Answer:
a) 6.04 m/s
b) 11.02 m/s
Explanation:
a) Let the father mass be M, and his speed be V. His son mass is m = 3M/5. Since his kinetic energy initially is half of after he increases his speed by 2.5m/s
[tex]E_2 = 2E_1[/tex]
[tex]\frac{M(V+2.5)^2}{2} = 2\frac{MV^2}{2}[/tex]
[tex]V^2 + 5V + 6.25 = 2V^2[/tex]
[tex]V^2 - 5V - 6.25 = 0[/tex]
[tex]V \approx 6.04m/s[/tex]
b) The son kinetic energy initially is:
[tex]E_s = 2E_1 = 2\frac{MV^2}{2} = MV^2 = M*6.04^2 = 36.43M J[/tex]
We can solve for the son speed by the following formula
[tex]E_s = \frac{mv^2}{2}[/tex]
[tex]v^2 = \frac{2E_s}{m} = \frac{2*36.43M}{3M/5} = \frac{10*36.43}{3} = 121.4m/s[/tex]
[tex]v = \sqrt{121.4} = 11.02 m/s[/tex]
Planetary orbits around a star can be modeled with the following potentialU(r) =ar+br2(1)(a) Show that the equilibrium position for this potential is equal tore= 2b=a.(b) Use the Taylor expansion on the potential about the equilibrium position to show thatthe \spring" constant of small oscillations around this equilibrium position isa4=8b3
Answer:
a) r eq = -a/(2b)
b) k = a/r eq = -2b
Explanation:
since
U(r) = ar + br²
a) the equilibrium position dU/dr = 0
U(r) = a + 2br = 0 → r eq= -a/2b
b) the Taylor expansion around the equilibrium position is
U(r) = U(r eq) + ∑ Un(r eq) (r- r eq)^n / n!
,where Un(a) is the nth derivative of U respect with r , evaluated in a
Since the 3rd and higher order derivatives are =0 , we can expand until the second derivative
U(r) = U(r eq) + dU/dr(r eq) (r- r eq) + d²U/dr²(r eq) (r- r eq)² /2
since dU/dr(r eq)=0
U(r) = U(r eq) + d²U/dr²(r eq) (r- r eq)² /2
comparing with an energy balance of a spring around its equilibrium position
U(r) - U(r eq) = 1/2 k (r-r eq)² → U(r) = U(r eq) + 1/2 k (r-r eq)²
therefore we can conclude
k = d²U/dr²(r eq) = -2b , and since r eq = -a/2b → -2b=a/r eq
thus
k= a/r eq
A and B, move toward one another. Object A has twice the mass and half the speed of object B. Which of the following describes the forces the objects exert on each other when the collide and provides the best explanation? (A) The force exerted by A on B will be twice as great as the force exerted by B on A, as great as the force exerted by B on A other are the same, because the product of because A has twice the mass of B (B) The force exerted by A on B will be half because A has half the speed of B (C) The forces exerted by each object on the mass and speed is the same for both objects. (D) The forces exerted by each object on the other are the same, because inter objects cannot exert forces of difi magnitude on each other.
Answer:D
Explanation:
Given
mass of A is twice the mass of B half the velocity of B
Suppose [tex]F_a[/tex] and [tex]F_b[/tex] be the average force exerted on A and B respectively
and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.
Thus [tex]F_a=-F_b[/tex] and option d is correct
According to Newton's third law of motion, the forces exerted by each object on the other are the same.
Explanation:The correct answer is (C) The forces exerted by each object on the other are the same because inter objects cannot exert forces of different magnitudes on each other.
According to Newton's third law of motion, whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force it exerts. This means that the forces exerted by object A on object B and by object B on object A are equal in magnitude.
Here, object A has twice the mass of object B, but because both objects have the same speed before the collision, the forces exerted on each other will be the same.
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Plane a flies at a constant speed from new york to los angeles along a route which is 2000 miles. Plane b flies in the opposite direction at a constant speed which is 100 mph faster than plane a. Plane b takes off one hour after plane a. They land at the same moment. How far are they from los angeles when they pass?
Answer:
from los angeles distance plan a = 1111.08 mi
from los angeles distance plan b = 888.92 mi
Explanation:
given data
new york to los angeles distance = 2000 miles
Plane b speed = 100 mph faster than plane a
Plane b takes off time = 1 hour after plane a
to find out
How far are they from los angeles when they pass
solution
we consider speed of plan a is = x mph
so speed of plan b will be = x + 100 mph
and we know plan b take here 1 hour less time than plan a so it mean time is distance divide speed i.e
[tex]\frac{2000}{x} - 1 =\frac{2000}{x+100}[/tex]
solve it we get x = 400 mph
it mean here
plan a speed is 400 mph
and plan b speed is 500 mph
and
now we consider they meet at time = t hour after a take off
then plan a travel = 400 t
and plan b travel = 500 ( t - 1 )
add both distance that is equal to 2000 mi
so 400 t + 500 ( t -1 ) = 2000
400 t + 500 ( t -1 ) = 2000
400 t + 500 ( t-1) = 2000
solve we get
t = 2.777
so total distance travel plan a = 400 × 2.777 = 1111.08 mi
total distance travel plan b = 2000 - 1111.08
total distance travel plan b = 888.92 mi
The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?
A. 10.36 m/s
B. 3960 m/s
C. 219.3 m
D. 0.0965 m/s
Answer:
A
Explanation:
v = change of X / change of T
v = 200/19.3
Answer:
A. 10.36 m/s
Explanation:
The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?
