Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, [tex]\sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}[/tex]................ii
form i and ii we can write
[tex]v^2= \frac{1}{2} u^2[/tex]
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially at rest. Assuming the collision takes place in nearly empty space (that is no external forces are involved so momentum will be conserved), what are the velocities of the proton and helium after the collision?
Answer:
Velocity of the helium nuleus = 1.44x10⁴m/s
Velocity of the proton = 2.16x10⁴m/s
Explanation:
From the conservation of linear momentum of the proton collision with the He nucleus:
[tex] P_{1i} + P_{2i} = P_{1f} + P_{2f] [/tex] (1)
where [tex]P_{1i}[/tex]: is the proton linear momentum initial, [tex]P_{2i}[/tex]: is the helium nucleus linear momentum initial, [tex]P_{1f}[/tex]: is the proton linear momentum final, [tex]P_{2f}[/tex]: is the helium nucleus linear momentum final
From (1):
[tex] m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f} [/tex] (2)
where m₁ and m₂: are the proton and helium mass, respectively, [tex]v_{1i}[/tex] and [tex]v_{2i}[/tex]: are the proton and helium nucleus velocities, respectively, before the collision, and [tex]v_{1f}[/tex] and [tex]v_{2f}[/tex]: are the proton and helium nucleus velocities, respectively, after the collision
By conservation of energy, we have:
[tex] K_{1i} + K_{2i} = K_{1f} + K_{2f} [/tex] (3)
where [tex]K_{1i}[/tex] and [tex]K_{2i}[/tex]: are the kinetic energy for the proton and helium, respectively, before the colission, and [tex]K_{1f}[/tex] and [tex]K_{2f}[/tex]: are the kinetic energy for the proton and helium, respectively, after the colission
From (3):
[tex] \frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2} [/tex] (4)
Now we have two equations: (2) ad (4), and two incognits: [tex]v_{1f}[/tex] and [tex]v_{2f}[/tex].
Solving equation (2) for [tex]v_{1f}[/tex], we have:
[tex] v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f} [/tex] (5)
From getting (5) into (4) we can obtain the [tex]v_{2f}[/tex]:
[tex] v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0 [/tex]
[tex] v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0 [/tex]
From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:
[tex] v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s} [/tex] (6)
Now, by introducing (6) into (5) we get the proton velocity after the collision:
[tex] v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4} [/tex]
[tex] v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s} [/tex]
The negative sign means that the proton is moving in the opposite direction after the collision.
I hope it helps you!
Final answer:
In an elastic collision between a proton and a helium nucleus, their final velocities can be found using the principles of conservation of momentum and kinetic energy.
Explanation:
In an elastic collision, both momentum and kinetic energy are conserved. To find the final velocities of the proton and helium nucleus after the collision, we can use the principles of conservation of momentum and kinetic energy.
Since the helium nucleus is initially at rest, its momentum before the collision is zero. The momentum before the collision can be calculated as:
Initial momentum proton = proton mass × proton velocity
Final momentum proton + Final momentum helium nucleus = proton mass × proton final velocity + helium nucleus mass × helium nucleus final velocity
Using the conservation of momentum, we can solve for the final velocities of the proton and helium nucleus. The final kinetic energy of the helium nucleus can also be calculated using the equation:
Final kinetic energy helium nucleus = 0.5 × helium nucleus mass × helium nucleus final velocity^2
An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction µ between the object and the surface. The object has speed v when it reaches x=0 and encounters a spring. The object compresses the spring, stops, and then recoils and travels in the opposite direction. When the object reaches x=0 on its return trip, it stops.
1. Find k, the spring constant.
2. Express k in terms of µ, m, g, and v.
Answer:
[tex]k= \frac{8\mu^2mg^2}{v^2}[/tex]
Explanation:
The mass of object is m
The coefficient of friction is μ
the speed of object at x= 0 is v
when the object compress the springs stops so,
The kinetic energy of the spring is equal to and friction work is done
[tex]\frac{1}{2}mv^2 = \frac{1}{2}kx^2+\mu(mg)x[/tex]...........
after that it returns from the rest and finally comes to rest after coming to x=0
position.
