The debris speed on hitting the surface of the Earth can be found by equating potential and kinetic energy, yielding the formula v = √(2GM/R), where G is the gravitational constant, M is Earth's mass, and R is Earth's radius.
Explanation:The key to solving for the final speed of this piece of debris is utilizing the concepts of energy conservation and gravitational potential energy. When the debris is at a distance equal to half the Earth's radius (R/2) above the surface, all of its energy is potential. At the point it hits the surface, all this potential energy would have been converted to kinetic energy, with negligible loss to thermal or any other types of energy as we are ignoring air resistance and gravitational pull from the moon.
The total energy when the debris is still in the air is given by its gravitational potential energy Pe = -GMm/(R/2), where G is the gravitational constant, M is the Earth’s mass, m is the debris mass, and R is the Earth's radius.
At the point the debris hits the Earth, all its potential energy would have been converted to kinetic energy Ke = ½mv², where v is its velocity on impact.
Equating the initial and final energies, we get -GMm/(R/2) = ½mv². Solving this equation for v (the magnitude of the velocity), we get v = √(2GM/R). This represents the speed of the debris when it hits the Earth's surface.
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Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance of 2.7×1011 m from the center of the sun, what is its speed when at a distance of 4.8×1010 m?
Answer:
v₂ = 7.6 x 10⁴ m/s
Explanation:
given,
speed of comet(v₁) = 1.6 x 10⁴ m/s
distance (d₁)= 2.7 x 10¹¹ m
to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m
v₂ = ?
speed of planet can be determine using conservation of energy
K.E₁ + P.E₁ = K.E₂ + P.E₂
[tex]\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}[/tex]
[tex]\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}[/tex]
[tex]v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}[/tex]
[tex]v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}[/tex]
[tex]v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}[/tex]
v₂ = 7.6 x 10⁴ m/s
The speed when at a distance of [tex]4.8*10^{10} m[/tex] is mathematically given as
vf = 6.92 * 10^(4) m/s
What is the speed when at a distance of [tex]4.8*10^{10} m[/tex]?Generally, the equation for the conservation of energy is mathematically given as
E = (1/2)mv^2 - GmM/r
Where
E_i = E_f
Hence
(1/2)mv_i^2 - [tex]\frac{GmM}{(r_i)}[/tex] = (1/2)mv_f^2 - [tex]\frac{GmM}{(r_f)}[/tex]
[tex]v_f = \sqrt{[(v_i)^2 + 2MG((1/r_f) - (1/r_i))]}[/tex]
Therefore
[tex]v_f = \sqrt{(2.1*10^4)^2 + 2(1.9891 * 10^{31})*(6.67 * 10^{-11})(\frac{1}{4.9} * 10^{10}) - (\frac{1}{2.5} * 10^{11}))} *20.408 *10^{12}[/tex]
[tex]v_f = \sqrt{(441000000) + 435.38 * 10^{7}}[/tex]
vf = 6.92 * 10^(4) m/s
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At what speed, as a fraction of c, must a rocket travel on a journey to and from a distant star so that the astronauts age 15 years while the Mission Control workers on earth age 130 years?
Answer:
The velocity as a fraction of c is 0.986 c m/s
Solution:
As per the question:
Time measured by the astronaut, t = 15 yrs
Time measured in the frame of mission control, t' = 130 yrs
Now,
Using the formula of time dilation:
[tex]t' = \frac{t}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex]
Substituting appropriate values in the above eqn:
[tex]130 = \frac{15}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex]
[tex]\sqrt{1 - \frac{v^{2}}{c^{2}}}= \frac{15}{130}[/tex]
Squaring both the sides we get:
[tex]1 - \frac{v^{2}}{c^{2}}= (\frac{15}{130})^{2}[/tex]
[tex]\frac{v^{2}}{c^{2}} = 1 - (\frac{15}{130})^{2}[/tex]
v = 0.986 c m/s
The rocket must travel at about 99.45% of the speed of light for the astronauts aboard to age 15 years while those on Earth age 130 years, according to the theory of special relativity and the phenomenon called time dilation.
Explanation:The question asks about the speed a rocket needs to travel as a fraction of the speed of light (c) for the astronaut aboard to age 15 years while those back on Earth age 130 years - an application of time dilation in special relativity. The solution to this is in the theory of relativity proposed by Albert Einstein, which states that time passes at different speeds for people depending on their relative motion.
The time dilation formula is given by Δt = γΔτ, where Δτ is the 'proper time' experienced by the moving astronaut, Δt is the time experienced by the stationary observers on Earth, and γ is the Lorentz factor, defined as γ = 1 / sqrt(1 - (v^2/c^2)).
Here, Δt is 130 years, Δτ is 15 years, and we are asked to solve for v/c. Plugging in the given values and solving for v/c, we find that v/c = sqrt(1 - (15/130)^2), approximately equals to 0.9945 or 99.45% of the speed of light. Therefore, the rocket must travel at about 99.45% of the speed of light for the astronaut to age 15 years while those on Earth age 130 years.
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In a double-slit interference experiment, a special lamp emitting yellow light from heated sodium atoms is used to produce an interference pattern on a screen located 1.49 m from a pair of slits separated by 0.12 mm. If the distance between adjacent bright regions in the resulting pattern is 7.32 mm, what is the wavelength (in nm) of the sodium light?
Answer:
λ=589nm
Explanation:
The wavelength of the sodium light can be calculated using the next equation:
[tex] \lambda = \frac{\Lambda D}{L} [/tex]
where Λ: is the distance between the adjacent bright regions, λ: is the wavelength of the sodium light, L: is the distance between slits and screen, and D: is the distance between slits.