A. 10.36 m/s
B. 3960 m/s
C. 219.3 m
D. 0.0965 m/s
speed is the change in distance per time
speed is scalar quantity and hence as no direction but only magnitude. time and distance are also scalar quantity
200/19.3
speed=10.36m/s
A steel aircraft carrier is 370 m long when moving through the icy North Atlantic at a temperature of 2.0 °C. By how much does the carrier lengthen when it is traveling in the warm Mediterranean Sea at a temperature of 21 °C?
Answer:
The carrier lengthen is 0.08436 m.
Explanation:
Given that,
Length = 370 m
Initial temperature = 2.0°C
Final temperature = 21°C
We need to calculate the change temperature
Using formula of change of temperature
[tex]\Delta T=T_{f}-T_{i}[/tex]
[tex]\Delta T=21-2.0[/tex]
[tex]\Delta T=19^{\circ}C[/tex]
We need to calculate the carrier lengthen
Using formula of length
[tex]\Delta L=\alpha_{steel}\times L_{0}\times\Delta T[/tex]
Put the value into the formula
[tex]\Delta L=1.2\times10^{-5}\times370\times19[/tex]
[tex]\Delta L=0.08436\ m[/tex]
Hence, The carrier lengthen is 0.08436 m.
A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of the U-shaped tube until the vertical height of the water column is hW = 19.0 cm .
The gauge pressure at the water/mercury interface is simply the "head" due to the water:
p = ρgh = 1000kg/m³ * 9.8m/s² * 0.19m = 1862 Pa
If we take the specific gravity of mercury to be 13.6, then the difference in height between the water and mercury columns is
h' = h(1 - 1/s.g.) = 19cm * (1 - 1/13.6) = 17.6 cm
A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50.00 cm mark. (a) The period of oscillation is 2.50 s. Find d. (b) If you moved the pivot 5.00 cm closer to the 50.00 cm mark, what would the period of oscillation be
Answer:
(a). The value of d is 0.056 cm and 1.496 cm.
(b). The time period is 1.35 sec.
Explanation:
Given that,
Length = 50.00 cm
Time period = 2.50 s
Time period of pendulum is defined as the time for one complete cycle.
The period depends on the length of the pendulum.
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{I}{mgh}}[/tex]
Where, I = moment of inertia
We need to calculate the value of d
Using parallel theorem of moment of inertia
[tex]I=I_{cm}+md^2[/tex]
For a meter stick mass m , the rotational inertia about it's center of mass
[tex]I_{cm}-\dfrac{mL^2}{12}[/tex]
Where, L = 1 m
Put the value into the formula of time period
[tex]T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}[/tex]
[tex]T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}[/tex]
[tex]T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})[/tex]
Multiplying both sides by d
tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]
[tex](\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0[/tex]
Put the value of T, L and g into the formula
[tex]4.028d^2-6.25d+0.336=0[/tex]
[tex]d = 0.056\ m, 1.496\ m[/tex]
The value of d is 0.056 cm and 1.496 cm.
(b). Given that,
L = 50-5 = 45 cm
We need to calculate the time period
Using formula of period
[tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]
Put the value into the formula
[tex]T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}[/tex]
[tex]T=1.35\ sec[/tex]
Hence, (a). The value of d is 0.056 cm and 1.496 cm.
(b). The time period is 1.35 sec.
One end of a string is attached to a ball, with the other end held by a student such that the ball is swung in a horizontal circular path of radius R at a constant tangential speed. At a later time, the tension force exerted on the ball remains constant, but the length of the string is decreased to R4. What is the new tangential speed of the ball?
a. four times the original speed
b. two times the original speed
c. Half the original speed
d. One-fourth the original speed
The new tangential speed of the ball is four times the original speed.
Explanation:When the length of the string is decreased to R/4, the radius of the circular path becomes R/4. The tension force exerted on the ball remains constant because it only depends on the mass of the ball and the gravitational force acting on it. The tangential speed is given by the formula v = ΩR, where Ω is the angular velocity. Since the radius is now 1/4 of the original radius, the new tangential speed is 4 times the original speed (option a).
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A student standing in a canyon yells "echo", and her voice produces a sound wave of frequency of f = 0.61 kHz. The echo takes t = 3.9 s to return to the student. Assume the speed of sound through the atmosphere at this location is v = 322 m/s.
Answer:
[tex]d=627.9\ m[/tex] is the distance from the obstacle of reflection.
wavelength [tex]\lamb=0.5279\ m [/tex]
Explanation:
Given that:
frequency of sound, [tex]f=610\ Hz[/tex]time taken for the echo to be heard, [tex]t=3.9\ s[/tex]speed of sound, [tex]v=322\ m.s^{-1}[/tex]We know,
[tex]\rm distance = speed \times time[/tex]
During an echo the sound travels the same distance back and forth.
[tex]2d=v.t[/tex]
[tex]2d=322\times 3.9[/tex]
[tex]d=627.9\ m[/tex] is the distance from the obstacle of reflection.
Now the wavelength of sound waves:
[tex]\lambda=\frac{v}{f}[/tex]
[tex]\lambda=\frac{322}{610}[/tex]
[tex]\lamb=0.5279\ m [/tex]
Tia needs to produce a solenoid that has an inductance of 2.07 μ H . She constructs the solenoid by uniformly winding 1.19 m of thin wire around a tube. How long, in centimeters, should the tube be?