So, now the potential energy of the spring is equal to the work done by the friction force.
[tex]\frac{1}{2}kx^2= \mu mgx[/tex]
[tex]\frac{1}{2}kx= \mu mg[/tex]
[tex]x= \frac{2\mu mg}{x}[/tex]
substitute the x value in equation 1
mv^2 = kx^2+2\mu (mg)x
[tex]mv^2= k\frac{2\mu mg}{k}+ 2\mu mg \frac{2\mu mg }{k}[/tex]
solving this we get
[tex]mv^2= \frac{8m^2\mu^2g^2}{k}[/tex]
therefore the force constant
[tex]k= \frac{8\mu^2mg^2}{v^2}[/tex]
Spring Constant is the measure of a spring's stiffness. The value of the spring constant k in terms of µ, m, g, and v is [tex]\dfrac{8m \mu ^2 g^2}{v^2 }[/tex].
What is the spring constant?Spring Constant is the measure of a spring's stiffness.
Given to us
Mass of object = m
The coefficient of friction = μ
The speed of the object at (x= 0) = v
When the object compresses the spring after a distance the spring and the object both stop, therefore, the Kinetic energy of the object transforms to the kinetic energy of the spring during compression and the potential energy when fully compressed, therefore,
The kinetic energy of the object during the compression,
[tex]\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x[/tex]
The kinetic energy of the object after complete compression,
[tex]\dfrac{1}{2}kx^2 = \mu m g x\\\\\dfrac{1}{2}kx = \mu m g \\\\x = \dfrac{2 \mu m g}{k}[/tex]
Substitute the value of x in the kinetic energy of the object during the compression,
[tex]\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x\\\\\dfrac{1}{2} mv^2 = \dfrac{1}{2}k(\dfrac{2 \mu m g}{k})^2 + \mu (mg)(\dfrac{2 \mu m g}{k})\\\\\\mv^2 = \dfrac{8m^2 \mu ^2 g^2}{k}\\\\k= \dfrac{8m^2 \mu ^2 g^2}{mv^2 }\\\\k= \dfrac{8m \mu ^2 g^2}{v^2 }[/tex]
Hence, the value of the spring constant k in terms of µ, m, g, and v is [tex]\dfrac{8m \mu ^2 g^2}{v^2 }[/tex].
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You a building a circuit that requires a 150 pF capacitor, but you only have a drawer full of 100 pF capacitors.
(How) can you make a 150 pF equivalent capacitor?
To develop this problem it is necessary to apply the concepts of sum of capacitors in a circuit, either in parallel or in series.
When capacitors are connected in series, they consecutively add their capacitance.
However, when they are connected in parallel, the sum that is made is that of the inverse of the capacitance, that is [tex]\frac {1} {C}[/tex] where, C is the capacitance.
For the given case, it is best to connect two of these capacitors in series and one in parallel, so
We have three 100 pF capacitors, then
[tex]C_1 = 100 pF\\C_2 = 100 pF\\C_3 = 100 pF\\[/tex]
Here we can see how two capacitors 1 and 2 are in series and the third in parallel.
[tex]\frac{1}{C_{serie}} = \frac{1}{100pF}+\frac{1}{100pF}[/tex]
[tex]\frac{1}{C_{serie}} = \frac{1}{50pF}[/tex]
Investing equality
[tex]C_{serie} = 50pF[/tex]
Adding it in parallel, then
[tex]C_{total} = C_{serie} +C_3[/tex]
[tex]C_{total} = 50pF+100pF[/tex]
[tex]C_{total} = 150pF[/tex]
A bullet is shot straight up in the air with an initial velocity of v=1000 feet per second. If the equation describing the bullets height from the ground is given by f(t)=−16t2+1000t+7, over what interval is the bullet speeding up, and over what interval is it slowing down?
