------
↓ Δ 5.59 cm | ↓
W1 W W2
Take moment about the pivot;
W1(0.715 - 0.0559) = W(0.0559) + W2(0.715 + 0.0559)
W1(0.659) = W(0.0559) + W2(0.771)
Since the weight of the first bucket balances the weight of the rod and theb second bucket, the weight of the first bucket is the greatest.
Thus, the bucket that holds more water is the bucket that is 65.9 cm from the balance point.
Learn more about principle of moment here: https://brainly.com/question/26117248
The bucket that holds more water is on the side of the rod that is closer to the balance point. This is because the weight of the water is causing a greater torque on that side, pulling the balance point towards it.
Explanation:This problem relates to the principle of torque and balance. The total torque on an object is the sum of each individual torque, and when a system is in equilibrium, the total torque is zero. The torque caused by a force is calculated by multiplying the force by the distance it is from the pivot.
In this scenario, the man is balancing a long rod with two non-identical water-filled buckets. The rod is balancing off-center, indicating that the system is unbalanced and more weight is on one side. The bucket that holds more water is on the side of the rod that is closer to the balance point because more weight is pulling down on that side.
Given that the balance point is 5.59 cm from the midpoint of the 1.43m rod, the side that has the balance point nearer to the mid-point is experiencing a higher force - meaning it contains the bucket with more water.
Learn more about Torque here:https://brainly.com/question/25708791
#SPJ3
A substance can absorb heat energy by the process of
Answer:
conduction,convection,radiation
Based on the processes of heat transfer, substance can absorb heat energy by the process of conduction, convection and radiation.
What is heat energy?Heat energy is the energy due to temperature difference between two bodies.
Heat energy always flow from hotter to colder bodies.
The processes of heat transfer are as follows:
conduction convection radiationTherefore, a substance can absorb heat energy by the process of conduction, convection and radiation.
Learn more about heat energy at: https://brainly.com/question/19666326
The approximate distance between Mars and Earth is about 1.34 × 108 miles. How many seconds would it take for TV pictures transmitted from the Viking space vehicle on Mars's surface to reach Earth? Enter your answer in scientific notation.
Answer:
Time taken from transmission = 719.61 seconds or 11.99 minutes
Explanation:
Given data:
The distance between the mars and the earth = 1.34 × 10⁸ miles
Now, converting the distance into meters
1 mile = 1.61 km = 1610 m
thus, we have
1.34 × 10⁸ miles = 1610 × 1.34 × 10⁸ m = 2.1574 × 10¹¹ m
Now, the picture will travel with the speed of light i.e 2.998 × 10⁸ m/s
also,
Time = Distance/speed
thus we have,
Time taken from transmission = (2.1574 × 10¹¹ m)/(2.998 × 10⁸ m/s)
or
Time taken from transmission = 719.61 seconds or (719.61/60 = 11.99 minutes)
The time it takes for TV pictures transmitted from the Viking space vehicle on Mars's surface to reach Earth is approximately 7.17 × 10^2 seconds.
Explanation:To calculate the time it takes for TV pictures transmitted from the Viking space vehicle on Mars's surface to reach Earth, we need to convert the distance between Mars and Earth from miles to meters. The approximate distance between Mars and Earth is 1.34 × 108 miles, which is approximately 2.15 × 1011 meters. The speed of light is approximately 3 × 108 meters per second. Dividing the distance by the speed of light, we can calculate the time it takes: 2.15 × 1011 meters / (3 × 108 meters per second) = 7.17 × 102 seconds.
Learn more about Time taken for TV pictures to travel from Mars to Earth here:https://brainly.com/question/12999483
#SPJ3
The electronegativity of nonmetals is relatively __________ as compared to the electronegativity of metals.
We have that Metals are elements that mostly produce ions of Positive charges eg [tex]Na^{2+}[/tex]
While Non metals tends to have negatively charged ions as like [tex]O^-[/tex]
We have that The electro negativity of nonmetals is relatively High as compared to the electro negativity of metals which is Low
From the question we are told
The electro negativity of nonmetals is relatively __________ as compared to the electro negativity of metals.
Generally
Metals are elements that mostly produce ions of Positive charges eg [tex]Na^{2+}[/tex]
While Non metals tends to have negatively charged ions as like [tex]O^-[/tex]
Therefore
From the definition above
Metals are elements that mostly produce ions of Positive charges eg [tex]Na^{2+}[/tex]
While Non metals tends to have negatively charged ions as like [tex]O^-[/tex]
We have that The electro negativity of nonmetals is relatively High as compared to the electro negativity of metals which is Low
For more information on this visit
https://brainly.com/question/17756498
A source emits a sound and is represented by the red dot in this map. Four people are located around the source, and the circles represent the sound wave.