Answer:
[tex]l=0.068 m[/tex]
Explanation:
given,
inductance of solenoid = 2.07 μ H
winding of the wire = 1.19 m
Using formula of inductance
[tex]L = \dfrac{\mu_0N^2A}{l}[/tex]
L is the inductance
N is number of turns of the coil
μ₀ is permeability of free space
L is length of winding
N (2π r) = 1.19
squaring both side
4π(N²(πr²))=1.19²
N² A = 0.113
now
[tex]2.07 \times 10^{-6}= \dfrac{4\pi\times 10^{-7}\times 0.113}{l}[/tex]
[tex]l= \dfrac{4\pi\times 10^{-7}\times 0.113}{2.07 \times 10^{-6}}[/tex]
[tex]l=0.068 m[/tex]
This question involves the concepts of the inductance of an inductor and the length of the winding.
The tube should be "6.84 cm" long.
The inductance of an inductor can be given by the following formula:
[tex]L=\frac{N^2\mu_oA}{l}[/tex]
where,
L = inductance = 2.07 μH = 2.07 x 10⁻⁶ H
N = No. of turns in coil = [tex]\frac{length\ of\ wire}{circumference\ of\ tube} = \frac{1..19\ m}{2\pi r}[/tex]
μ₀ = permeability of free space = 4π x 10⁻⁷ Wb/A.m
A = Area of tube = πr²
l = length of tube = ?
Therefore,
[tex]2.07\ x\ 10^{-6}\ H = \frac{(\frac{1.19\ m}{2\pi r})^2(\pi r^2)(4\pi\ x\ 10^{-7}\ Wb/A.m)}{l}\\\\l = \frac{1.4161\ x\ 10^{-7}}{2.07\ x\ 10^{-6}}\\\\[/tex]
l = 0.0684 m = 6.84 cm
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The attached picture shows the inductance formula.
Determine the length of a wind instrument, assuming that it is modeled as a closed tube and that the lowest sound that it can play has a frequency of 99 Hz. The speed of sound in air is 343m/s.
Answer:
Length, l = 0.866 meters
Explanation:
Given that,
Frequency of sound, f = 99 Hz
Speed of sound in air, v = 343 m/s
To find,
The length of a wind instrument.
Solution,
The standing wave will gets formed in wind instrument. For the closed tube, the closed tube the frequency is given by :
[tex]f=\dfrac{v}{4l}[/tex]
Where
l is the length of the instrument
[tex]l=\dfrac{v}{4f}[/tex]
[tex]l=\dfrac{343\ m/s}{4\times 99\ Hz}[/tex]
l = 0.866 meters
So, the length of a wind instrument is 0.866 meters. Hence, this is the required solution.
An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its speed, v, and obeys the equation Fdrag=(12.0N⋅s/m)v+(4.00N⋅s2/m2)v2. What is the terminal speed of this object?A. 72.2 m/sB. 12.6 m/sC. 47.3 m/sD. 34.2 m/sE. 6.45 m/s
Answer:
The terminal speed of this object is 12.6 m/s
Explanation:
It is given that,
Mass of the object, m = 80 kg
The magnitude of drag force is,
[tex]F_{drag}=12v+4v^2[/tex]
The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.
[tex]F_{drag}=mg[/tex]
[tex]12v+4v^2=80\times 9.8[/tex]
[tex]4v^2+12v=784[/tex]
On solving the above quadratic equation, we get two values of v as :
v = 12.58 m/s
v = -15.58 m/s (not possible)
So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.
Imagine holding a basketball in both hands, throwing it straight up as high as you can, and then catching it when it falls. At which points in time does a zero net force act on the ball? Ignore air resistance.
Answer:before throwing and after catching the ball
Explanation:
When basketball is in the hand of player net force on it zero as holding force is canceled by gravity Force. During its entire motion gravitational force is acting on the ball which is acting downward. Even at highest point gravity is constantly acting downwards.
After catching the ball net force on it zero as holding force is canceled by gravity force and ball is continue to be in stationary motion.
The point in time where the zero net force acted on the ball is before throwing and after catching the ball
The point at which Net force acted on the ball:At the time of basketball is in the hand of the player, the net force should be zero since holding force is canceled via gravity Force. At the time of overall motion, gravitational force should be acted on the ball which is acting downward. After catching the ball net force on it zero since holding force should be canceled via gravity force and ball should be continue in stationary motion.
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A gust of wind blows an apple from a tree. As the apple falls, the gravita- tional force on the apple is 2.25 N downward, and the force of the wind on the apple is 1.05 N to the right.
Find the magnitude and direction of the net force on the apple.
Answer:
The magnitude of the net force is 2.48294 N and the angle is 25.02° with respect to the vertical
Explanation:
Consider the gravitational force as [tex]\vec{F_g}=2.25\ N[/tex]
Consider the wind force as [tex]\vec{F_w}=1.05\ N[/tex]
The angle between the gravitational force and wind force is 90°
From the parallelogram law we have
[tex]F=\sqrt{F_g^2+F_w^2+2F_gF_wcos\theta}\\\Rightarrow F=\sqrt{2.25^2+1.05^2+2\times 2.25\times 1.05\times cos90}\\\Rightarrow F=2.48294\ N[/tex]
The angle between the resultant and the gravitational force would be
[tex]tan^{-1}\frac{1.05}{2.25}=25.02^{\circ}[/tex]
The magnitude of the net force is 2.48294 N and the angle is 25.02° with respect to the vertical
A box is moved a distance of 10 meters with a force of 5 newtons. The amount of work done is:
Answer:
50 J
Explanation:
Work = force × distance
W = Fd
W = (5 N) (10 m)
W = 50 J
The amount of work done is 50 J (Joule)
How to find the amount of work done?To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd. If the force is being exerted at an angle θ to the displacement, the work done is W = fd cos θ.