Answer:
It slows down from 0-31.25s
It speeds up from 31.25-62.507s
Explanation:
If we find the maximum of the equation, we will know the moment when it changes direction. It will slow down on the first interval, and speed up on the second one.
[tex]Y = -16t^2+1000t+7[/tex]
[tex]Y' = -32t+1000 = 0[/tex] Solving for t:
t = 31.25s
The bullet will slow down in the interval 0-31.25s
Let's now find the moment when it hits ground:
[tex]Y = 0 = -16t^2+1000t+7[/tex] Solving for t:
t1 = -0.0069s t2 = 62.507s
The bullet will speed up in the interval 31.25-62.507s
Unpolarized light is passed through an optical filter that is oriented in the vertical direction. 1) If the incident intensity of the light is 90 W/m2, what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)
Answer:
45 W/m^2
Explanation:
Intensity of light, Io = 90 W/m^2
According to the law of Malus
[tex]I=I_{0}Cos^{2}\theta[/tex]
The average value of Cos^θ is half
So, I = Io/2
I = 90 /2
I = 45 W/m^2
Unpolarized light, when passed through a polarizer, reduces its intensity by half. So, the intensity if the light that emerges from a vertical filter will be 45 W/m².
Explanation:Given that the incident intensity of the unpolarized light is 90 W/m², when passed through a vertically oriented optical filter, the emerging light will be polarized and will have its intensity halved as it's the property of a polarizing filter to decrease the intensity of unpolarized light by a factor of 2. The formula used in this process is I = Io cos² θ. In the case of unpolarized light passing through a single polarizer, θ is 0. So, the formula simplifies to I = Io/2.
Therefore, the intensity of the light that emerges from the vertically oriented optical filter is: I = 90 W/m² / 2 = 45 W/m².
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A vacuum-filled parallel plate capacitor has an energy density of 0.1 J/m^3 and the plates are separated by 0.2 mm.
An electron is released from rest near negativity charge plate of the capacitor.
How fast is the electron moving when it reaches the positively charged plate?
Answer:
The speed of electron is [tex]3.2\times10^{6}\ m/s[/tex]
Explanation:
Given that,
Energy density = 0.1 J/m³
Separation = 0.2 mm
We need to calculate the potential difference
Using formula of energy density
[tex]J=\dfrac{1}{2}\epsilon_{0}E^2[/tex]
[tex]J=\dfrac{1}{2}\epsilon_{0}\dfrac{V^2}{d^2}[/tex]
[tex]V^2=\dfrac{0.1\times(0.2\times10^{-3})^2\times2}{8.85\times10^{-12}}[/tex]
[tex]V^2=\sqrt{903.95}[/tex]
[tex]V=30.06\ V[/tex]
We need to calculate the speed of electron
Using energy conservation
[tex]U=eV=\dfrac{1}{2}mv^2[/tex]
Put the value into the formula
[tex]1.6\times10^{-19}\times30.06=\dfrac{1}{2}\times9.1\times10^{-31}\times v^2[/tex]
[tex]v^2=\dfrac{1.6\times10^{-19}\times30.06\times2}{9.1\times10^{-31}}[/tex]
[tex]v=\sqrt{1.057\times10^{13}}[/tex]
[tex]v=3.2\times10^{6}\ m/s[/tex]
Hence, The speed of electron is [tex]3.2\times10^{6}\ m/s[/tex]
A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationary nitrogen nucleus of mass 14m. Find the velocities of the proton and the nitrogen nucleus after the collision.