As the source moves west, which person hears the highest pitch?
-person A
-person B (not correct)
-person C
-person D
Answer:
person A
Explanation:
because the sound is moving twords them
In the context of the Doppler Effect, person A will hear the highest pitch if they are located to the west of the moving sound source. The correct answer is a.
Explanation:As the source of the sound moves west, the observer who would hear the highest pitch is the one located to the west of the source, i.e., in the direction of the source's movement.
Since the source is moving towards the west, according to the Doppler Effect, the sound waves are compressed, and the wavelength is reduced in the direction of motion (west). This results in a higher frequency, which corresponds to a higher pitch, being heard by the observer in that direction.
If person A is to the west of the source, then person A will hear the highest pitch. It's important to note that person B has been indicated as not correct, hence, if person A is in the correct position relative to the westward movement of the source, person A will experience the higher pitch sound as compared to the others positioned differently.
Which of the following most logically completes the argument below?
Although the number of large artificial satellites orbiting the Earth is small compared to the number of small pieces of debris in orbit, the large satellites interfere more seriously with telescope observations because of the strong reflections they produce. Because many of those large satellites have ceased to function, the proposal has recently been made to eliminate interference from nonfunctioning satellites by exploding them in space. This proposal, however, is ill conceived, since _______.
(A) many nonfunctioning satellites remain in orbit for years
(B) for satellites that have ceased to function, repairing them while they are in orbit would be prohibitively expensive
(C) there are no known previous instances of satellites’ having been exploded on purpose
(D) the only way to make telescope observations without any interference from debris in orbit is to use telescopes launched into extremely high orbits around the Earth
(E) a greatly increased number of small particles in Earth’s orbit would result in a blanket of reflections that would make certain valuable telescope observations impossible
Answer:
(E) a greatly increased number of small particles in Earth’s orbit would result in a blanket of reflections that would make certain valuable telescope observations impossible
Explanation:
The trade is one strong reflection for many weak reflections (and more dangerous near-Earth space travel).
None of the answer choices except the last one has anything to do with the effect of exploding a satellite. When you are arguing that exploding a satellite is ill conceived, you need to address specifically the effects of exploding the satellite.
Two buses leave a station at the same time and travel in opposite directions. One bus travels 16 km/h slower than the other. If the two buses are 848 km apart after 4 hours, what is the rate of each bus?
Answer:
Speed of one bus is [tex]114 \frac{km}{h}[/tex] and speed of another bus is [tex] 98 \frac{km}{h}[/tex]
Explanation:
Let the speeds of two buses be [tex]v_1\, and\, v_2[/tex] respectively
Since speed of one bus is 16 km/h slower than the other
Therefore [tex]v_1= v_2+16\frac{km}{h}[/tex]
Since buses are traveling in opposite directions so relative speed of buses
[tex]v_{rel}=v_1+v_2= v_2+16\frac{km}{h}+v_2=2v_2+16\frac{km}{h}[/tex]
Distance traveled , [tex]d=v_{rel}t[/tex]
=>[tex]848=4(2v_2+16)[/tex]
=>[tex]v_2=98 \frac{km}{h}[/tex]
[tex]\therefore v_1=v_2+16\frac{km}{h}=(98+16) \frac{km}{h}=114 \frac{km}{h}[/tex]
=>[tex]v_1=114 \frac{km}{h}[/tex]
Thus speed of one bus is [tex]114 \frac{km}{h}[/tex] and speed of another bus is [tex] 98 \frac{km}{h}[/tex]
A team of sled dogs starts pulling a sled at the start of a dog-sledding race. The total mass of the sled is
205 kg, including the driver and load, and the net force acting on the sled is 744 N to the north. What is
the sled's acceleration?
Newton's 2nd law: F = m a
Divide each side by m :. a = F/m
Plug in the 2 given numbers:
a = 744N / 205kg
Acceleration = 3.63 m/s^2 north
Using the formula for acceleration (Newton's second law), the sled's acceleration is calculated to be approximately 3.63 m/s² to the north.
Explanation:The question is asking about acceleration, which in Physics is defined as the rate of change of velocity per unit of time. This is a concept in Newton's second law of motion which can be expressed with the formula F = ma. Here, F is the net force on the object, m is the mass of the object, and a is the acceleration of the object.
Therefore, we can solve the given problem by rearranging the formula to a = F/m.
In this scenario, the net force (F) is 744 N and the mass (m) is 205 kg. Substituting the values into the formula we have:
a = F/m = 744 N / 205 kg = approximately 3.63 m/s² (rounded to two decimal places).
Therefore, the sled's acceleration is approximately 3.63 m/s², directed to the north.