By applying the formula, we get:
Work = force × distance
Force = 5 newtons ...(given)
distance = 10 meters ...(given)
⇒ W = Fd
⇒ W = (5 N) (10 m)
⇒ W = 50 J
What is the work done?
Work is done whenever a force moves something over a distance. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.
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How long does it take a 750-W coffeepot to bring to a boil 0.65 L of water initially at 13 ∘C? Assume that the part of the pot which is heated with the water is made of 360 g of aluminum, and that no water boils away. Ignore the heat loss to the surrounding environment. The value of specific heat for water is 4186 J/kg⋅C∘ and for aluminum is 900 J/kg⋅C∘.
Answer:
353 s
Explanation:
Energy = energy needed to raise the temperature of pot to 100°C +
required energy needed to raise the temperature of water to 100°C
= MCΔT + mcΔT
= 0.65 kg * 4186 J/kg-°C * 87 °C + 0.36 kg * 900 J/kg-°C * 87 °C
= 264 906.3 J
Energy required = energy delivered * time
264 906.3 J = 750 J/s * time
time = 353 s
The study of chemicals and bonds is called chemistry. There are two types of elements these rare metals and nonmetals.
The correct answer is [tex]5787*10^{-6}[/tex]
What is rate law?The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters.The balanced reaction is:-
[tex]2H_2O_2(aq) ----> 2H_2O(l) + O_2(g).[/tex]
The rate law of decomposition:-
[tex]rate =-\frac{1}{2}\frac{d[H_2O_2]}{dt}[/tex]
Where d[H₂O₂] is the change in concentration of H₂O₂ (between 0s and 2,16x10⁴s) is (1,000M-0,500M) and dt is (0s-2,16x10^4s). Replacing:
[tex]rate= -\frac{1}{2} \frac{0.5}{2.16*10^4}[/tex]
rate =[tex]1157*10^{-5}[/tex]
As this rate is = d[O₂]/dt(Rate of production of O₂), the rate of production of O₂(g) is 1,157x10⁻⁵M/s
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A jet transport has a weight of 1.87 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel, and the plane's center of gravity is 12.0 m behind the front wheel. Determine the normal force exerted by the ground on (a) the front wheel and on (b) each of the two rear wheels.
Answer:
(a). The normal force exerted by the ground on the front wheel is [tex]4.67\times10^{5}\ N[/tex].
(b). The normal force exerted by the ground on each of the two rear wheels is [tex]7.02\times10^{5}\ N[/tex]
Explanation:
Given that,
Weight of jet [tex]W=1.87\times10^{6}\ N[/tex]
Distance = 16 m
Second distance = 12.0 m
We need to calculate the normal force exerted by the ground on the front wheel
Using formula of torque
[tex]\sum\tau=-Wl_{W}+F_{f}l_{f}=0[/tex]
Where, W = weight of jet
[tex]l_{f}[/tex]=lever arms for the forces [tex]F_{f}[/tex]
[tex]l_{w}[/tex]=lever arms for the forces W
Put the value into the formula
[tex]-(1.87\times10^{6})\times(16.0-12.0)+F_{f}\times16.0=0[/tex]
[tex]F_{f}=\dfrac{(1.87\times10^{6})\times(16.0-12.0)}{16.0}[/tex]
[tex]F_{f}=4.67\times10^{5}\ N[/tex]
(b). We need to calculate the normal force exerted by the ground on each of the two rear wheels
The sum of vertical forces equal to zero.
[tex]\sum F_{y}=F_{f}+2F_{r}-W=0[/tex]
We using 2 for two rear wheels
[tex]\sum F_{y}=0[/tex]
[tex]F_{f}+2F_{r}-W=0[/tex]
[tex]F_{r}=\dfrac{F_{f}-W}{2}[/tex]
Put the value into the formula
[tex]F_{r}=\dfrac{-4.67\times10^{5}+1.87\times10^{6}}{2}[/tex]
[tex]F_{r}=7.02\times10^{5}\ N[/tex]
Hence, (a). The normal force exerted by the ground on the front wheel is [tex]4.67\times10^{5}\ N[/tex].
(b). The normal force exerted by the ground on each of the two rear wheels is [tex]7.02\times10^{5}\ N[/tex]
An archer puts a 0.4 kg arrow to the bowstring. An average force of 190.4 N is exerted to draw the string back 1.47 m. The acceleration of gravity is 9.8 m/s². Assuming no frictional loss, with what speed does the arrow leave the bow? Answer in units of m/s. If the arrow is shot straight up, how high does it rise? Answer in units of m.
Answer:
v = 37.4 m/s , h = 71.39m
Explanation:
To find the velocity given:
m = 0.4 kg
F =190.4 N
d = 1.47 m
g = 9.8 m/s^2
So use the equation of work to solve the kinetic energy
W = F *d = 190.4 N * 1.47m
W = 279.88 J
Ke = 1 / 2 * m* v^2
v = √2*Ke / m =√ 2 *279.88 / 0.4 kg
v = 37.4 m/s
Now to find the high to rise can use the conserved law so:
Ke = Pe
279.88 = m*g*h
Solve to h'
h = 279.88 / 0.4kg * 9.8m/s^2
h =71.39 m
Final answer:
The arrow leaves the bow with a speed of approximately 37.49 m/s and, when shot straight up, rises to a maximum height of about 71.4 m.
Explanation:
To determine the speed with which the arrow leaves the bow, we apply the work-energy principle, which states that work done on the arrow is converted into its kinetic energy. The work done W by the bow can be calculated by multiplying the force F exerted by the distance d over which the force is applied: W = F × d. It is given that F is 190.4 N and d is 1.47 m; thus, W = 190.4 N × 1.47 m = 279.888 J.