Answer:
[tex]V_p = 267.258 m/s[/tex]
[tex]V_n = 38.375 m/s[/tex]
Explanation:
using the law of the conservation of the linear momentum:
[tex]P_i = P_f[/tex]
where [tex]P_i[/tex] is the inicial momemtum and [tex]P_f[/tex] is the final momentum
the linear momentum is calculated by the next equation
P = MV
where M is the mass and V is the velocity.
so:
[tex]P_i = m(270 m/s)[/tex]
[tex]P_f = mV_P + M_nV_n[/tex]
where m is the mass of the proton and [tex]V_p[/tex] is the velocity of the proton after the collision, [tex]M_n[/tex] is the mass of the nucleus and [tex]V_n[/tex] is the velocity of the nucleus after the collision.
therefore, we can formulate the following equation:
m(270 m/s) = m[tex]V_p[/tex] + 14m[tex]V_n[/tex]
then, m is cancelated and we have:
270 = [tex]V_p[/tex] + [tex]14V_n[/tex]
This is a elastic collision, so the kinetic energy K is conservated. Then:
[tex]K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2[/tex]
and
Kf = [tex]\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2[/tex]
then,
[tex]\frac{1}{2}m(270)^2[/tex] = [tex]\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2[/tex]
here we can cancel the m and get:
[tex]\frac{1}{2}(270)^2[/tex] = [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex]
now, we have two equations and two incognites:
270 = [tex]V_p[/tex] + [tex]14V_n[/tex] (eq. 1)
[tex]\frac{1}{2}(270)^2[/tex] = [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex]
in the second equation, we have:
36450 = [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex] (eq. 2)
from this last equation we solve for [tex]V_n[/tex] as:
[tex]V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }[/tex]
and replace in the other equation as:
270 = [tex]V_p +[/tex] 14[tex]\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }[/tex]
so,
[tex]V_p = -267.258 m/s[/tex]
Vp is negative because the proton go in the -i hat direction.
Finally, replacing this value on eq. 1 we get:
[tex]V_n = \frac{270+267.258}{14}[/tex]
[tex]V_n = 38.375 m/s[/tex]
An audio engineer takes decibel readings at distances of r1 = 13 m and r2 = 24 m from a concert stage speaker during a sound check. When he is r1 from the speaker, the engineer registers a decibel level of β1 = 101 dB on his loudness meter.
Answer
given,
Distance for decibel reading
r₁ = 13 m
r₂ = 24 m
When the engineer is at r₁ reading is β₁ = 101 dB
now, Calculating the Intensity at r₁
Using formula
[tex]\beta = 10 log(\dfrac{I}{I_0})[/tex]
I₀ = 10⁻¹² W/m²
now inserting the given values
[tex]101= 10 log(\dfrac{I}{10^{-12}})[/tex]
[tex]10.1= log(\dfrac{I}{10^{-12}})[/tex]
[tex]\dfrac{I}{10^{-12}}= 10^{10.1}[/tex]
[tex]I =10^{10.1} \times 10^{-12}[/tex]
I = 0.01258 W/m²
now, calculating power at r₁
P₁ = I₁ A₁
P₁ = 0.01258 x 4 π r²
P₁ = 0.01258 x 4 π x 13²
P₁ = 26.72 W
Final answer:
The question deals with acoustics in physics, where an audio engineer uses the decibel scale to measure sound pressure levels and requires an understanding of the inverse square law and logarithmic measurements of sound intensity.
Explanation:
The student's question involves the concept of sound intensity levels and decibels, which are part of the field of acoustics, a topic in physics. Specifically, the student is dealing with the measurement of sound pressure levels at various distances from a sound source, which is an application of the inverse square law in acoustics. The decibel (dB) scale is a logarithmic unit used to measure sound intensity, with 0 dB representing the threshold of human hearing at an intensity of 10-12 W/m².
The loudness experienced by the audio engineer at various distances can be computed using the formula for sound intensity level (SIL), which is SIL = 10 log(I/I0), where I is the sound intensity and I0 is the reference intensity of 10-12 W/m².
The acceleration due to gravity on the Moon's surface is known to be about one-sixth the acceleration due to gravity on the Earth.
Given that the radius of the Moon is roughly one-quarter that of the Earth, find the mass of the Moon in terms of the mass of the Earth.