Learn more about acceleration here:https://brainly.com/question/11789833
#SPJ11
A battery-operated car utilizes a 12.0 V system. Find the charge (in C) the batteries must be able to move in order to accelerate the 850 kg car from rest to 25.0 m/s, make it climb a 2.30 ✕ 102 m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 4.00 ✕ 102 N force for an hour.
Answer:
The required charge is 3.175 × 10⁶ C
Explanation:
Given:
Potential difference, ΔV = 12.0 V
mass of the car (m) = 850 kg
velocity of the car, v = 25.0 m/s
Height of the hill upto which car traveled, h = 2.3 × 10² m = 230 m
Now, the energy required to accelerate the car from rest to 25 m/s will be
E₁ = kinetic energy of the car = [tex]\frac{1}{2}mv^2[/tex]
on substituting the values we get
E₁ = [tex]\frac{1}{2}850\times 25^2[/tex]
or
E₁ = 2.65 × 10⁵ J
Now, the potential energy gained by the car at the height 'h' will be
E₂ = mgh = 850 × 9.8 × 230 = 1.91 × 10⁶ J
Also, the energy required to make the car travel at the constant speed of 25.0 m/s with the force of 4 × 10² N for an hour will be
W = Force × displacement
now, the displacement will be = velocity × time = 25 m/s × 1hr = 25 × 3600 = 90000 m [as 1 hr = 3600 seconds]
substituting in the above equation, we get
W = 4.0 × 10² N × 90000 m = 36 × 10⁶ J
Now, the electric potential energy (ΔU) is given as:
ΔU = qΔV
where, q is the charge
Now this electric potential energy is required to do all the work in the above cases
thus,
ΔU = E₁ + E₂ + W
or
qΔV = E₁ + E₂ + W
on substituting the values in the above equation, we get
q × 12 = 2.65 × 10⁵ J + 1.91 × 10⁶ J + 36 × 10⁶ J
or
q = (3.81 × 10⁷)/12
or
q = 3.175 × 10⁶ C
Hence, the required charge is 3.175 × 10⁶ C
An artificial heart valve was tested for its ability to function under extreme conditions, to a maximum flow rate of 4.00 x 10-4 m3/s. What speed would this correspond to for an average red blood cell within a blood vessel of cross-sectional area 5.00 x 10-6 m2?
Answer:
80 m/s
Explanation:
Rate of flow of blood, Volume per second = 4 x 10^-4 m^3/s
Area of blood vessel, A = 5 x 10^-6 m^2
So, Volume per second = area x velocity of blood
velocity of blood = (4 x 10^-4) / (5 x 10^-6) = 80 m/s
The speed of blood is 80 m/s.
In a follow-up experiment, two identical gurneys are placed side-by-side on a ramp with their wheels locked to eliminate spinning. Gurney 1 has a dummy placed on it to give it a total mass of 200 kg, while Gurney 2 is loaded with a dummy that makes it only 50 kg overall. If the ramp has a coefficient of friction of μs, which gurney is more likely to slide down the ramp?
Given the same ramp angle and coefficient of friction, the lighter gurney (50kg) is more likely to slide down the ramp since it requires less friction force to stay stationary as compared to the heavier gurney (200kg).
Explanation:The likelihood of the gurneys sliding down the slope depends upon the balance of forces on each. The force due to gravity on each gurney is mg sin θ, where m is mass, g is acceleration due to gravity, and θ is the angle of the ramp. The force of static friction on each gurney is μs N = μs mg cos θ, where μs is the coefficient of static friction, and N is the normal force, which equals mg cos θ.
In order for the gurneys not to slide, the friction force (μs mg cos θ) must be equal to or greater than the force due to gravity (mg sin θ). For a larger mass (like gurney 1 with the 200 kg dummy), the friction force is greater, so it is more likely to stay put. On the other hand, the smaller mass (gurney 2 with the 50 kg dummy) suffers less friction force due to its lesser weight. Therefore, as long as the angle of the ramp and the coefficient of friction are the same for both gurneys, gurney 2 (50 kg) is more likely to slide down the ramp.