The kinetic energy KE of the arrow can be given by KE = ½ mv², where m is the mass of the arrow and v is its velocity. Since work done equals the kinetic energy, we get 279.888 J = ½ × 0.4 kg × v². Solving for v gives us a velocity of approximately 37.49 m/s.
To find how high the arrow goes if it's shot straight up, we use the conservation of energy, where the initial kinetic energy is converted to gravitational potential energy at the highest point. The potential energy PE at maximum height can be given by PE = mgh, where g is the acceleration due to gravity (9.8 m/s²) and h is the height. Setting KE equal to PE, we have 279.888 J = 0.4 kg × 9.8 m/s² × h. Solving for h gives us a maximum height of approximately 71.4 m.
A 49 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 0.7 m/s. The acceleration of gravity is 9.8 m/s2. Find her altitude as she crosses the bar. Neglect air resistance, as well as any energy absorbed by the pole.
Answer:
Height will be 5.127 m
Explanation:
We have given mass m = 49 kg
Speed over the bar v = 10 m /sec
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
Kinetic energy on the ground [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 49\times 10^2=2450J[/tex]
Potential energy on the ground [tex]=mgh=0[/tex] (as height will be zero )
Speed above the bar = 0.7 m /sec
So kinetic energy above the bar [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 49\times 0.7^2=12J[/tex]
Potential energy above the bar = mgh
From energy conservation
Total kinetic energy = total potential energy
So [tex]2450+12=0+49\times 9.8\times h[/tex]
[tex]h=5.127m[/tex]
Using the principle of conservation of energy, we can calculate the pole vaulter's height above the bar as being approximately 5.02 meters.
In this question, we're dealing with conservation of energy. The energy of the pole vaulter is kinetic energy when she is running (1/2*mass*speed^2), and as she goes over the bar, her energy becomes potential energy (mass*gravity*height) and a little kinetic energy (1/2*mass*speed^2).
Let's use the following formula: Kinetic Energy initial + Potential Energy initial = Kinetic Energy final + Potential Energy final.
Initially, her kinetic energy is 1/2 * 49 kg * (10 m/s)^2 = 2450 J. She has no potential energy, because she's on the ground (height=0). When she's above the bar, her kinetic energy is 1/2 * 49 kg * (0.7 m/s)^2 = 12.075 J. We don't yet know her final potential energy, because we're trying to find her height. So, we'll call her final potential energy m*g*h, or 49 kg * 9.8 m/s^2 * h.
We then plug these values into our energy equation: 2450 J + 0 J = 12.075 J + 49 kg * 9.8 m/s^2 * h. Solving for h gives us h = (2450 J - 12.075 J) / (49 kg * 9.8 m/s^2) = 5.02 meters, so this would be her altitude as she crosses the bar.
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A passenger on a balloon drops a baseball over the side of the gondola. As the baseball falls faster, the drag force from air resistance increases.
Which of these describes what happens to the motion of the ball from the time the ball is dropped to the time when the drag force becomes equal to the force of gravity?
A The acceleration of the ball remains constant.
B The speed of the ball decreases.
C The acceleration of the ball decreases.
D The speed of the ball remains constant.
Answer:
The answer is C.
Explanation:
For acceleration to be achieved so to speak there must be a force acting on it. The only force on the ball before the air drag increases is gravity. As the air resistance increases the force resisting gravity increases. This means that the forces start to cancel out. Therefore the acceleration must get smaller.
These forces will continue to cancel until it reaches terminal velocity.
What happens to the motion of the ball is, the acceleration of the ball decreases and will become zero when drag force on the ball equals the force of gravity.
The downward motion of the ball is reduced by frictional force opposing the motion. The frictional force opposing the motion is the drag force of the air or air resistance.
The net downward force on the ball is given as;
[tex]W -F_D = ma\\\\[/tex]
when the drag force on the ball equals force of gravity, the acceleration of the ball will be zero.
[tex]W - W = ma\\\\(W = F_D)\\\\0 = ma\\\\a = 0[/tex]
Thus, we can conclude that what happens to the motion of the ball is, the acceleration of the ball decreases and will become zero when drag force on the ball equals the force of gravity.
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Sometimes, when the wind blows across a long wire, a low-frequency "moaning" sound is produced. The sound arises because a standing wave is set up on the wire, like a standing wave on a guitar string. Assume that a wire (linear density = 0.0180 kg / m ) sustains a tension of 350 N because the wire is stretched between two poles that are 17.43 m apart. The lowest frequency that an average, healthy human ear can detect is 20.0 Hz. What is the lowest harmonic number n that could be responsible for the "moaning" sound?
Answer:
N = 5 harmonics
Explanation:
As we know that frequency of the sound is given as
[tex]f = \frac{N}{2L}\sqrt{\frac{T}{\mu}}[/tex]
now we have
[tex]T = 350 N[/tex]
[tex]\mu = 0.0180 kg/m[/tex]
L = 17.43 m
now we have
[tex]f = \frac{N}{2(17.43)}\sqrt{\frac{350}{0.0180}}[/tex]
[tex]f = 4 N[/tex]
if the lowest audible frequency is f = 20 Hz
so number of harmonics is given as
[tex]20 = 4 N[/tex]
N = 5 harmonics
The velocity function (in meters per second) is given for a particle moving along a line. v(t) = 5t − 9, 0 ≤ t ≤ 3 (a) Find the displacement. -9/2 Correct: Your answer is correct. m (b) Find the distance traveled by the particle during the given time interval.
Answer:
a) The displacement is -4.5 m.
b) The traveled distance is 11.7 m.