Answer:
Explanation:
If the acceleration due to gravity on the Moon's surface is [tex]a_M[/tex] and the acceleration due to gravity on the Earth is [tex]a_E[/tex], we can write that:
[tex]a_M=\frac{1}{6}a_E[/tex]
[tex]\frac{a_M}{a_E}=\frac{1}{6}[/tex]
If the radius of the Moon is [tex]r_M[/tex] and the radius of the Earth is [tex]r_E[/tex], we can write that:
[tex]r_M=\frac{1}{4}r_E[/tex]
[tex]\frac{r_M}{r_E}=\frac{1}{4}[/tex]
By Newton's 2nd Law we know that F=ma and using Newton's law of universal gravitation we can calculate the gravitational force an object with mass m experiments from a planet with mass M being at a distance r from it. We will assume our object is on the surface so this distance will be the radius of the planet.
Since the force the object experiments is the force of gravitation we can write, for Earth:
[tex]F=ma_E=\frac{GM_Em}{r_E^2}[/tex]
which means:
[tex]a_E=\frac{GM_E}{r_E^2}[/tex]
[tex]M_E=\frac{a_Er_E^2}{G}[/tex]
And for the Moon:
[tex]F=ma_M=\frac{GM_Mm}{r_M^2}[/tex]
which means:
[tex]a_M=\frac{GM_M}{r_M^2}[/tex]
[tex]M_M=\frac{a_Mr_M^2}{G}[/tex]
We can then write the fraction:
[tex]\frac{M_M}{M_E}=\frac{a_Mr_M^2}{G}\frac{G}{a_Er_E^2}=\frac{a_M}{a_E}(\frac{r_M}{r_E})^2=\frac{1}{6}(\frac{1}{4})^2=0.01[/tex]
Which means:
[tex]M_M=0.01M_E[/tex]
Final answer:
To find the mass of the Moon in terms of the mass of the Earth, we can use the formula for gravitational acceleration and the given ratios between the Moon and Earth's acceleration and radii. Using these values, we can solve for the mass of the Moon in terms of the mass of the Earth.
Explanation:
To find the mass of the Moon in terms of the mass of the Earth, we can use the formula for gravitational acceleration: g = GM/r². Given that the acceleration due to gravity on the Moon is one-sixth that of Earth and the radius of the Moon is one-quarter that of Earth, we can set up the following equation:
(1/6) * (9.8 m/s²) = GM/(1/4 * R)²
Simplifying, we get: 1/6 * 9.8 = GM/(1/16)
Now, we can solve for the mass of the Moon (M) in terms of the mass of the Earth (m):
M = (1/6 * 9.8 * R²)/(1/16 * G)
Substituting the values for R and G, we get:
M = (1/6 * 9.8 * (1/4)²)/(1/16 * 6.67 × 10⁻¹¹)
M ≈ 0.123m
A circular coil 17.7 cm in diameter and containing 18 loops lies flat on the gound. The Earth's magnetic field at this location has magnitude 5.5E-5 T and points into the Earth at an angle of 66.0 degrees below the horizontal line and aimed due north. Determine the torque if a 7.4 A clockwise current passes through the coil.
Answer:
Explanation:
Diameter of the circular coil d = 17.7 cm = 0.177m
Current in the coil, I = 7.4 A
Number of loops in the coil, N = 18
Magnetic field, B = 5.5 x 10-5 T
Angle below line, (θ) = 66°
Write the expression for the torque in the coil
Torque = N x I x B x A x sin (θ)
Where A is the area of the coil and θ° is the angle between the magnetic field and the coil face.
The cross- sectional area of the coil = (pie)(d/2)²
A = (π)(0.177/2)² = 0.0246 m²
The Earth’s magnetic field points into the earth at an angle 66 below the line. The line points towards the north
Hence to find the angle between the magnetic field and the coil face we need to subtract the given angle by 90°
Theta = 90 – 66 = 24°
Substitute the value to find out torque
Torque = 18 x 7.4 x 5.5 x 10⁻⁵ x 0.0246 x sin(24) = 7.33 x 10⁻⁵ N-m
The torque on the coil is 7.33 x 10⁻⁵ N-m
On a frozen pond, a 8.54-kg sled is given a kick that imparts to it an initial speed of ð£0=1.87 m/s. The coefficient of kinetic friction between sled and ice is ðð=0.087. Use work and energy concepts to find the distance the sled moves before coming to rest.