Learn more about Friction here:https://brainly.com/question/13000653
#SPJ12
A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum displacement from equilibrium of 0.204 m. (a) What is the spring constant? N/m (b) What is the kinetic energy of the system at the equilibrium point? J (c) If the maximum speed of the block is 3.45 m/s, what is its mass? kg (d) What is the speed of the block when its displacement is 0.160 m? m/s (e) Find the kinetic energy of the block at x = 0.160 m. J (f) Find the potential energy stored in the spring when x = 0.160 m. J (g) Suppose the same system is released from rest at x = 0.204 m on a rough surface so that it loses 12.2 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant? m
(a) 2446 N/m
When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:
[tex]E=U=\frac{1}{2}kA^2[/tex]
where
U is the elastic potential energy
k is the spring constant
A is the maximum displacement (the amplitude)
Here we have
U = E = 50.9 J
A = 0.204 m
Substituting and solving the formula for k,
[tex]k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m[/tex]
(b) 50.9 J
The total mechanical energy of the system at any time during the motion is given by:
E = K + U
where
K is the kinetic energy
U is the elastic potential energy
We know that the total mechanical energy is constant: E = 50.9 J
We also know that at the equilibrium point, the elastic potential energy is zero:
[tex]U=\frac{1}{2}kx^2=0[/tex] because x (the displacement) is zero
Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:
[tex]K=E=50.9 J[/tex]
(c) 8.55 kg
The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when
K = 50.9 J (at the equilibrium position)
Kinetic energy can be written as
[tex]K=\frac{1}{2}mv^2[/tex]
where m is the mass
Solving the equation for m, we find the mass:
[tex]m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg[/tex]
(d) 2.14 m/s
When the displacement is
x = 0.160 m
The elastic potential energy is
[tex]U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J[/tex]
So the kinetic energy is
[tex]K=E-U=50.9 J-31.3 J=19.6 J[/tex]
And so we can find the speed through the formula of the kinetic energy:
[tex]K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s[/tex]
(e) 19.6 J
The elastic potential energy when the displacement is x = 0.160 m is given by
[tex]U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J[/tex]
And since the total mechanical energy E is constant:
E = 50.9 J
the kinetic energy of the block at this point is
[tex]K=E-U=50.9 J-31.3 J=19.6 J[/tex]
(f) 31.3 J
The elastic potential energy stored in the spring at any time is
[tex]U=\frac{1}{2}kx^2[/tex]
where
k = 2446 N/m is the spring constant
x is the displacement
Substituting
x = 0.160 m
we find the elastic potential energy:
[tex]U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J[/tex]
(g) x = 0
The postion at that instant is x = 0, since it is given that at that instant the system passes the equilibrium position, which is zero.
The response provides calculations for determining the spring constant, the kinetic energy at the equilibrium point, and the mass of the block. It explains the relevant formulas and step-by-step solutions for each part of the question.
Spring Constant Calculation: Using the formula for total mechanical energy, E = [tex]1/2kA^2[/tex], where A is the maximum displacement, the spring constant is calculated as k = [tex]2E / A^2[/tex]. Substituting the given values, k = 2 * 50.9 / [tex](0.204)^2[/tex] = 500 N/m.
Kinetic Energy at Equilibrium: At the equilibrium point, all energy is in the form of potential energy; hence, the kinetic energy is zero.
Mass Calculation: Using the formula for maximum speed vmax = sqrt(E / m), where E is the total mechanical energy and m is the mass, the mass is calculated as m = E / [tex]vmax^2[/tex]. Substituting the values, m = 50.9 / [tex](3.45)^2[/tex]= 4.82 kg.
The mass of a string is 5.5 × 10-3 kg, and it is stretched so that the tension in it is 230 N. A transverse wave traveling on this string has a frequency of 160 Hz and a wavelength of 0.66 m. What is the length of the string?
Answer:
The length of the string is 0.266 meters.
Explanation:
It is given that,
Mass of the string, [tex]m=5.5\times 10^{-3}\ kg[/tex]
Tension in the string, T = 230 N
Frequency of wave, f = 160 Hz
Wavelength of the wave, [tex]\lambda=0.66\ m[/tex]
We need to find the length of the string. Let l is the length of the string. The speed of a transverse wave is given by :
[tex]v=\sqrt{\dfrac{T}{M}}[/tex]
M is the mass per unit length, M = m/l
[tex]v=\sqrt{\dfrac{lT}{m}}[/tex]
[tex]l=\dfrac{v^2m}{T}[/tex]
The velocity of a wave is, [tex]v=\nu\times \lambda[/tex]
[tex]l=\dfrac{(\nu\times \lambda)^2m}{T}[/tex]
[tex]l=\dfrac{(160\ Hz\times 0.66\ m)^2\times 5.5\times 10^{-3}\ kg}{230\ N}[/tex]
l = 0.266 meters
So, the length of the string is 0.266 meters. Hence, this is the required solution.