Explanation:
Hi there!
a)The velocity of the particle is the derivative of the displacement function, x(t):
v(t) = dx/dt = 5t - 9
Separating varibles:
dx = (5t - 9)dt
Integrating both sides from x = x0 to x and from t = 0 to t.
x - x0 = 1/2 · 5t² - 9t
x = 1/2 · 5t² - 9t + x0
If we place the origin of the system of reference at x = x0, the displacement at t = 3 will be x(3):
x(3) = 1/2 · 5 · (3)² - 9(3) + 0
x(3) = -4.5
The displacement at t = 3 s is -4.5 m. It means that the particle is located 4.5 m to the left from the origin of the system of reference at t = 3 s.
b) When the velocity is negative, the particle moves to the left. Let´s find the time at which the velocity is less than zero:
v = 5t - 9
0 > 5t - 9
9/5 > t
1.8 s > t
Then until t = 1.8 s, the particle moves to the left from the origin of the reference system.
Let´s find the position of the particle at that time:
x = 1/2 · 5t² - 9t
x = 1/2 · 5(1.8 s)² - 9(1.8 s)
x = -8.1 m
From t = 0 to t = 1.8 s the traveled distance is 8.1 m. After 1.8 s, the particle has positive velocity. It means that the particle is moving to the right, towards the origin. If at t = 3 the position of the particle is -4.5 m, then the traveled distance from x = -8.1 m to x = -4.5 m is (8.1 m - 4.5 m) 3.6 m.
Then, the total traveled distance is (8.1 m + 3.6 m) 11.7 m.
The distance travelled by the particle during the given time interval is 45/2 meters.
Explanation:The distance travelled by the particle can be found by integrating the absolute value of the velocity function over the given time interval:
The distance is, therefore, the integral of |5t - 9| from 0 to 3:
D = ∫(5t - 9)dt = [5(t^2/2) - 9t] from 0 to 3 = (45/2) meters
So, the distance travelled by the particle during the given time interval is 45/2 meters.
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Your grandfather clock's pendulum has a length of 0.9930 m. Part A If the clock runs slow and loses 19 s per day, how should you adjust the length of the pendulum? Note: due to the precise nature of this problem you must treat the constant g as unknown (that is, do not assume it is equal to exactly 9.80 m/s2).
To compensate for the clock's slowness, you should adjust the length of the pendulum to [tex]\( 0.9939 \)[/tex] meters.
To calculate the necessary adjustment to the length of the pendulum to compensate for the clock's slowness, we can start by determining the time period of the pendulum using the given length and the unknown acceleration due to gravity, denoted as [tex]\( g \).[/tex] The time period [tex]\( T \)[/tex] of a simple pendulum is given by the formula:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
where [tex]\( L \)[/tex] is the length of the pendulum.
Given that the clock loses 19 seconds per day, we convert this to seconds per period, as [tex]\( 1 \) day has \( 24 \) hours, \( 60 \)[/tex] minutes per hour, and [tex]\( 60 \)[/tex] seconds per minute:
[tex]\[ \text{Seconds lost per period} = \frac{19 \, \text{s}}{24 \times 60 \times 60 \, \text{s/day}} \][/tex]
Let's calculate the time period [tex]\( T \)[/tex] and the adjustment needed.
[tex]\[ T = 2\pi \sqrt{\frac{0.9930 \, \text{m}}{g}} \][/tex]
Now, we'll solve for [tex]\( g \):[/tex]
[tex]\[ T = 2\pi \sqrt{\frac{0.9930 \, \text{m}}{g}} \]\[ T^2 = 4\pi^2 \frac{0.9930 \, \text{m}}{g} \]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{T^2} \][/tex]
Substitute [tex]\( T \) with \( 24 \, \text{hours} \)[/tex] converted to seconds:
[tex]\[ T = 24 \times 60 \times 60 \, \text{s} = 86,400 \, \text{s} \]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{(86,400 \, \text{s})^2} \][/tex]
Calculate [tex]\( g \):[/tex]
[tex]\[ g = 4\pi^2 \frac{0.9930 \, \text{m}}{7.48224 \times 10^9 \, \text{s}^2} \]\[ g = 0.0000016512 \, \text{m/s}^2 \][/tex]
Now that we have [tex]\( g \)[/tex], we can find the adjusted length of the pendulum. The new time period [tex]\( T' \)[/tex] can be calculated as follows:
[tex]\[ T' = 2\pi \sqrt{\frac{L'}{g}} \][/tex]
Where [tex]\( L' \)[/tex] is the adjusted length. Rearrange to solve for [tex]\( L' \):[/tex]
[tex]\[ L' = \frac{T'^2 \times g}{4\pi^2} \][/tex]
Given that the clock loses [tex]\( 19 \)[/tex] seconds per day and we want it to lose [tex]\( 0 \)[/tex] seconds, the new time period [tex]\( T' \) is \( 86,400 \)[/tex] seconds. Substitute into the formula:
[tex]\[ L' = \frac{(86,400 \, \text{s})^2 \times 0.0000016512 \, \text{m/s}^2}{4\pi^2} \][/tex]
[tex]\[ L' = 0.9939 \, \text{m} \][/tex]
So, to compensate for the clock's slowness, you should adjust the length of the pendulum to [tex]\( 0.9939 \)[/tex] meters.
Complete Question:
Your grandfather clock's pendulum has a length of 0.9930 m. Part A If the clock runs slow and loses 19 s per day, how should you adjust the length of the pendulum? Note: due to the precise nature of this problem you must treat the constant g as unknown (that is, do not assume it is equal to exactly 9.80 m/s2).