We can find the distance the sled travels before it comes to rest by equating the work done by the frictional force with the initial kinetic energy of the sled. We first calculate the initial kinetic energy with (1/2)mv². The force of friction is given by μmg. The work done by friction is equal to this force times the distance traveled, and we set this equal to the negative initial kinetic energy to account for the sled stopping.
Explanation:The subject of this question is about the concept of work and energy in a physics problem involving a sled on a frozen pond. To solve this, we would use the principle of work and energy where the work done on the sled is equal to the change in its kinetic energy according to the work-energy theorem. Since the sled comes to rest, the final kinetic energy is zero.
First, we calculate the initial kinetic energy of the sled using the formula (1/2)mv², where m is the mass (8.54 kg), and v is the velocity (1.87 m/s). Next, we know that the work done by the friction on the sled is equal to the force of friction times the distance traveled which is equal to the change in kinetic energy. The force of friction can be calculated by multiplying the mass of the sled, the acceleration due to gravity (9.8 m/s²), and the coefficient of kinetic friction (0.087). Set this work equal to the negative initial kinetic energy (as it acts to bring the sled to a stop) and solve for the distance.
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To find the distance the sled moves before coming to rest, we can use the work and energy concepts. Work done by the friction force is equal to the change in kinetic energy. By setting the work done by friction equal to the change in kinetic energy and using the given values, we can solve for the distance the sled moves.
Explanation:To find the distance the sled moves before coming to rest, we will use the work and energy concepts. The work done on an object is equal to the change in its kinetic energy, so we can use the work-energy principle to solve the problem. The work done by the friction force is equal to the force of friction multiplied by the distance the sled moves. Setting the work done by friction equal to the change in kinetic energy, we can solve for the distance the sled moves.
The work done by the friction force is given by the equation:
W = F * d * cos(180°)
where W is the work done, F is the force of friction, d is the distance, and cos(180°) is the angle between the force of friction and the displacement. The work done by the friction force is equal to the negative change in kinetic energy of the sled, which can be calculated as:
ΔKE = -(1/2) * m * v²
where ΔKE is the change in kinetic energy, m is the mass of the sled, and v is the final velocity of the sled.
Setting the work done by friction equal to the negative change in kinetic energy and solving for the distance d:
W = -(1/2) * m * v²
F * d * cos(180°) = -(1/2) * m * v²
Using the given values for the mass of the sled (m = 8.54 kg), the initial velocity (v = 1.87 m/s), and the coefficient of kinetic friction (µ = 0.087), we can solve for the distance d.
First, we need to calculate the magnitude of the friction force using the equation:
F = µ * N
where µ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the sled, which can be calculated as:
N = m * g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the values into the equations:
F = 0.087 * (8.54 kg * 9.8 m/s²) = 7.94 N
Now, we can substitute the values for the force of friction and the change in kinetic energy into the equation:
F * d * cos(180°) = -(1/2) * (8.54 kg) * (1.87 m/s)²
Using the cosine of 180° as -1:
7.94 N * d * (-1) = -(1/2) * (8.54 kg) * (1.87 m/s)²
Simplifying the equation:
-7.94 N * d = -(1/2) * (8.54 kg) * (1.87 m/s)²
Dividing both sides of the equation by -7.94 N:
d = [(1/2) * (8.54 kg) * (1.87 m/s)²] / 7.94 N
Calculating the distance d:
d = 2.008 m
Therefore, the sled moves a distance of 2.008 meters before coming to rest.
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PART ONE: A gust of wind blows an apple from a tree.
As the apple falls, the force of gravity on the
apple is 9.39 N downward, and the force of
the wind on the apple is 1.50 N to the right.
What is the magnitude of the net external
force on the apple?
Answer in units of N.