Answer:
L = 0.275 m
Explanation:
velocity of transverse wave in a stretched string is given as
[tex]v =\sqrt \frac{T}{\mu}[/tex]
where T = tension = 230N
μ = linear density
[tex]μ = \frac[m}{L}[/tex]
where length L is in meters
Velocity = [tex]n\lambda[/tex]
so we have after equating both value of velocity
[tex]\sqrt \frac{T}{\mu} = n\lambda[/tex]
[tex]\frac{T}{\mu} =(n\lambda)^{2}[/tex]
μ = [tex]\frac{T}{(n\lambda)^{2}}[/tex]
μ = [tex] \frac{230}{(160*0.66)^{2}}[/tex]
μ = 0.020 kg/m
but μ = [tex]\frac[m}{L}[/tex]
so length of string is
L = [tex]\frac{5.5*10^{-3}}{0.020}[/tex]
L = 0.275 m
A block of mass m attached to a horizontally mounted spring with spring constant k undergoes simple harmonic motion on a frictionless surface. How would the maximum speed of the block be affected if the spring constant was increased by a factor of 4 while holding the amplitude of oscillation constant?
Answer:
The maximum speed would double
Explanation:
The maximum speed of a mass-spring system is given by:
[tex]v=\omega A[/tex]
where
[tex]\omega[/tex] is the angular frequency
A is the amplitude of the motion
The angular frequency in a spring-mass system is
[tex]\omega=\sqrt{\frac{k}{m}}[/tex]
where
k is the spring constant
m is the mass
Substituting inside the first equation,
[tex]v=\sqrt{\frac{k}{m}} A[/tex]
In this problem:
- the spring constant is increased by a factor 4: k' = 4 k
- the amplitude remains constant: A' = A
So the new maximum speed would be
[tex]v'=\sqrt{\frac{4k}{m}} A= \sqrt{4} (\sqrt{\frac{k}{m}}A)= 2v[/tex]
So, the maximum speed would double.
The maximum speed of the block in simple harmonic motion is not affected by changes in the spring constant when the amplitude of oscillation is held constant.
Explanation:The maximum speed of the block in simple harmonic motion is not affected by changes in the spring constant when the amplitude of oscillation is held constant.
The maximum speed of the block is determined by the amplitude of oscillation. As long as the amplitude remains the same, increasing the spring constant by a factor of 4 will not change the maximum speed.
Therefore, the maximum speed of the block would not be affected if the spring constant was increased by a factor of 4 while holding the amplitude of oscillation constant.
Learn more about maximum speed of the block here:https://brainly.com/question/9906458
#SPJ3
A grinding wheel is in the form of a uniform solid disk of radius 7.00 cm and mass 2.00 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.600 N,m that the motor exerts on the wheel. (a) How long does the wheel take to reach its final operating speed of 1 200 rev/min? (b) Through how many revolutions does it turn while accelerating?
Answer:
Part a)
[tex]t = 1.03 s[/tex]
Part b)
[tex]N = 10.3 rev[/tex]
Explanation:
Mass of the disc is given as
m = 2.00 kg
radius of the disc is given as
r = 7 cm
so moment of inertia of the disc is given as
[tex]I = \frac{1}{2}mr^2[/tex]
[tex]I = \frac{1}{2}(2)(0.07)^2[/tex]
[tex]I = 4.9 \times 10^{-3} kg m^2[/tex]
Now given that torque on the disc is
[tex]\tau = 0.600 Nm[/tex]
so here the angular acceleration is given as
[tex]\alpha = \frac{\tau}{I}[/tex]
[tex]\alpha = \frac{0.600}{4.9 \times 10^{-3}}[/tex]
[tex]\alpha = 122.45 rad/s^2[/tex]
Part a)
if disc start from rest and then achieve final speed as 1200 rpm then
[tex]f = 1200/60 = 20 rev/s[/tex]
so final speed is
[tex]\omega = 2\pi(20) = 40\pi rad/s[/tex]
now the time taken to reach this speed is given as
[tex]\alpha t = \omega[/tex]
[tex](122.45) t = 40\pi[/tex]
[tex]t = 1.03 s[/tex]
Part b)
Number of revolution in the same time is given as
[tex]N = \frac{\omega_o + \omega}{4\pi} t[/tex]
[tex]N = \frac{40\pi + 0}{4\pi}(1.03) = 10.3 rev[/tex]
A charged particle that is moving in a static uniform magnetic fieldA) may experience a magnetic foce, but its speed will not changeB) may experience a magnetic foce, but its direction of motion will not changeC) may experience a magnetic force which will cause its speed to changeD) will always experience a magnetic force, regardless of its direction of motionE) none of the above statements are true
Answer:
A) may experience a magnetic foce, but its speed will not change
Explanation:
A charged particle moving in an uniform magnetic field experiences a force given by:
[tex]F=qvB sin \theta[/tex]
where
q is the charge of the particle
v is the velocity of the particle
B is the magnitude of the magnetic field
[tex]\theta[/tex] is the angle between the direction of v and B
From the formula we notice that:
- if the particle is moving parallel to the field, [tex]\theta=0[/tex], so force experienced by the particle is zero --> so we can exclude option D)
Moreover, the magnetic force experienced by the particle is perpendicular to the direction of motion. This means that:
- The direction of motion of the particle will change --> so we can exclude option B)
In addition, we can notice that since the magnetic force is perpendicular to the direction of motion of the particle, the work done by the force on the particle is zero: this means that the kinetic energy of the particle does not change, so the speed of the particle does not change. So we can exclude also option C, therefore the only correct option is
A) may experience a magnetic foce, but its speed will not change
A wire of resistivity ρ must be replaced in a circuit by a wire of the same material but four times as long. If, however, the total resistance is to remain as before, the diameter of the new wire mustA- be four times the original diameter.B- be one-fourth the original diameter.C- be the same as the original diameter.D- be one-half the original diameter.E- be two times the original diameter.