On your first trip to Planet X you happen to take along a 210 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the acceleration due to gravity on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 26.9 cm . You then pull the mass down 7.00 cm and release it. With the stopwatch you find that 9.00 oscillations take 14.3 s .Can you now satisfy your curiosity?what is the new g
Answer:
4.20m/s^2
Explanation:
The mass of 210g stretch the spring by 26.9cm, 9 oscillation took 14.3 s.
Period (T) = time of oscillation / number of oscillation = 14.3/ 9 = 1.59s
Using the formula for period of a spring;
T = 2π√(m/k)
Divide both side by 2π
T/2π = √(m/k)
Square both side
T^2/ (2π)^2 = m/k
Make k subject of the formula
K = 4π^2 * m/ T^2
Also using Hooke's law
Since the spring was suspended from the ceiling,
F (mg) = k * DL ( where g is the new gravity in m/s^2, k is the force constant of the spring in N/m and DL is the change in length in meter
Make k subject of the formula
Mg/DL = k
Since both equation equal to K then
4π^2 * m/ T^2 = mg/DL
Cancel m on both side
4π^2 * DL / T^2 = g
DL = 26.9cm = 26.9/100 m = 0.269 m
T = 1.59s and π = 3.142
Substitute these into the equation
g = ( 4 * 3.142 * 3.142* 0.269) / (1.59^2) = 4.20 m/s^2
Final answer:
To determine the acceleration due to gravity on Planet X, the period T of the spring-mass system's oscillations is used along with the mass m and the spring constant k, which is calculated using the initial stretch of the spring.
Explanation:
To find the acceleration due to gravity (g) on Planet X, we can use the properties of simple harmonic motion for a spring-mass system. The formula for the period T of a mass m attached to a spring with spring constant k is T = 2π√(m/k). Since the spring stretches 26.9 cm when the 210 g mass is attached, the force due to gravity must be equal to the force the spring exerts at that stretch, which is F = kx, where x is the stretch distance. Thus, mg = kx, and we can solve for k using k = mg/x (with g as the gravity on Earth, 9.81 m/s²). Then we can calculate the period using the measured time of oscillations. From the period, we can solve for the gravity on Planet X.
Considering 9 oscillations took 14.3 seconds, the period T for one oscillation is 14.3 s / 9. With T and m known, one can solve for k and then use that to find g. Since the initial stretch was caused by the force of gravity, the spring constant k can be found using Earth's gravity, and subsequently used in the equation T = 2π√(m/g) rearranged to solve for g on Planet X.
Oppositely charged objects attract each other. This attraction holds electrons in atoms and holds atoms to one another in many compounds. However, Ernest Rutherford’s model of the atom failed to explain why electrons were not pulled into the atomic nucleus by this attraction. What change to the atomic model helped solve the problem seen in Rutherford’s model?
Answer:
A) Bohr’s work with atomic spectra led him to say that the electrons were limited to existing in certain energy levels, like standing on the rungs of a ladder.
Explanation:
The change to the atomic model that helped solve the problem seen in Rutherford's model was the discovery of the strong nuclear force.
Explanation:Rutherford's model required that the electrons be in motion. Positive and negative charges attract each other, so stationary electrons would fall into the positive nucleus. However, Rutherford's model failed to explain why electrons were not pulled into the atomic nucleus by this attraction. The change to the atomic model that helped solve this problem was the discovery of the strong nuclear force, which is much stronger than electrostatic interactions and holds the protons and neutrons together in the nucleus.
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A player passes a 0.600-kg basketball downcourt for a fast break. The ball leaves the player’s hands with a speed of 8.30 m/s and slows down to 7.10 m/s at its highest point.
a. Ignoring air resistance, how high above the release point is the ball when it is at its maximum height?
b. How would doubling the ball’s mass affect the result in part a.? Explain.
Answer:
Explanation:
Decrease in kinetic energy of the ball
= 1/2 m ( v² - u² )
= 1/2 x .6 ( 8.3² - 7.1² )
= 5.544 J
This energy will be converted into potential energy
mgH = 5.544 ( H is maximum height attained )
H = 5.544 /( .6 x 9.8 )
= .942 m
= 94.2 cm
b ) In a ) we used the relation
mgH = 1/2 m ( v² - u² )
H = ( 1/2g ) x ( v² - u² )
Here we see H or height attained does not depend upon mass as m cancels out on both sides.
Express the angular velocity ω of the wheel in terms of the displacement d, the magnitude F of the applied force, and the moment of inertia of the wheel Iw, if you've found such a solution. Otherwise, following the hints for this part should lead you to express the angular velocity ω of the wheel in terms of the displacement d, the wheel's radius r, and α.
The angular velocity (ω) of a wheel can be expressed in terms of the force applied (F), wheel's radius (r), and its moment of inertia (I), all of which factor into the wheel's torque and resulting angular acceleration. Through the rotational analog of Newton's second law and kinematic equations, one can derive a formula for ω that connects these physical quantities
To express the angular velocity (ω) of the wheel in terms of displacement (d), magnitude of the force (F), and the moment of inertia of the wheel (Iw), we use physical principles of rotational motion. We can derive this using the definition of torque and the rotational version of Newton's second law, which states that torque (τ) is equal to the moment of inertia (I) times the angular acceleration (α): τ = Iα. For a wheel, torque is also equal to the force applied (F) times the radius (r): τ = Fr.
By equating the two expressions for torque, we get Fr = Iα, and therefore α = F × r/I. With the angular acceleration and knowing the time the force is applied, we can find the final angular velocity using the kinematic equation for rotational motion: ωfinal = ωinitial + α × t. Assuming that initial angular velocity (ωinitial) is zero and time (t) can be deduced from the linear displacement (d) and the linear velocity (v), given that v = ω × r, we can express the final angular velocity in terms of displacement (d), wheel's radius (r), and angular acceleration (α).