[tex] \lambda = \frac{7.32mm \cdot 0.12mm}{1490mm} = 5.89\cdot 10^{-4}mm = 589nm [/tex]
Therefore, the λ of the sodium light is 589 nm.
I hope it helps you!
While painting the top of an antenna 275 m in height, a worker accidentally lets a 1.00 L water bottle fall from his lunchbox. The bottle lands in some bushes at ground level and does not break.
Answer:
The force of impact of the water bottle is F = 13,475 N
Explanation:
Given data,
The height of the antenna, h = 275 m
The mass of the 1 L water bottle, m = 1 kg
Let the bottle moves distance immediately after the impact is, d = 0.2 m
The force exerted by the bottle on the bushes at the ground is given by the formula,
F = mgh / d
Substituting the values
F = 1 x 9.8 x 275 / 0.2
= 13,475 N
The value of the force of impact can be reduced by increasing the value of d, it is like the lowering the hand along with the motion of the ball to catch it thereby reduce the force of impact.
The force of impact of the water bottle is F = 13,475 N
The height of a cylinder is decreasing at a constant rate of 8 inches per minute, and the volume is decreasing at a rate of 161 cubic inches per minute. At the instant when the height of the cylinder is 66 inches and the volume is 919 cubic inches, what is the rate of change of the radius? The volume of a cylinder can be found with the equation V=\pi r^2 h.V=πr 2 h. Round your answer to three decimal places.
Answer:
0.056 inches per minute
Explanation:
dh/dt = 8 inches per minute
dV/dt = 161 cubic inch per minute
h = 66 inches
V = 919 cubic inch
dr/dt = ? rate of change of radius
The volume of cylinder is given by
V = πr²h
where, r be the radius of cylinder
Differentiate both sides with respect to t
dV/dt = πr² x dh/dt + πh x 2r dr/dt .... (1)
When h = 66 inches, V = 919 cubic inches
So, 919 = 3.14 x r² x 66
r = 2.11 inch
Substitute the values in equation (1)
161 = 3.14 x 2.11 x 2.11 x 8 + 3.14 x 66 x 2 x 2.11 x dr/dt
dr/dt = 0.056 inches per minute
The rate of change of the radius of the cylinder, given the provided rates of change of volume and height and current volume and height, is approximately -0.123 inches/minute.
Explanation:
This problem is a classic example of related rates in calculus, specifically focusing on the cylinder. Given the volume of a cylinder is V = πr^2h, we know that the rate at which the volume is changing (dV/dt) is related to the rate at which the radius is changing (dr/dt) and the rate at which the height is changing (dh/dt). In this problem, we are given dV/dt (-161 cubic inches/minute) and dh/dt (-8 inches/minute). Using these given rates, the current volume and height, we can differentiate the volume formula with respect to time to find dr/dt.
Applying the chain rule, we get dV/dt = πr^2 dh/dt + 2πrh dr/dt. Substituting all given values and solving for dr/dt, we find that the rate of change of the radius when the height of the cylinder is 66 inches comes out to be approximately -0.123 inches/minute (rounded to three decimal places).
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Sound waves are created by _____.
a. the compression of air molecules
b. the rarefaction of air molecules.
c. undulating displacement of air molecules caused by pressure changes.
d. None of the answers is correct.
Answer:
C. undulating displacement of air molecules caused by pressure changes.
Explanation:
Sound is a mechanical wave arising from the movement back and forth of the media objects through which the sound wave travels.If a sound wave moves through air from left to right, then particles of air will move both to the right and to the left as the sound wave's energy passes through it.
The sound wave is formed by air molecules that are displaced due to pressure changes.
Therefore the answer is C.
Sound waves are caused by the undulating displacement of air molecules due to pressure changes. Hence, option(c) is correct.
1. When an object like a speaker cone vibrates, it alternately compresses and expands the air around it.
2. The regions where air molecules are pushed together are called compressions, and the regions where they are spread apart are called rarefactions.
3. Sound waves travel through the air as longitudinal waves, consisting of alternating high-pressure (compression) and low-pressure (rarefaction) regions.
4. These pressure changes propagate at a speed of approximately 340 m/s in air, creating the disturbance known as a sound wave.
Compression: Area of high pressure where molecules are close together.Rarefaction: Area of low pressure where molecules are spread apart.In a Compton scattering experiment, a beam of x-rays of wavelength 0.07 nm scatters of nearly free electrons. The scattered radiation is investigated as a function of the angle theta between it and the primary beam. The scattered radiation contains__________.
a) a component shifted in wavelength by an amount that decreases as the scattering angle increases.
b) a component shifted in wavelength by an amount that is independent of the scattering angle.
c) a component shifted in wavelength by an amount that is independent of the wavelength of the primary beam.
d) two components with wavelength shifted up and down by an equal amount.
e) no wavelength other than that of the primary beam.
Answer:
Option (e).
Explanation:
The Compton's equation is:
[tex] \lambda^{'} = \frac{h}{m_{e} c} \cdot (1 - Cos(\theta)) + \lambda = \lambda_{c} (1 - Cos(\theta)) + \lambda [/tex] (1)
where λ': is the wavelength scattered, λ: is the initial wavelength, h: is Planck's constant, [tex] m_{e} [/tex]: is the electron rest mass, c: speed of light, Θ: scattering angle (between λ and λ'), and [tex] \lambda_{c} [/tex]: is a constant known as the Compton wavelength of the electron = 0.00243 nm.
