Explanation:
We know the equation
[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]
where l is length of pendulum, g is acceleration due to gravity and T is period.
Rearranging
[tex]g= \frac{4\pi^2l}{T^2}[/tex]
Length of pendulum in Tokyo = 0.9923 m
Length of pendulum in Cambridge = 0.9941 m
Period of pendulum in Tokyo = Period of pendulum in Cambridge = 2s
We have
[tex]\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}= \frac{\frac{4\pi^2 l_{\texttt{Tokyo}}}{ T_{\texttt{Tokyo}}^2}}{\frac{4\pi^2 l_{\texttt{Cambridge}}}{ T_{\texttt{Cambridge}}^2}}\\\\\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}=\frac{\frac{0.9923}{2^2}}{\frac{0.9941}{2^2}}=0.998[/tex]
Ratio of free fall acceleration of Tokyo to Cambridge = 0.998
A 440.0-Hz tuning fork is sounded together with an out-of-tune guitar string, and a beat frequency of 3 Hz is heard. When the string is tightened, the frequency at which it vibrates increases, and the beat frequency is heard to decrease. What was the original frequency of the guitar string?
Answer:
The original frequency of the guitar string is 437.0 Hz
Explanation:
It is an interesting phenomenon that when two waves of slightly different frequencies (f1 and f2) interfere they produce beats, and the frequency (fb) of those beats is:
[tex] f_{b}=\mid f_{1}-f_{2}\mid [/tex] (1)
Note that because the equation takes the absolute value of the difference between f1 and f2 we have two possible results of [tex] f_{1}-f_{2} [/tex] that satisfy the equation, those are:
[tex] 440.0-f_{2}=3\Longrightarrow f_{2}=437.0\,Hz [/tex]
[tex] -(440.0-f_{2})=3\Longrightarrow f_{2}=443.0\,Hz [/tex]
Because the frequency of the beat is heard to decrease when the frequency of the guitar string is increased, that means the guitar string was at first under the pitch, so the original frequency of the guitar string is 437.0 Hz
Final answer:
The original frequency of the guitar string was 437 Hz, since the string was out of tune and produced a beat frequency of 3 Hz with a 440 Hz tuning fork, and the beat frequency decreased when the string was tightened.
Explanation:
The original frequency of the guitar string can be determined using the concept of beat frequency. The beat frequency is the absolution difference between the frequencies of two sounds. In this case, the beat frequency was 3 Hz when the guitar string was sounded with the 440.0 Hz tuning fork.
Since the beat frequency decreased when the string tension was increased, the original frequency of the string must have been lower than the tuning fork. Therefore, the original frequency of the guitar string could be either 437 Hz or 443 Hz. However, since tightening the string increases its frequency and resulted in a decrease in the beat frequency, this indicates that the original frequency was 437 Hz (440 Hz - 3 Hz).
An object is hanging from a rope. When it is held in air the tension in the rope is 8.86 N, and when it is submerged in water the tension is 7.84 N. What is the density of the the object
Answer:
8684.2 kg/m³
Explanation:
Tension in the rope as a result of the weight = 8.86 N
Tension in the rope when submerge in water = 7.84
upthrust = 8.86 - 7.84 =1.02 N = mass of water displaced × acceleration due to gravity
Mass of water displaced = 1.02 / 9.81 = 0.104 kg
density of water = mass of water / volume of water
make volume subject of the formula
volume of water displaced = mass / density ( 1000) = 0.104 / 1000 = 0.000104 m³
volume of the object = volume of water displaced
density of the object = mass of the object / volume of the object = (8.86 / 9.81) / 0.000104 = 0.9032 / 0.000104 = 8684.2 kg/m³
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).
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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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.
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.
The ability to move or change an object or what a wave carries
Waves can transfer energy and momentum from one place to another without a permanent shift in the mass. They propagate via different oscillation modes and can impact the propagation direction when they encounter an interface between two media.
Explanation:The ability to move or change an object or what a wave carries, essentially describes the energy transfer capability of waves. According to Big Idea 6, waves can transfer energy and momentum from one place to another without permanently relocating the mass. This aspect characterizes waves and differentiates them from other phenomena.