Answer:
E. two times the original diameter
Explanation:
Resistance of a wire is:
R = ρ L/A
where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area.
For a round wire with diameter d:
R = ρ L / (¼ π d²)
The two wires must have the same resistance, so:
ρ₁ L₁ / (¼ π d₁²) = ρ₂ L₂ / (¼ π d₂²)
The wires are made of the same material, so ρ₁ = ρ₂:
L₁ / (¼ π d₁²) = L₂ / (¼ π d₂²)
The new length is four times the old, so 4 L₁ = L₂:
L₁ / (¼ π d₁²) = 4 L₁ / (¼ π d₂²)
1 / (¼ π d₁²) = 4 / (¼ π d₂²)
Solving:
1 / (d₁²) = 4 / (d₂²)
(d₂²) / (d₁²) = 4
(d₂ / d₁)² = 4
d₂ / d₁ = 2
So the new wire must have a diameter twice as large as the old wire.
A projectile is defined as: (A) any object that is in flight.(B) any object that moves through air or space solely under the influence of gravity.(C) any object that applies an upward force to oppose gravity as it travels through air or space.
Answer:
B. Any object that moves through air or space solely under the influence of gravity.
Explanation:
Once it is dropped it continues in motion by it's own inertia and is influenced only by the downward force of gravity.
Final answer:
The correct definition of a projectile is any object that moves through air or space solely under the influence of gravity. Projectile motion involves an object launched into the air, only affected by gravity, with its horizontal and vertical motions being independent. A projectile's path is termed its trajectory, with its peak called the maximum height, and the distance covered called the range.
Explanation:
A projectile is best defined as (B) any object that moves through air or space solely under the influence of gravity. This definition eliminates any objects that are powered by engines or propellers after launch, such as rockets, and instead includes objects like balls thrown in sports, bullets fired from a gun, or rocks thrown into the air.
Once a projectile is in motion, it is subject to what is known as projectile motion. To clarify, projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The initial force that causes the projectile to move is disregarded after launch, and the only force acting on the projectile is the force of gravity pulling it downwards.
Properties of projectile motion are important when studying the physics behind these objects. When a projectile is in flight, its horizontal and vertical motions are independent of one another. This means that the projectile's horizontal velocity remains constant (ignoring air resistance), while the vertical velocity changes due to the acceleration of gravity. The trajectory of a projectile is its path through the air, and is typically a curved shape due to the influence of gravity. The maximum height reached by the projectile is the highest point in its trajectory, and the range is the maximum horizontal distance it covers.
A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples with a 50,000-kg second car, initially at rest and with brakes released. What is the speed of the two cars after coupling?
Answer:
The speed of the two cars after coupling is 0.46 m/s.
Explanation:
It is given that,
Mass of car 1, m₁ = 15,000 kg
Mass of car 2, m₂ = 50,000 kg
Speed of car 1, u₁ = 2 m/s
Initial speed of car 2, u₂ = 0
Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
[tex]V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}[/tex]
V = 0.46 m/s
So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.
The speed of the two cars after coupling is 0.857 m/s.
Explanation:To find the speed of the two cars after coupling, we can apply the law of conservation of momentum. The momentum of both cars before the collision is equal to the momentum of both cars after the collision. Since the second car is initially at rest, its initial momentum is zero. The momentum of the first car is given by its mass times its speed. By setting up an equation with the initial momentum of the first car and the final momentum of both cars combined, we can solve for the final speed of the two cars after coupling.
Using the equation:
mass1 * v1i = (mass1 + mass2) * vf
Substituting the given values:
(15,000 kg) * (2.0 m/s) = (15,000 kg + 50,000 kg) * vf
Simplifying the equation gives:
vf = 0.857 m/s
Learn more about collision here:https://brainly.com/question/30636941
#SPJ3
A conducting bar rests on two parallel horizontal rails 50 cm apart forming a square area. The resistance of the bar and the rails is assumed to be constant and equal to 0.05 Ω. A uniform magnetic field 0.5 T is perpendicular to the plane of the rails. The value of the field is uniformly reduced to zero in a time of 0.1 s. What is the maximum value of the force acting on the bar during that time?