From equation (1), the scattered radiation is directly proportional to the scattering angle, having the maximum value when Θ=90°:
[tex] \theta = 90 ^{ \circ} \rightarrow \lambda^{'} = \lambda_{c} (1-Cos(90)) + \lambda = \lambda_{c} + \lambda [/tex]
And the minimum value when Θ=0°:
[tex] \theta = 0 ^{ \circ} \rightarrow \lambda^{'} = \lambda_{c} (1-Cos(0)) + \lambda = \lambda [/tex]
Hence, the options (a) and (b) are incorrect.
Similarly, we can see from equation (1) that the scattered radiation depends also on the wavelength of the primary beam and no wavelength other than that (since [tex] \lambda_{c} [/tex] is a constant), so the correct option is (e).
Have a nice day!
how many newtons of force will 2 pieces of paper have if one has a charge of .2 Coilombs and the other has a charge of .3 Coulombs and they are .005 meters apart?
What will happen if they are the same charge (+ or - ) Opposite charges ?
1) The electrostatic force between the two pieces of paper is [tex]2.16\cdot 10^{13} N[/tex]
2)
If the charges of the two pieces of paper are both + or both -, they repel each otherIf the charges of one piece of paper is positive and the other one is negative, they attract each otherExplanation:
1)
The electrostatic force between two charged objects is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges
r is the separation between the two charges
In this problem, we have the following situation:
[tex]q_1 = 0.2 C[/tex] is the charge on the first piece of paper
[tex]q_2 = 0.3 C[/tex] is the charge on the second piece of paper
[tex]r=0.005 m[/tex] is their separation
Substituting into the equation, we find the magnitude of the electrostatic force between them:
[tex]F=k\frac{q_1 q_2}{r^2}=(8.99\cdot 10^9) \frac{(0.2)(0.3)}{(0.005)^2}=2.16\cdot 10^{13} N[/tex]
2)
The electrostatic force can be either attractive of repulsive, depending on the relative sign of the charges of the objects involved.
In particular, we have:
If the two charges have same sign (both positive or both negative), the force between them is repulsiveIf the two charges have opposite sign (one positive and one negative), the force between them is attractiveTherefore, in this case:
If the charges of the two pieces of paper are both + or both -, they repel each otherIf the charges of one piece of paper is positive and the other one is negative, they attract each otherLearn more about electric force:
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The red light from a helium-neon laser has a wavelength of 727.7 nm in air. Find the speed, wavelength, and frequency of helium-neon laser light in air, water, and glass. (The glass has an index of refraction equal to 1.50.)
Answer:
speed 2.997 10⁸, 2,254 10⁸, 1,999 10⁸ m / s
[tex]\lambda_{n}[/tex] 727.7, 547, 485 nm
Explanation:
The index of refraction is defined as the relationship between the speed of light in a vacuum and in a material medium
n = c / v
Let's calculate the speed of light in the media
Air
The refractive index is very close to that of the vacuum n = 1,00029
In most experiments they are considered equal
v = c/ n
v = 2,998 10⁸ / 1,00029
v = 2,997 10⁸ m / s
Water
n = 1.33
v = 2.998 10⁸ / 1.33
v = 2,254 10⁸ m / s
Glass
n = 1.50
v = 2,998 10⁸ / 1,50
v = 1,999 10⁸ m / s
Frequency and wavelength are related by the equation
c = λ f
When a beam with a given frequency hits excites the electrons of the material and induces forced oscillations, which has the same frequency of the incident, so the frequency of the beam does not change when passing from one medium to the other.
As speed changes the only way that equality is maintained is that the wavelength changes
[tex]\lambda_{n}[/tex] = λ₀ / n
Air
[tex]\lambda_{n}[/tex] =727.7 nm
Water
[tex]\lambda_{n}[/tex] = 727.7 / 1.33
[tex]\lambda_{n}[/tex] = 547.14 nm
Glass
[tex]\lambda_{n}[/tex] = 727.7 / 1.50
[tex]\lambda_{n}[/tex] = 485.13 nm
A vessel at rest explodes, breaking into three pieces. two pieces, having equal mass, fly off perpendicular to one another with the same speed of 60 m/s. One goes along the negative x-axis and the other along the negative y-axis. The third piece has three times the mass of each other piece. What are the direction and magnitude of its velocity immediately after the explosion? What is the angle with respect to the x-axis? What is the magnitude of its velocity?
Answer:
v₃ = 28.2842 m/s
∅ = 45°
Explanation:
Given info
vi = 0 m/s
m₁ = m₂ = m
m₃ = 3m
mi = m₁ + m₂ + m₃ = m + m + 3m = 5m
v₁x= - 60 m/s
v₁y= 0 m/s
v₂x= 0 m/s
v₂y= - 60 m/s
We can apply the Principle of Conservation of Momentum as follows
pix = pfx ⇒ mi*vix = m₁*v₁x + m₂v₂x + m₃*v₃x
⇒ 5m*(0) = m*(-60) + m*(0) + 3m*v₃x
⇒ 0 = -60*m + 3m*v₃x ⇒ v₃x = 20 m/s (→) (I)
piy = pfy ⇒ mi*viy = m₁*v₁y + m₂v₂y + m₃*v₃y
⇒ 5m*(0) = m*(0) + m*(-60) + 3m*v₃y
⇒ 0 = -60*m + 3m*v₃y ⇒ v₃y = 20 m/s (↑) (II)
then
v₃ = √(v₃x² + v₃y²)
⇒ v₃ = √((20 m/s)² + (20 m/s)²) = 20√2 m/s = 28.2842 m/s
is the magnitude of its velocity immediately after the explosion
∅ = tan⁻¹(v₃y / v₃x)
⇒ ∅ = tan⁻¹(20 m/s / 20 m/s) = tan⁻¹(1) = 45°
is the direction of its velocity immediately after the explosion (the angle with respect to the x-axis)
The third piece of the exploding vessel moves along the positive x and y axes due to the conservation of momentum. The magnitude of its velocity can be derived by applying the Pythagorean theorem to its momentum components and its angle with respect to the x-axis is 45 degrees.