Waves can propagate through various oscillation modes, such as transverse and longitudinal. For instance, when you shake one end of a rope, the perturbation moves in the form of a wave up to the other end of the rope, thus, facilitating energy transfer without a physical shift of the rope's parts.
Another important aspect to remember about waves is their ability to affect the path of propagation when they encounter an interface between two different media, as articulated by Enduring Understanding 6.E. This property explains various natural phenomena, such as the bending of light when it moves from air to water.
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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 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.
Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500ºC rock must be placed in 4.00 kg of 15.0ºC water to bring its temperature to 100ºC, if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings and take the average specific heat of the rocks to be that of granite. (answer in kg)
Answer:
m = 4.65 kg
Explanation:
As we know that the mass of the water that evaporated out is given as
[tex]m = 0.0250 kg[/tex]
so the energy released in form of vapor is given as
[tex]Q = mL[/tex]
[tex]Q = (0.0250)(2.25 \times 10^6)[/tex]
[tex]Q = 56511 J[/tex]
now the heat required by remaining water to bring it from 15 degree to 100 degree
[tex]Q_2 = ms\Delta T[/tex]
[tex]Q_2 = (4 - 0.025)(4186)(100 - 15)[/tex]
[tex]Q_2 = 1.41\times 10^6J[/tex]
total heat required for above conversion
[tex]Q = 56511 + 1.41 \times 10^6 = 1.47 \times 10^6 J[/tex]
now by heat energy balance
heat given by granite = heat absorbed by water
[tex]m(790)(500 - 100) = 1.47 \times 10^6[/tex]
[tex]m = 4.65 kg[/tex]
A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 245 Hz. A person on the platform right next to the speaker detects the sound waves reflected off the wall and those emitted by the speaker.
How fast should the platform move, vp, for the person to detect a beat frequency of 3.00 Hz?
Answer:
Explanation:
by Doppler effect
F apparent = F real x (Vair +- Vobserver) / (Vair +- Vsource)
case1 :if observer is approaching a source then put a + in numerator and - in denominator
case 2 : if observer is going away from source then do opposite.
it is case1 cos observer is moving toward wall . actually wall is acting as a source for observer
so
F apparent = 240 x (344 + vp ) / (344 - vp )
now F apparent - Freal = beat frequency of 6.00 Hz
so {240 x (344 + vp ) / (344 - vp )} - 230 = 6
240 x (344 + vp -344 + vp) / (344 - vp ) = 6
240 x 2 x vp = 6 x (344 - vp )
480 vp = 2064 - 6 x vp
486 x vp = 2064
vp = 4.24 m/s
What will a spring scale read for the weight of a 58.0-kg woman in an elevator that moves in free fall?
The answer is 0N but could someone pls explain why?
Answer:
0 N
Explanation:
The elevator is under free fall. So, when a body is under free fall, the acceleration is only due to gravity. So, the acceleration of the elevator or the woman inside it, is acceleration due to gravity in the downward direction.
The spring scale gives the value of the normal force acting on the woman and doesn't give the exact weight of the woman. Under normal conditions, when the spring scale is at rest, then the upward normal force equals the weight and hence weight of a body is equal to the normal force acting on the body.
But, here, the body is not at rest. Weight\tex](mg)[/tex] acts in the downward direction and normal force[tex](N)[/tex] acts in the upward direction. The woman is moving down with acceleration equal to acceleration due to gravity[tex](g)[/tex]
So, we apply Newton's second law on the woman.
[tex]\textrm{Net force} = \textrm{mass}\times \textrm{acceleration}\\F_{net}=ma[/tex]
Net force is equal to the difference of the downward force and upward force.
[tex]F_{net}=mg-N[/tex]
Now, [tex]F_{net}=ma[/tex]
[tex]mg-N=mg\\N=mg-mg=0[/tex]
Therefore, the reading on the spring scale is 0 N.
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.
A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the horizontal, frictionless surface of a frozen lake. ????(x) is the only horizontal force on the box. If the box is initially at rest at x = 0, what is its speed after it has traveled 14.0 m?