Answer:
Force = 6.25 N
Explanation:
Given:
Distance between the rails, l = 50cm = 0.5m
since it is forming a square area. thus, area = 0.5² m²
Resistance of the bar = 0.05Ω
Magnetic field, B = 0.5T
Time = 0.1s
Now,
The value for EMF is given as
EMF = (ΔArea x Field) /time
substituting the values in the above equation we get
EMF = (0.5² x 0.5) /0.1 = 1.25 Volts
now,
Current, i = EMF/ Resistance
substituting the values in the above equation we get
i = 1.25/0.05 = 25 A
Now the force is given as:
Force = i x l x B
Force = 25 x 0.5 x 0.5
Force = 6.25 N
A rectangular field is 300 meters long and 300 meters wide. What is the area of the field in square kilometers? Do not round your answer. Be sure to include the correct unit in your answer.
Answer:
Area of the rectangular field in kilometers is 0.09 [tex]km^2[/tex]
Explanation:
We know that 1 kilometers = 1000 meters
since we need to find the area in unit of kilometers
therefore converting length and width into kilometers
1000 meters = 1 kilometers
300 meters =[tex]\frac{300}{1000} = 0.3[/tex]
Likewise width = 0.3 km
Area = length x width
= 0.3km x 0.3 km
= 0.09 [tex]km^2[/tex]
At 1 atm pressure, the heat of sublimation of gallium is 277 kJ/mol and the heat of vaporization is 271 kJ/mol. To the correct number of significant figures, how much heat is required to melt 2.50 mol of gallium at 1 atm pressure?
Answer:15 KJ
Explanation:
We have given
at 1 atm
The heat of sublimation of Gallium =277 KJ/mol
heat of vaporization of Gallium =271 KJ/mol
[tex]\Delta H=277-271=6KJ/mol[/tex]
So heat required to melt 2.50 mol of gallium at 1 atm
Heat Required=[tex]2.5\times 6=15 KJ[/tex]
So heat required to melt 2.5 mol of gallium is 15 KJ
Final answer:
The heat required to melt 2.50 mol of gallium at 1 atm pressure is 692.5 kJ.
Explanation:
To find the heat required to melt 2.50 mol of gallium, we can use the heat of fusion (also known as the heat of sublimation) of gallium. The heat of fusion is the energy required to change a substance from a solid to a liquid at its melting point. In this case, we are given that the heat of sublimation of gallium is 277 kJ/mol.
Since gallium sublimes at 1 atm pressure, it bypasses the liquid phase and turns directly from a solid to a gas. Therefore, the heat required to melt 2.50 mol of gallium is the same as the heat of sublimation.
Therefore, to the correct number of significant figures, the heat required to melt 2.50 mol of gallium at 1 atm pressure is 2.50 mol x 277 kJ/mol = 692.5 kJ.
Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.0 kg⋅m2 and for arms and legs in is 0.90 kg⋅m2 . If she starts out spinning at 5.2 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?
Answer:
Her angular speed (in rev/s) when her arms and one leg open outward is [tex]1.56\frac{rev}{s}[/tex]
Explanation:
Initial moment of inertia when arms and legs in is [tex]I_i=0.90 kg.m^{2}[/tex]
Final moment of inertia when her arms and on leg open outward, [tex]I_f=3.0 kg.m^{2}[/tex]
Initial angular speed [tex]w_i=5.2\frac{rev}{s}[/tex]
Let the final angular speed be [tex]w_f[/tex]
Since external torque on her is zero so we can apply conservation of angular momentum
[tex]\therefore L_f=L_i[/tex]
=>[tex]I_fw_f=I_iw_i[/tex]
=>[tex]w_f=\frac{I_iw_i}{I_f}=\frac{0.9\times5.2 }{3.0}\frac{rev}{s}=1.56\frac{rev}{s}[/tex]
Thus her angular speed (in rev/s) when her arms and one leg open outward is [tex]1.56\frac{rev}{s}[/tex]
The power (P) required to run a motor is equal to the voltage (E) applied to that motor times the current (I) supplied to the motor. If the motor data says the motor uses 180 watts of power and the voltage applied to the motor is 120 volts, how much current will the motor require?
Final answer:
The motor will require a current of 1.5 amperes to operate at 180 watts with an applied voltage of 120 volts, based on the formula for electric power P = IV.