Explanation:This question involves concepts from physics specifically dealing with the conservation of momentum in two-dimensional motion. As the vessel explodes, the total initial momentum of the vessel (at rest) should be equal to the total final momentum, given that no external force is acting on the system.
Since the two pieces with equal mass fly off perpendicular to each other at the same speed, the x and y-components of their momentum will cancel each other out as they are equal in magnitude but opposite in direction. Consequently, the third piece with thrice the mass must account for the total momentum. Therefore, its direction of motion will be along the positive x and y axes.
To figure out the magnitude of velocity, we use the Pythagorean theorem on the x and y momentum components (which are equal in this scenario). We then derive the velocity of the last piece by the ratio of its momentum to its mass. The angle with respect to the x-axis would be 45 degrees as it is moving equally along the positive x and y axes.
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A typical car has 16 L of liquid coolant circulating at a temperature of 95 ∘C through the engine's cooling system. Assume that, in this normal condition, the coolant completely fills the 2.0 −L volume of the aluminum radiator and the 14.0 −L internal cavities within the steel engine. When a car overheats, the radiator, engine, and coolant expand and a small reservoir connected to the radiator catches any resultant coolant overflow. Estimate how much coolant overflows to the reservoir if the system is heated from 95 ∘C to 105 ∘C. Model the radiator and engine as hollow shells of aluminum and steel, respectively. The coefficient of volume expansion for coolant is β=410⋅10−6/C∘
Answer:
[tex]\Delta V=0.0592\ L[/tex]
Explanation:
Given:
Initial temperature of the coolant, [tex]T_f=95^{\circ}C[/tex]final temperature of the coolant, [tex]T_f=105^{\circ}C[/tex]total volume of the coolant, [tex]V_c=16\ L[/tex]coefficient of volume expansion for coolant, [tex]\beta_c=410\times 10^{-6}\ ^{\circ}C^{-1}[/tex]volume of Al radiator, [tex]V_a=2\ L[/tex]volume of steel radiator, [tex]V_s=14\ L[/tex]We have:
coefficient of volume expansion for Aluminium, [tex]\beta_a=75\times 10^{-6}\ ^{\circ}C^{-1}[/tex]
coefficient of volume expansion for steel, [tex]\beta_a=35\times 10^{-6}\ ^{\circ}C^{-1}[/tex]
Now, change in volume of the coolant after temperature rises:
[tex]\Delta V_c=V_c.\beta_c.\Delta T[/tex]
[tex]\Delta V_c=16\times 410\times 10^{-6}\times (105-95)[/tex]
[tex]\Delta V_c=0.0656\ L[/tex]
Now, volumetric expansion in Aluminium radiant:
[tex]\Delta V_a=V_a.\beta_a.\Delta T[/tex]
[tex]\Delta V_a=2\times 75\times 10^{-6}\times (105-95)[/tex]
[tex]\Delta V_a=0.0015\ L[/tex]
Now, volumetric expansion in steel radiant:
[tex]\Delta V_s=V_s.\beta_s.\Delta T[/tex]
[tex]\Delta V_s=14\times 35\times 10^{-6}\times (105-95)[/tex]
[tex]\Delta V_s=0.0049\ L[/tex]
∴Total extra accommodation volume created after the expansion:
[tex]V_X=\Delta V_s+\Delta V_a[/tex]
[tex]V_X=0.0049+0.0015[/tex]
[tex]V_X=0.0064\ L[/tex]
Hence, the volume that will overflow into the small reservoir will be the volume of coolant that will be extra after the expanded accommodation in the radiator.
[tex]\Delta V=\Delta V_c-\Delta V_X[/tex]
[tex]\Delta V=0.0656-0.0064[/tex]
[tex]\Delta V=0.0592\ L[/tex]
When the system is heated from 95 °C to 105 °C, approximately 0.066 liters of coolant would overflow to the reservoir, using the coefficient of volume expansion for the coolant.
Explanation:To estimate how much coolant overflows due to thermal expansion when the system is heated from 95 °C to 105 °C, we use the coefficient of volume expansion for coolant (given as β = 410×10-6/°C). Calculate the change in volume (ΔV) with the formula:
ΔV = βVΔ0ΔT
where:
β is the coefficient of volume expansionVΔ0 is the original volume of the coolant (16 L)ΔT is the change in temperature (105 °C - 95 °C = 10 °C)Plugging in the values, we get:
ΔV = (410×10-6/°C)(16 L)(10 °C)
ΔV = 0.066 L
Therefore, approximately 0.066 liters of coolant would overflow to the reservoir. Since metals like aluminum and steel also expand, this is an approximation as we are not factoring in the expansion of the radiator and engine housing.
What was: The Big Bang (Science)
ANSWER FOR 10 POINTS!!
Answer:The universe sprang into existence as singularity around 13.7 billion years ago.
Explanation:
Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg · m². The father exerts a force on the merry-go-round perpendicular to its 1.50 m radius to achieve a torque of 375 N · m.