To further involve the moment of inertia in this expression, we would substitute the earlier derived expression for α into our final angular velocity equation: ω = (F × r/I) × t. Now, if we relate time (t) to the linear displacement (d) such that the linear velocity (v) = d/t, and since v = ω×r, we can isolate ω to derive a relationship that includes the moment of inertia.
A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the stick a distance D from the middle of the stick. Before the collision, the puck was moving with speed v0 in a direction perpendicular to the stick, as indicated in the figure. The collision is completely inelastic, and the puck remains attached to the stick after the collision.
What is the angular momentum Lcm of the system before the collision, with respect to the center of mass of the final system?
Express Lcm in terms of the given variables.
The angular momentum of the hockey stick-puck system before the collision, with respect to the center of mass of the final system, is given by Lcm = mp * v0 * D, where mp is the puck's mass, v0 is its initial velocity, and D is the distance from the center of the stick.
Angular Momentum of Hockey Stick-Puck System
To find the angular momentum Lcm of the system before the collision with respect to the center of mass of the final system, we follow these steps:
Before the collision, the puck is moving with velocity v0. Since the collision is completely inelastic, the stick and puck move as a single system after the collision.The angular momentum L before the collision is given by: This expression simplifies to:Therefore, the angular momentum of the system before the collision, with respect to the center of mass of the final system, is Lcm = mp * v0 * D.
At a rock concert, a dB meter registered 128 dB when placed 2.9 m in front of a loudspeaker on the stage. The intensity of the reference level required to determine the sound level is 1.0×10−12W/m2. Determine the following:
A. What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?
B. How far away would the sound level be 82 dB?
Answer:
A)
665.5 W
B)
575.5 m
Explanation:
A)
[tex]S[/tex] = Sound level registered = 128 dB
[tex]I_{o}[/tex] = Intensity at reference level = [tex]1\times10^{-12}[/tex] Wm⁻²
[tex]I[/tex] = Intensity at the location of meter
Sound level is given as
[tex]S = 10 log_{10}\left ( \frac{I}{I_{o}} \right )[/tex]
[tex]128 = 10 log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\12.8 = log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\10^{12.8} = \frac{I}{1\times10^{-12}} \right \\I = 6.3[/tex]
[tex]P[/tex] = power output of the speaker
[tex]r[/tex] = distance from the speaker = 2.9 m
Power output of the speaker is given as
[tex]P = I(4\pi r^{2} )\\P = (6.3) (4) (3.14) 2.9^{2}\\P = 665.5 W[/tex]
B)
[tex]S[/tex] = Sound level = 82 dB
[tex]I_{o}[/tex] = Intensity at reference level = [tex]1\times10^{-12}[/tex] Wm⁻²
[tex]I[/tex] = Intensity at the location of meter
Sound level is given as
[tex]S = 10 log_{10}\left ( \frac{I}{I_{o}} \right )[/tex]
[tex]82 = 10 log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\8.2 = log_{10}\left ( \frac{I}{1\times10^{-12}} \right )\\10^{8.2} = \frac{I}{1\times10^{-12}} \right \\I = 1.6\times10^{-4}Wm^{-2}[/tex]
Power of the speaker is given as
[tex]P = I(4\pi r^{2} )\\665.5 = (1.6\times10^{-4}) (4) (3.14) r^{2}\\r = 575.5 m[/tex]
The power output of the speaker at the rock concert was approximately 137 Watts. The sound level would be 82 dB at a distance of about 103 meters from the speaker.
Explanation:This problem involves the use of the formula for calculating sound levels in decibels (dB). The formula is L = 10 log(I/I0) where L is the sound level in dB, I is the intensity of the sound wave, and I0 is the reference intensity. Now, let's solve both parts.
A. At the dB meter location, reverse calculate the intensity I: I = I0 * 10^(L/10) = 1.0 10^−12 * 10^(128/10) = 1.26 W/m^2. Since the sound spreads spherically, the power P is the intensity times the area of the sphere: P = I * 4πr^2 = 1.26 * 4π*(2.9)^2 = 137 Watts.
B. To find the distance where the sound would register at 82 dB, we first find the intensity I at that level: I = I0*10^(82/10) = 1.0x10^−12 * 10^8.2 = 6.31x10^-5 W/m^2. With I, we can solve for r: r = sqrt(P/(4πI)) = sqrt(137/(4π*(6.31x10^-5))) = 103 meters.
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Two asteroids in outer space collide, and stick together. The mass of each asteroid, and the velocity of each asteroid before the impact, are known. To find the momentum of the stuck-together asteroids after the impact, what approach would be useful?
(1) Use the Energy Principle.
(2) Use the Momentum Principle.
(3) It depends on whether or not the speed of the asteroids was near the speed of light.
(4) Use the relationship among velocity, displacement, and time.
(5) It depends on whether the collision was elastic or inelastic.
Answer: (2) Use the Momentum Principle.
Explanation:
In fact, it is called the Conservation of linear momentum principle, which establishes the initial momentum [tex]p_{i}[/tex] of the asteroids before the collision must be equal to the final momentum [tex]p_{f}[/tex] after the collision, no matter if the collision was elastic or inelastic (in which the kinetic energy is not conserved).
In this sense, the linear momentum [tex]p[/tex] of a body is defined as:
[tex]p=mV[/tex]
Where [tex]m[/tex] is the mass and [tex]V[/tex] the velocity.
Therefore, the useful approach in this situation is option (2).