PART TWO:What is the direction of the net external force
on the apple (measured from the downward
vertical, so that the angle to the right of
downward is positive)?
Answer in units of ◦
Explanation:
Part 1.
The force of gravity on the apple, [tex]F_1=9.39\ N[/tex] (downward)
The force of the wind on the apple, [tex]F_2=1.5\ N[/tex] (right)
Let F is the magnitude of the net force acting on it. The resultant of two vectors is given by :
[tex]F=\sqrt{F_1^2+F_2^2}[/tex]
[tex]F=\sqrt{9.39^2+1.5^2}[/tex]
F = 9.509 N
Part 2.
[tex]tan\theta=\dfrac{F_2}{F_1}[/tex]
[tex]tan\theta=\dfrac{1.5}{9.39}[/tex]
[tex]\theta=9.07^{\circ}[/tex]
So, the direction of the net external force is 9.07 degrees wrt vertical. Hence, this is the required solution.
About once every 30 minutes, a geyser known as Old Faceful projects water 11.0 m straight up into the air. Use g = 9.80 m/s^2, and take atmospheric pressure to be 101.3 kPa. The density of water is 1000 kg/m^3. What is the speed of the water when it emerges from the ground?
Answer:
The speed of the water is 14.68 m/s.
Explanation:
Given that,
Time = 30 minutes
Distance = 11.0 m
Pressure = 101.3 kPa
Density of water = 1000 kg/m³
We need to calculate the speed of the water
Using equation of motion
[tex]v^2=u^2+2gs[/tex]
Where, u = speed of water
g = acceleration due to gravity
h = height
Put the value into the formula
[tex]0=u^2-2\times9.8\times11.0[/tex]
[tex]u=\sqrt{2\times9.8\times11.0}[/tex]
[tex]u=14.68\ m/s[/tex]
Hence, The speed of the water is 14.68 m/s.
A boy exerts a force of 11.8 N at 28.0° above the horizontal on a 6.15 kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.75 m, assuming the sled starts with an initial speed of 0.370 m/s and slides horizontally without friction.
Explanation:
It is given that,
Force exerted by a boy, F = 11.8 N
Angle above the horizontal, [tex]\theta=28^{\circ}[/tex]
Mass of the sled, m = 6.15 kg
Distance moved, d = 2.75 m
Initial speed, u = 0.37 m/s
Let W is the work done by the boy. Using the expression for the work done to find it as :
[tex]W=Fd\ cos\theta[/tex]
[tex]W=11.8\times 2.75\ cos(28)[/tex]
W = 28.65 joules
Let v is the final speed of the sled. Using the work energy theorem to find it. It states that the work done is equal to the change in kinetic energy of an object. It is given by :
[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]
[tex]v=\sqrt{\dfrac{2W}{m}+u^2}[/tex]
[tex]v=\sqrt{\dfrac{2\times 28.65}{6.15}+(0.37)^2}[/tex]
v = 3.07 m/s
So, the final speed of the sled after it moves 2.75 m is 3.07 m/s. Hence, this is the required solution.
The final speed of the sled is approximately 3.073 m/s and the the work done by the boy on the sled is 28.64 J.
To solve this problem, we need to calculate the work done by the boy on the sled and then use that to find the final speed of the sled.