(a) Calculate the rotational kinetic energy (in J) in the merry-go-round plus child when they have an angular velocity of 14.8 rpm. J
(b) Using energy considerations, find the number of revolutions the father will have to push to achieve this angular velocity starting from rest. revolutions
(c) Again, using energy considerations, calculate the force (in N) the father must exert to stop the merry-go-round in four revolutions. N
Answer:
Part a)
[tex]KE = 101.4 J[/tex]
Part b)
[tex]N = 0.043 revolution[/tex]
Part c)
F = 2.7 N
Explanation:
Part a)
As we know that the rotational kinetic energy of the merry go round is given as
[tex]KE = \frac{1}{2}I\omega^2[/tex]
[tex]KE = \frac{1}{2}84.4(\omega^2)[/tex]
here we know that
[tex]\omega = 2\pi(\frac{14.8}{60})[/tex]
[tex]\omega = 1.55 rad/s[/tex]
Now we have
[tex]KE = \frac{1}{2}(84.4)(1.55^2)[/tex]
[tex]KE = 101.4 J[/tex]
Part b)
Now we know that work done due to torque = change in kinetic energy
[tex]W = KE_f - KE_i[/tex]
[tex]\tau (2N\pi) = 101.4 - 0[/tex]
[tex]375(2\pi N) = 101.4[/tex]
[tex]N = 0.043 revolution[/tex]
Part c)
In order to stop it in four revolutions we have
[tex]\tau(2\pi N) = \Delta KE[/tex]
[tex]FR(2\pi N) = 101.4[/tex]
[tex]F(1.5)(2\pi \times 4) = 101.4[/tex]
F = 2.7 N
The rotational kinetic energy is 100.64 J. The father needs to push approximately 0.0426 revolutions to start the merry-go-round. The required force to stop the merry-go-round after 4 revolutions is 2.68 N.
Explanation:To answer this question, we first need to understand the concept of rotational kinetic energy, which is given by the equation K.E. = 0.5 * I * ω², where I is the moment of inertia and ω is the angular velocity. Next, we must familiarize ourselves with the concept of torque, which relates to the force applied to create rotational motion.
For the rotational kinetic energy when they have an angular velocity of 14.8 rpm, we first convert the angular velocity to rad/s using the formula ω = 2πN/60 where N is in rpm. Plugging in the given values, we get ω = 1.55 rad/s. We then plug I = 84.4 kg·m² and ω = 1.55 rad/s into the kinetic energy equation to get K.E. = 0.5 * 84.4 kg·m² * (1.55 rad/s)² = 100.64 J. For the number of revolutions required to achieve this angular velocity, we first calculate the work done using the formula Work = Torque * Θ, where Θ is the angular displacement in rad. We equate the work done to the kinetic energy we found in the previous part, resulting in 100.64 J = 375 N·m * Θ. Solving for Θ gives us Θ = 0.268 rad. Finally, we convert this to revolutions using the formula 1 rev = 2π rad, yielding 0.0426 revolutions. To calculate the required force to stop the merry-go-round, we use the same Work = Torque * Θ, but this timewe replaceg Θ with the angular displacement for 4 revolutions (4 * 2π rad = 25.12 rad). Equating the work (100.64 J) to the torque gives us the required force F = 100.64 J / (1.50 m * 25.12 rad) = 2.68 N.Learn more about Rotational Motion here:https://brainly.com/question/37888082
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You are asked to construct a mobile with four equal m = 141 kg masses, and three light rods of negligible mass and equal lengths. The rods are of length 55 cm. (a) At what location on the level 1 rod should the free end of rod 2 be attached?
Answer:
The free end must be attached at a distance of 27.5 cm
Solution:
Mass, m = 141 kg
Length of the rods, L= 55 cm
Now,
As clear from fig. 1:
The free end of the rod 2 must be attached at:
F = 2 W
[tex]WL = W(55 - L)[/tex]
2L = 55
L = 27.5 cm
One way to keep the contents of a garage from becoming too cold on a night when a severe subfreezing temperature is forecast is to put a tub of water in the garage. If the mass of the water is 148 kg and its initial temperature is 22.3°C, how much energy must the water transfer to its surroundings in order to freeze completely?
The energy [tex]6.311 \times 10^{7} \mathrm{J}[/tex] is transferred to the surroundings from water in order to freeze completely.
Explanation:
Water will transfer to surrounding will come from cooling energy from 22.3°C to 0°C and then freezing energy is
[tex]\mathrm{E}_{\mathrm{t}}=\mathrm{E}_{\text {cooling }}+\mathrm{E}_{\text {freezing }}[/tex]
[tex]\mathrm{E}_{\mathrm{t}}=\mathrm{M}\left(\mathrm{C}_{\mathrm{w}} \Delta \mathrm{t}+\mathrm{L}_{\mathrm{f}}\right)[/tex]
We know that,
\mathrm{C}_{\mathrm{W}}=4190 \mathrm{J} / \mathrm{kgk}
[tex]\mathrm{L}_{\mathrm{f}}=333 \times 10^{3} \mathrm{J} / \mathrm{kg}[/tex]
As per given question,
M = 148 kg
[tex]\Delta \mathrm{t}=22.3^{\circ} \mathrm{C}[/tex]
Substitute the values in the above formula,
[tex]\mathrm{E}_{\mathrm{t}}=148\left(4190 \times 22.3+333 \times 10^{3}\right)[/tex]
[tex]E_{t}=148\left(93437+333 \times 10^{3}\right)[/tex]
[tex]E_{t}=148 \times 426437[/tex]
[tex]\mathrm{E}_{\mathrm{t}}=6.311 \times 10^{7} \mathrm{J}[/tex]
The energy [tex]6.311 \times 10^{7} \mathrm{J}[/tex] is transferred to the surroundings from water in order to freeze completely.