Answer:
[tex]v_f=8.17\frac{m}{s}[/tex]
Explanation:
First, we calculate the work done by this force after the box traveled 14 m, which is given by:
[tex]W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m[/tex]
Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:
[tex]W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}[/tex]
The box is initially at rest, so [tex]v_i=0[/tex]. Solving for [tex]v_f[/tex]:
[tex]v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}[/tex]
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 spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface of length 2 m with a coefficient of friction 0.16, then compresses the second spring of constant 2 N/m. The acceleration of gravity is 9.8 m/s2.
How far will the second spring compress in order to bring the mass to a stop?
Answer:
[tex]X_2=25.27m[/tex]
Explanation:
Here we will call:
1. [tex]E_1[/tex]: The energy when the first spring is compress
2. [tex]E_2[/tex]: The energy after the mass is liberated by the spring
3. [tex]E_3[/tex]: The energy before the second string catch the mass
4. [tex]E_4[/tex]: The energy when the second sping compressed
so, the law of the conservations of energy says that:
1. [tex]E_1 = E_2[/tex]
2. [tex]E_2 -E_3= W_f[/tex]
3.[tex]E_3 = E_4[/tex]
where [tex]W_f[/tex] is the work of the friction.
1. equation 1 is equal to:
[tex]\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2[/tex]
where K is the constant of the spring, x is the distance compressed, M is the mass and [tex]V_2[/tex] the velocity, so:
[tex]\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2[/tex]
Solving for velocity, we get:
[tex]V_2[/tex] = 65.319 m/s
2. Now, equation 2 is equal to:
[tex]\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd [/tex]
where M is the mass, [tex]V_2[/tex] the velocity in the situation 2, [tex]V_3[/tex] is the velocity in the situation 3, [tex]U_k[/tex] is the coefficient of the friction, N the normal force and d the distance, so:
[tex]\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2) [/tex]
Volving for [tex]V_3[/tex], we get:
[tex]V_3 = 65.27 m/s[/tex]
3. Finally, equation 3 is equal to:
[tex]\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2[/tex]
where [tex]K_2[/tex] is the constant of the second spring and [tex]X_2[/tex] is the compress of the second spring, so:
[tex]\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2[/tex]
solving for [tex]X_2[/tex], we get:
[tex]X_2=25.27m[/tex]
The second spring will compress by 25.27 meters to bring the mass to a stop.
To solve this problem, we will go step-by-step considering energy conservation and the work done by friction.
1) Calculate the potential energy stored in the first spring:
The spring constant k is 20 N/m and the compression x is 8 m. The potential energy (PE) in the spring is calculated using the formula:
PE = (1/2) * k * x²
PE = (1/2) * 20 N/m * (8 m)²= 640 J
When the spring is released, this potential energy will convert to kinetic energy (KE) of the mass:
KE = 640 J
Calculate the work done by friction (Wf):
The frictional force (f) is calculated by:
f = μ * m * g
f = 0.16 * 0.3 kg * 9.8 m/s² = 0.4704 N
The work done by friction over a distance (d) of 2 m is:
Wf = f * d
Wf = 0.4704 N * 2 m = 0.9408 J
The kinetic energy after overcoming friction is:
Remaining KE = 640 J - 0.9408 J = 639.0592 J
Calculate the compression of the second spring (x₂):
The remaining kinetic energy will convert into potential energy in the second spring with a constant of 2 N/m.
PE = (1/2) * k₂ * x₂²
We set the remaining KE equal to this potential energy:
639.0592 J = (1/2) * 2 N/m * x₂²
639.0592 J = x₂²
x₂ = √639.0592 = 25.27 m
Therefore, the second spring will compress by 25.27 meters.
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]
An electroscope is a fork-shaped device commonly used to detect the presence of charge. The tin leaves of an electroscope will separate when charged negatively. Over time, however, these leaves will cease separating, even without being touched by another object. Explain why this may occur.
Answer:
All these is caused by the repulsion force.
Explanation:
The electroscope produces a series of electric charges that produce a repulsion force when is putted in contact with a electric charged object.
As the physics law mentions, two different forces are repealed, the electrocospe is charged negatively and the object positively, causing a repulsion force that avoids that both objects touch the other.
A heavy-duty stapling gun uses a 0.147-kg metal rod that rams against the staple to eject it. The rod is pushed by a stiff spring called a "ram spring" (k = 32800 N/m). The mass of this spring may be ignored. Squeezing the handle of the gun first compresses the ram spring by 4.44 cm from its unstrained length and then releases it. Assuming that the ram spring is oriented vertically and is still compressed by 1.53 cm when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.