Explanation:
To calculate the current required by the motor, we will use the formula for electric power, which is P = IV, where P is the power in watts, I is the current in amperes, and V is the voltage in volts.
Given:
The power consumed by the motor (P) = 180 watts
The voltage applied to the motor (V) = 120 volts
We need to solve for I (current):
I = P / V
I = 180 W / 120 V
I = 1.5 A
Therefore, the motor will require 1.5 amperes of current when operating at 180 watts of power with a voltage of 120 volts.
If my cylinder of air lasts 60 minutes while I am at the surface breathing normally, assuming all else is the same, how long will it last at 20 metres/66 feet breathing normally? 60 minutes
Answer:
Explanation:
Let the volume of air be V. at atmospheric pressure, that is 10⁵ Pa
At 20 m below surface pressure will be
atmospheric pressure + hdg
10⁵ + 20 x 9.8 x 1000 = 2.96 x 10⁵Pa
At this pressure volume V becomes V/ 2.96
This volume will last 1/2.96 times time that is 60/2.96 = 20.27 minutes.
Since air is a mixture, it does not have a "molar mass." However, for calculation purposes, it is possible to speak of its "effective molar mass." (An effective molar mass is a weighted average of the molar masses of a mixture's components.) If air at STP has a density of 1.285 g/L, its effective molar mass is ________ g/mol.
Answer:
The effective molar mass of air at STP is 28.82 g/mol.
Explanation:
At STP, the value of pressure is 1 atm.
At STP, the temperature is equal to 273.15 K
P = 1atm, T = 273.15 K
Density of the gas at STP ,d= 1.285 g/L
[tex]PV=nRT[/tex] (Ideal gas equation)
[tex]PV=\frac{\text{Mass of air}}{\text{Molar mass of air(M)}}RT[/tex]
[tex]Density=\frac{Mass}{Volume}[/tex]
[tex]P\times M=d\times RT[/tex]
[tex]M=\frac{1.285 g/L}{1 atm}\times 0.0821 atm L/mol K\times 273.15 K[/tex]
M = 28.81 g/mol
The effective molar mass of air at STP is 28.82 g/mol.
Final answer:
The effective molar mass of air at standard temperature and pressure (STP), given its density of 1.285 g/L, is approximately 28.8 g/mol. This value is a weighted average based on the major components of air, mainly nitrogen and oxygen.
Explanation:
If air at STP has a density of 1.285 g/L, its effective molar mass is determined using the density and the ideal gas law. Given in the discussion, the effective molar mass of dry air is approximately 28.8 g/mol. This effective molar mass is a weighted average of the major components of air, primarily nitrogen (N₂) and oxygen (O₂), which account for about 80% and 20% of air's composition, respectively. At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. Therefore, the molar mass can be related to the density of the gas at these conditions to find its effective molar mass.
Using the information provided and considering the average composition of air, the calculation confirms that the effective molar mass of air at STP, considering its density of 1.285 g/L, is closely approximated to be 28.8 g/mol. This value is essential for various calculations in chemistry and environmental science, including converting between the mass of air and the number of moles of air, which is useful in stoichiometric calculations and understanding the behavior of gases under different conditions.
A climber using bottled oxygen accidentally drops the oxygen bottle from an altitude of 4500 m. If the bottle fell straight down this entire distance, what is the velocity of the 3-kg bottle just prior to impact at sea level? (Note: ignore air resistance)
Answer:
300 m/s
Explanation:
As the cylinder drops off so its initial velocity is zero.
h = 4500 m, g = 10 m/s^2, u = 0
Use third equation of motion
v^2 = u^2 + 2 g h
v^2 = 0 + 2 x 10 x 4500
v^2 = 90000
v = 300 m /s
Final answer:
The velocity of the oxygen bottle just before impact at sea level, having fallen straight down from an altitude of 4500 m ignoring air resistance, is approximately 297.4 m/s.
Explanation:
To calculate the velocity of the oxygen bottle just before impact, we can use the principles of physics, specifically the conservation of energy or kinematics under the influence of gravity. Assuming air resistance is negligible, all potential energy (PE) of the bottle at the altitude of 4500 m will be converted into kinetic energy (KE) just before impact.
The potential energy (PE) at the height is given by PE = m*g*h, where m is mass, g is the acceleration due to gravity (approximately 9.81 m/s2), and h is the height from which it falls. The kinetic energy (KE) just before impact is given by KE = 0.5 * m * v2, where v is the velocity.
Setting PE equal to KE, we have m*g*h = 0.5 * m * v2. Solving for v (and noting that the mass cancels out), we find v = sqrt(2*g*h). Plugging in the values, we get v = sqrt(2*9.81*4500), which calculates to a velocity of approximately 297.4 m/s.