To calculate the energy required to freeze a tub of water completely, we need to account for the energy needed to both cool the water to its freezing point and then to convert it to ice. This involves the specific heat capacity of water and the latent heat of fusion.
We use the formula Q = mc extDelta T to calculate the energy to change the temperature of water to 0°C, where:
Q is the heat transfer,m is the mass of the water,c is the specific heat capacity of water (4.18 kJ/kg°C), andextDelta T is the temperature change (22.3°C to 0°C).Then, we calculate the energy needed for the phase change (freezing) using the formula Q = mLf, where:
Lf is the latent heat of fusion of water (333.55 kJ/kg).The total energy is the sum of the energy to cool the water and the energy to freeze it.
Calculations:
To cool to 0°C: Q = (148 kg) * (4.18 kJ/kg°C) * (22.3°C) = 13854 kJTo freeze: Q = (148 kg) * (333.55 kJ/kg) = 49365 kJTotal energy: Q_total = 13854 kJ + 49365 kJ = 63219 kJA(n) _____ generator collects charges on a charge-carrying belt, transfers the charges onto a large metal sphere, and discharges the charges, thereby releasing up to 5,000,000 volts of electricity or more.
Answer:
Explanation:
This generator is called Vande Graff generator.
It collects the charge from the belts and accumulate on a large sphere.
It is used to accelerate the charge particles and based on the principle of corona discharge and charge resides on the outer surface.
A 96.1 kg man sits on the stern of a 6 m long boat. The prow of the boat touches the pier, but the boat isn’t tied. The man notices his mistake, stands up and walks to the boat’s prow, but by the time he reaches the prow, it’s moved 3.66 m away from the pier. Assuming no water resistance to the boat’s motion, calculate the boat’s mass (not counting the man). Answer in units of kg.
Info:
weight: 96.1 kg
Length: 6 m
moved: 3.66 m
Answer:
m L = m d + M d
m L − m d = M d
m (L − x) = M x
M = m (L − x) / x
M = 96.1 kg (6 m − 3.66 m) / 3.66 m
M = 61.44098361 kg
The boats mass.
A 96.1 kg of man sits on a stem of a 6-meter longboat. The boat ouches the pier and but is not tied. The man notices a mistake and walks to the prow of the boat he moves a distance of 3.66 meters away from the pier. Thus assuming the given information the bot offers no resistance.
The mass of the boat will be equal to 61.4 kg.
As all the given information states that the boat is not tied and is at a distance f 3.66 m. The mass of boat will be m L = m d + M dm L − m d = M d m (L − x) = M xM = m (L − x) / x then the M = 96.1 kg (6 m − 3.66 m) / 3.66 mHence the M = 61.44098361 kg.Learn more about the man sits on the stern.
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The bicycle wheel is spinning counterclockwise when viewed from above (so as you view the picture, the right hand side of the wheel is moving away from you). The man and stool are initially at rest. If the man now rotates the wheel’s axis 180˚ so that his right hand is above the wheel, the wheel/man/stool system will:__________
Answer:
a) rotate counterclockwise as viewed from above.
Explanation:
The text tells us it is spinning counterclockwise.
An oscillating block-spring system has a mechanical energyof 1.00 J and amplitude of 10.0 cm and a maximum speed of 1.20m/s. Find (a) the spring constant and (b) the mass of theblock, and (c) the frequency of oscillation.
Answer:
Explanation:
Given
Mechanical Energy of Spring-Block system is 1 J
Maximum Amplitude is [tex]A=10 cm[/tex]
maximum speed [tex]v_{max}=1.2 m/s[/tex]
Suppose [tex]x=A\sin \omega t [/tex]be general equation of motion of spring-mass system
where A=max amplitude
[tex]\omega [/tex]=Natural frequency of oscillation
t=time
[tex]v_{max}=A\omega =1.2[/tex]
[tex]0.1\cdot \omega =1.2[/tex]
[tex]\omega =12 rad/s[/tex]
maximum kinetic Energy must be equal to total Mechanical Energy when spring is un deformed i.e. at starting Position
[tex]\frac{1}{2}mv_{max}^2=1[/tex]
[tex]m=\frac{2}{1.2^2}=\frac{2}{1.44}=1.38 kg[/tex]
Also [tex]\omega [/tex]is also given by
[tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]k=\omega ^2\cdot m[/tex]
where k= spring constant
[tex]k=12^2\cdot 1.38 [/tex]
[tex]k=200 N/m[/tex]
What are the major steps of solar system formation in the correct order?
Answer:(1) Pre-solar nebula
(2). Planet formation
Explanation:
Nebular hypothesis says that the Solar System formed from the gravitational collapse of a fragment of a giant molecular cloud.
One of the collapsing fragments called the pre-solar nebula formed what became the Solar System.
The planet can also be formed by accretion. Accretion is a process in which planets began as dust grains in orbit around the central protostar.
When the terrestrial planets were forming, they remained immersed in a disk of gas and dust.
(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?
(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?
(c) What is the average force of friction if the hill has a slope of 2.5º above the horizontal?
Answer:
Explanation:
a ) Let the height achieved be h .
We shall apply law of conservation of mechanical energy.