Answer:
v = 13.75 m/s
Explanation:
mass of rod (m) = 0.147 kg
spring constant (k) = 32800 N/m
initial compression (Y1) = 4.44 cm = 0.0444 m
final compression (Y2) = 1.53 cm = 0.0153 m
find the speed at the instant of contact
kinetic energy of a spring = elastic potential energy
0.5m[tex]v^{2}[/tex] = 0.5k[tex]x^{2}[/tex]
[tex]v^{2}[/tex] = [tex]\frac{0.5k[tex]x^{2}[/tex]}{0.5m} \[/tex]
where x = Y1 - Y2
x = 0.0444 - 0.0153 = 0.0291 m
[tex]v^{2}[/tex] = [tex]\frac{0.5 x 32800 x [tex]0.0291^{2}[/tex]}{0.5 x 0.147} \[/tex]
[tex]v^{2}[/tex] = 188.95
v = 13.75 m/s
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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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]
If 12,000,000 kilocalories of chicken can be produced per hectare, how much land is needed to provide someone with 2,000 kilocalories/day for a year?a. 0.01 hab.0.02 hac. 0.06 ha
Answer:
it will need 0.06 hectare of land
So option (c) is correct
Explanation:
We have given that 12000000 kilo calories of chicken can be produced per hectare
It is given that someone with 2000 kilo calories / day for a year
Total number of days in a year = 365
So total kilo calories of chicken = 365×2000 = 730000 kilo calories
As 12000000 kilo calories of chicken can be produced per hectare
So 730000 kilo calories need a land of [tex]=\frac{730000}{12000000}=0.06hectare[/tex]
So it will need 0.06 hectare of land
Final answer:
To provide someone with 2,000 kilocalories/day for a year, 0.06 hectares of land used for chicken production is needed, as the closest correct answer.
Explanation:
The question asks how much land is needed to provide someone with 2,000 kilocalories/day for a year if 12,000,000 kilocalories of chicken can be produced per hectare. First, calculate the total kilocalories needed for a year by multiplying 2,000 kcal/day by the number of days in a year (365), which equals 730,000 kcal/year. Then, find how much land is needed by dividing the total annual kilocalorie requirement (730,000 kcal) by the amount of kilocalories produced per hectare (12,000,000 kcal).
This calculation gives us:
730,000 kcal / 12,000,000 kcal/ha = 0.0608333 ha
Therefore, the closest answer is c. 0.06 ha.
A 32.5 g iron rod, initially at 21.8°C, is submerged into an unknown mass of water at 63.3°C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 58.9°C. What is the mass of the water?
Final Answer:
The mass of the water is approximately 48.4 grams.
Explanation:
1. To solve this problem, we can use the principle of conservation of energy, specifically the principle of heat exchange.
2. We can calculate the heat gained by the iron rod and the heat lost by the water, and equate them since there is no heat exchange with the surroundings.
3. The formula for heat exchange is given by: [tex]\[ Q = mc\Delta T \][/tex]
where Q is the heat exchanged, m is the mass, c is the specific heat capacity, and [tex]\( \Delta T \)[/tex] is the change in temperature.
4. We can set up the equation for heat exchange for the iron rod and the water:
[tex]\[ mc_{\text{iron}}\Delta T_{\text{iron}} = mc_{\text{water}}\Delta T_{\text{water}} \][/tex]
5. Rearranging the equation and solving for the mass of water:
[tex]\[ m_{\text{water}} = \frac{m_{\text{iron}}c_{\text{iron}}\Delta T_{\text{iron}}}{c_{\text{water}}\Delta T_{\text{water}}} \][/tex]
6. Substituting the given values, we can calculate the mass of water to be approximately 48.4 grams.
Therefore, the mass of the water is approximately 48.4 grams.
The mass of the water is approximately 199.8 grams.
Here's how to find the mass of the water:
1. Define the variables:
m_iron = mass of iron rod = 32.5 g
c_iron = specific heat capacity of iron = 0.49 J/g°C
T_iron_initial = initial temperature of iron rod = 21.8°C
T_water_initial = initial temperature of water = 63.3°C
T_final = final temperature of the mixture = 58.9°C
m_water = mass of water (unknown)
c_water = specific heat capacity of water = 4.186 J/g°C
2. Apply the principle of heat transfer:
Heat lost by iron = Heat gained by water
3. Set up the equation:
m_iron * c_iron * (T_iron_initial - T_final) = m_water * c_water * (T_final - T_water_initial)
4. Substitute the known values and solve for m_water:
32.5 g * 0.49 J/g°C * (21.8°C - 58.9°C) = m_water * 4.186 J/g°C * (58.9°C - 63.3°C)
5. Solve for m_water:
m_water ≈ 199.8 g
Therefore, the mass of the water is approximately 199.8 grams.
A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? (Assume that the aluminum and the water are thermally isolated from everything else.)
Final answer:
To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the hot metal is equal to the heat gained by the water. Using the given equation, we can solve for the specific heat capacity of the metal, which is found to be 3.75 J/g°C.
Explanation:
To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the hot metal is equal to the heat gained by the water. We can use the equation:m1c1(T1 - T) = m2c2(T - T2)
where:
m1 is the mass of the metal (34.5 g)
c1 is the specific heat capacity of the metal
T1 is the initial temperature of the metal (75°C)
T is the final temperature of the system (39°C)
m2 is the mass of the water (64.0 g)
c2 is the specific heat capacity of water (4.18 J/g°C)
T2 is the initial temperature of the water (25°C)
Plugging in the given values, we can solve for c1: 34.5g × c1 × (75°C - 39°C) = 64.0g × 4.18 J/g°C × (39°C - 25°C)
Simplifying the equation gives:34.5g × c1 × 36°C = 64.0g × 4.18 J/g°C × 14°C
Dividing both sides by 34.5g × 36°C gives:c1 = (64.0g × 4.18 J/g°C × 14°C) / (34.5g × 36°C)
Calculating the value gives:c1 = 3.75 J/g°C
The final temperature for both the aluminum and water at thermal equilibrium is 17.3 °C. This is calculated using the principle of conservation of energy. The heat lost by aluminum equals the heat gained by water.
To find the final temperature of a 32.5 g cube of aluminum and 105.3 g of water at thermal equilibrium, we use the principle of conservation of energy. The heat lost by the aluminum will be equal to the heat gained by the water.
Calculate the heat transfer equation:Let us simplify: (32.5 g) x (0.900 J/g°C) x (T_final - 45.8 °C) = - (105.3 g) x (4.186 J/g°C) x (T_final - 15.4 °C)
29.25 x (T_final - 45.8) = -440.8858 x (T_final - 15.4)19.25T_final - 1339.85 = -440.8858T_final + 6793.6422470.1358T_final = 8133.4922T_final = 17.3 °CTherefore, the final temperature of both substances at thermal equilibrium is **17.3 °C**.
What is the kinetic energy of a 0.500 kg ball thrown at 30.0 m/s?
Answer: The answer is 225 joules
A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 49.0-gram mass is attached at the 13.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?
Answer: M = 118.9g
Explanation:
Using the principle of moment to solve the equilibrium question,
Sum of clockwise moments is equal to the sum of anticlockwise moments
*Check attachment for the diagram
Note that the mass of the body is always located at the center i.e 50cm from the end.
To get M, taking moment about the fulcrum,
Since moment is force × perpendicular distance and according to the above principle, we have;
49×26.2= M×10.8
M= 49×26.2/10.8
M = 118.9g
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
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.
A bulb with a resistance of 4 Ohms is plugged into a 2V battery. It has a capacity of 3 amps-hours. How long will this bulb remain at a constant brightness? Show all your working.
Answer:
Bulb will remain at constant brightness for 6 hour
Explanation:
We have given resistance of the bulb R = 4 ohm
Potential difference of the battery V = 2 volt
So current [tex]i=\frac{V}{R}=\frac{2}{4}=0.5A[/tex]
Capacity of the battery is given as 3 amp hour
We have to find the time by which bulb will remain at constant brightness
So time will be [tex]=\frac{3}{0.5}=6hour[/tex]
So bulb will remain at constant brightness for 6 hour