1 /2 mv² = mgh
h = v² / 2g
v = 110 km/h
= 30.55 m /s
h = [tex]\frac{30.55\times30.55}{2\times9.8}[/tex]
h = 47.61 m
b )
Kinetic energy of car in the beginning
= 1/2 x 750 x (30.55)²
= 349988.43 J
Potential energy at 22 m height
= 750 x 9.8 x 22
= 161700 J
Energy lost due to frictional force
= 349988.43 - 161700
= 188288.43 J
c )
Distance covered along the slope
d = 22 / sin2.5
= 22 / 0.043619
d = 504.36 m
If F be average frictional force
work done by friction
F x d
= F x 504.36
so
F x 504.36 = 188288.43
F = 188288.43 / 504.36
= 373.32 N
Final answer:
Using the conservation of energy principle, the car's initial kinetic energy is converted into potential energy as it ascends the hill. The potential energy at the given height can be used to determine the thermal energy generated by friction. Calculating the average force of friction requires knowing this thermal energy and the distance traveled, considering the slope angle.
Explanation:
To calculate the height a car can coast up with negligible work done by friction, we can use the principle of conservation of energy. Specifically, the car's initial kinetic energy (due to its initial speed) will be converted into potential energy (due to gaining height) as it coasts uphill until it comes to a stop.
(a) The Height a Car Can Coast Up
Initial kinetic energy (KE) is given by the equation KE = \(\frac{1}{2}mv^2\), where m is the mass of the car and v is its speed. If we convert 110 km/h to meters per second (30.56 m/s), we can calculate the available kinetic energy. The potential energy (PE) at the height h is given by PE = mgh, where g is the acceleration due to gravity (9.81 m/s^2) and h is the height. Setting KE equal to PE, we can solve for h.
(b) Thermal Energy Generated by Friction
If the car actually reaches a height of 22.0 m, we can calculate the difference in the theoretical and actual potential energy to find the thermal energy due to friction. Subtracting the actual potential energy from the total initial kinetic energy gives us the energy lost to friction.
(c) Average Force of Friction
To calculate the average force of friction on a slope of 2.5°, we will use the energy lost to friction divided by the distance traveled along the slope and correct for the slope angle. This gives the component of the friction force that acts parallel to the hill's surface.
An insulated thermos contains 106.0 cm3 of hot coffee at a temperature of 80.0 °C. You put in 11.0 g of ice cube at its melting point to cool the coffee. What is the temperature of the coffee once the ice has melted and the system is in thermal equilibrium? Treat the coffee as though it were pure water. (Answer in °C)
Answer:
the final temperature is T f = 64.977 ° C≈ 65°C
Explanation:
Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:
Q coffee + Q ice = Q surroundings =0 (insulated)
We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).
The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)
The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)
therefore
m co * c co * (T fco - T ico) + m ice * L + m ice * c wat * (T fwa - T iwa) = 0
assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water
d co = dw = 1 gr/cm³
therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr
m co * c wat * (T f - T ico) + m ice * L + m ice * c wat * (T f - T iwa) = 0
m co * c wat * T f+ m ice * c wat * T f = m ice * c wat * T iwa + m co * c wat * Tico -m ice * L
T f = (m ice * c wat * T iwa + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )
replacing values
T f = (11 g * 4.186 J/g°C * 0°C + 106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C + 106 g * 4.186 J/g°C* ) = 64,977 ° C
T f = 64.977 ° C
Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz.
Final answer:
The question is about the angles of maxima and minima in a double-slit interference pattern for two identical loudspeakers placed 1.00 m apart.
Explanation:
The distance between two identical loudspeakers placed 1.00 m apart is given as 'd'.
In the given situation, the listener stands 4.00 m from the wall directly in front of one of the speakers. This distance can be considered as 'L'.
The formula to calculate the distance between adjacent maxima and minima in a double-slit interference pattern is:
d*sin(theta) = m*lambda
- Where 'd' is the distance between the speakers
- 'theta' is the angle at which the maxima or minima occur
- 'm' is the order of the maxima or minima
- 'lambda' is the wavelength of the sound
Applying this formula, we can calculate the angles at which the maxima and minima occur for the given situation.
The phase difference between the waves reaching the observer is approximately 0.66 radians. The frequency closest to 300 Hz for minimal sound due to destructive interference is approximately 286 Hz.
To solve the problem, we need to find the phase difference and the frequency closest to 300 Hz for minimal sound.
a.) The speed of sound, v, is typically about 343 m/s in air. Using the formula for wavelength, λ = v/f, we get:
λ = 343 m/s / 300 Hz λ = 1.143 mFor the listener standing 4.00 m from one speaker and 4.12 m from the second speaker, the difference in the path length, ΔL, is:
ΔL = sqrt((4.00 m)² + (1.00 m)²) - 4.00 m ΔL = 4.12 m - 4.00 m = 0.12 mThe phase difference, φ, is given by:
φ = (2π/λ) * ΔL φ = (2π/1.143 m) * 0.12 m φ ≈ 0.66 radiansb.) For destructive interference, the path difference should be half-integer multiples of the wavelength:
(2n + 1)λ/2 = ΔLSimplifying, we find the frequency:
0.12 m = (2n + 1) * (343 m/s) / (2f)Solving for f when n = 0 (closest to 300 Hz):
f = 343 m/s / (2 * 0.12 m) f ≈ 286 HzThus, the frequency closest to 300 Hz for minimal sound is approximately 286 Hz.
complete question:
Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz.
(a) What is the phase difference in radians between the waves from the speakers when they reach the observer? (Your answer should be between 0 and 2?.)
rad
(b) What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?in Hz
You need to move a 105 kg sofa to a different location in the room. It takes a force of 102 N to start it moving. What is the coefficient of static friction between the sofa and the carpet? (use g = to about -10 m/s/s)
Answer:
0.0971
Explanation:
we know that
[tex]f max = kn[/tex]
that fmax is maximum of static friction , k is coefficient of friction and n is surface reaction force
so we know that from newtons second law
mg=n
so
kmg = 102
k = 102/mg = 102/(10*105) = 0.0971
Final answer:
The coefficient of static friction (μs) between the sofa and carpet is calculated by dividing the force required to start the sofa moving by the normal force, yielding μs = 102 N / 1050 N.
Explanation:
The coefficient of static friction μs is the ratio of the force of static friction Fs to the normal force N. It can be calculated using the equation μs = Fs / N, where Fs is the force required to start moving the object and N is the weight of the object acting perpendicular to the surface. Here, the normal force is equivalent to the weight of the sofa, which is the mass of the sofa multiplied by the acceleration due to gravity (g). With a mass of 105 kg and g approximated as -10 m/s2, the normal force is N = 105 kg × 10 m/s2 = 1050 N. The provided force to start the sofa moving is 102 N, so the coefficient of static friction is μs = 102 N / 1050 N.
A person is using a rope to lower a 5.0-n bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the force exerted by the rope on the bucket?
Answer:
5 N
Explanation:
The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N
Answer:
Magnitude of force on the rope=5N
Explanation:
The bucket is lowered by a constant speed,therefore the tension on the rope must be equal to the weight of the bucket.
The neurons that select a particular motor program are the __________.
Answer:
upper motor neurons in the premotor cortex
Explanation:
Motor neurons are a type of nervous system cells that are located in the brain and in the spinal cord. They have the function of producing the stimuli that cause the contraction of the different muscle groups of the organism. They are therefore essential for daily activities that require muscle contraction: walking, talking, moving hands and in general all body movements.
A canister is released from a helicopter 500 m above the ground. The canister is designed to withstand an impact speed of up to 100 m/s.
(a) Ignoring air resistance, find an equation of the height at any time t.
(b) Find the impact speed of the canister.
(c) Your answer to (b) should be less than 100 m/s.
Rather than just releasing it, let’s see if we could break it by throwing it down with an initial velocity. Re-do parts (a) and (b) with an initial velocity v0, then find the value of v0 required to break the canister.
Answer:
(a) h = 500 - 4.9t²
(b) 98.99 m/s
(c) h = 500 - Vot - 4.9t²
14.14 m/s
Explanation:
(a) You can use the equation of linear motion s = vi.t + 0.5gt²
(b) [tex]v_{f} ^{2} = v_{i} ^{2} +2ax\\v = \sqrt{0^{2} + 2*9.8*500 } = 98.99 m/s[/tex]
(c) [tex]v_{f} ^{2} = v_{i} ^{2} + 2ax\\v_{i} = \sqrt{100^{2} -2*9.8*500}=14.14 m/s[/tex]
The initial velocity mucg be greater than 14.14 m/s to break the cannister
Draw a free-body diagram of the rod ab. Assume the contact surface at b is smooth.
Answer:
See attachment
Explanation:
Which of the following is true of Giovanni Pierluigi da Palestrina’s Pope Marcellus Mass?
a. It demonstrates the Protestant practice of simple congregational singing
b. The text is in Palestrina’s native language of Italian
c. All the parts were originally sung by men and boys
d. It was banned by the Council of Trent 1
Answer: c is true
Explanation: the boy choirstars sang the soprano and alto. Although Palestrina choir is made up of six voice part of soprano, alto, tenor, baritone and bass and the choir sings without instruments.
In the case of Giovanni Pierluigi da Palestrina’s Pope Marcellus Mass, the correct statement is that all the parts were originally sung by men and boys. It was not written in Italian, does not represent Protestant music tactics and was not banned by the Council of Trent.
Explanation:Out of the provided options, the statement that is true of Giovanni Pierluigi da Palestrina’s Pope Marcellus Mass is that all the parts were originally sung by men and boys (option c). The work is a significant one in the history of polyphonic choral music and it marked a departure from the complex style of the early Renaissance period. Instead of Italian, the mass is actually written in Latin, which was the language conventionally used in the liturgy of the Catholic Church (thus discrediting option b). Unlike Protestant music of the time (which negates option a), it does not emphasize congregational singing but instead focuses on the choir. It was also never banned by the Council of Trent. (disproving option d)
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A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is parallel to the ground. The mass of the stuntman is 119 kg, and the coefficient of kinetic friction between the road and him is 0.697. Find the tension in the cable.
Answer:
T = 812.8414 N
Explanation:
Using the law of newton we found the vertical(y) and horizontal(x) forces as:
∑[tex]F_x[/tex] = T - [tex]F_k[/tex] = ma
Where T is the tension, [tex]F_k[/tex] is the friction force, m is the mass of the stuntman and a is the aceleration of the stuntman.
but a is equal to 0 because he is moving at a constant velocity, so:
T - [tex]F_k[/tex] = 0
T = [tex]F_k[/tex]
Also,
[tex]F_k[/tex] = [tex]U_kN[/tex]
where [tex]U_k[/tex] is the coefficient of kinetic friction and N is the normal force.
For find N we use:
∑[tex]F_y =[/tex] N - mg = 0
N = mg
N = (119)(9.8)
N = 1166.2
Finally we solve for T as:
T = [tex]U_kN[/tex]
T = (0.697)(1166.2)
T = 812.8414 N