First, let's calculate the work done by the boy. Work (W) is defined as the product of the force (F) applied in the direction of motion and the distance (d) over which the force is applied. The force applied in the direction of motion is the horizontal component of the force exerted by the boy. We can calculate this using trigonometry:
[tex]\[ F_{\text{horizontal}} = F \cdot \cos(\theta) \][/tex]
[tex]\[ F_{\text{horizontal}} = 11.8 \, \text{N} \cdot \cos(28.0^\circ) \][/tex]
[tex]\[ F_{\text{horizontal}} \approx 11.8 \, \text{N} \cdot 0.8829 \][/tex]
[tex]\[ F_{\text{horizontal}} \approx 10.42 \, \text{N} \][/tex]
Now, we can calculate the work done:
[tex]\[ W = F_{\text{horizontal}} \cdot d \][/tex]
[tex]\[ W = 10.42 \, \text{N} \cdot 2.75 \, \text{m} \][/tex]
\[ W \approx 28.64 \, \text{J} \]
Next, we need to find the final speed of the sled. The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy (KE):
[tex]\[ W = \Delta KE \][/tex]
[tex]\[ W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \][/tex]
Given that the initial speed [tex]\( v_i \)[/tex] is 0.370 m/s and the mass [tex]\( m \)[/tex] of the sled is 6.15 kg, we can solve for the final speed [tex]\( v_f \)[/tex]:
[tex]\[ 28.64 \, \text{J} = \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot v_f^2 - \frac{1}{2} \cdot 6.15 \, \text{kg} \cdot (0.370 \, \text{m/s})^2 \][/tex]
[tex]\[ 28.64 \, \text{J} = 3.075 \, \text{kg} \cdot v_f^2 - 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ 28.64 \, \text{J} + 0.405975 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]
[tex]\[ 29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 3.075 \, \text{kg} \cdot v_f^2 \][/tex]
[tex]\[ v_f^2 = \frac{29.046 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{3.075 \, \text{kg}} \][/tex]
[tex]\[ v_f^2 \approx 9.444 \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ v_f \approx \sqrt{9.444 \, \text{m}^2/\text{s}^2} \][/tex]
[tex]\[ v_f \approx 3.073 \, \text{m/s} \][/tex]
The key to making a concise mathematical definition of escape velocity is to consider the energy. If an object is launched at its escape velocity, what is the total mechanical energy Etotal of the object at a very large (i.e., infinite) distance from the planet
Answer:
Zero
Explanation:
Total energy of the EARTH is the sum total of the all the kinetic and the gravitation potential energy of the earth.
Which can be written as
[tex]E_{total} = KE+U[/tex]
[tex]= \frac{1}{2}mv^2+ \frac{GMm}{R}[/tex]
At very large distances the kinetic as well as the gravitational potential energies become zero.
therefore, E_total= 0
When an object is launched at its escape velocity, its total mechanical energy at infinite distance from the planet is zero. The object loses all its kinetic energy, and its potential energy also becomes zero as it reaches infinity.
Escape velocity is the minimum initial velocity required for an object to escape the gravitational pull of a planet or other large body and reach an infinite distance where gravitational force becomes zero. When an object is launched at this escape velocity, its total mechanical energy (Etotal) at an infinite distance is zero. This is because the object gives up all its kinetic energy and the potential energy approaches zero as the distance (r) approaches infinity.
Using conservation of energy, we can illustrate this with the following steps:
Gravitational potential energy (U) at distance r is given by: U = -GMm/r.Kinetic energy (K) is given by: K = 1/2 mv2.At launch (surface of the body, r = R), the total mechanical energy is the sum of kinetic and potential energy: Etotal = K + U = 0 (since U is negative and K must add up to zero mechanical energy).At an infinite distance, the potential energy (U) approaches zero as r → ∞ and the velocity (v) approaches zero (so K also approaches zero with v). Thus, Etotal remains zero.Four point masses, each of mass 1.3 kg are placed at the corners of a rigid massles square of side 1.1 m. Find the moment of inertia of this system about an axis that is perpendicular to the plane of the square and passes through one of the masses.
Answer:
I= 6.292 kg.m²
Explanation:
Given that
m = 1.3 kg
Side of square a= 1.1 m
The distance r
[tex]r=\sqrt{{a^2}+{a^2}}[/tex]
[tex]r={a}{\sqrt 2}[/tex]
[tex]r={1.1}{\sqrt 2}[/tex]
The moment of inertia I
The axis passes through one of the mass then the distance of the that mass from the axis will be zero.
I = m a² + m a² + m r²
By putting the values
I = m a² + m a² + m r²
I =m( 2 a² + r²)
I =1.3( 2 x 1.1² + 2 x 1.1²)
I = 1.3 x 4 x 1.1² kg.m²
I= 6.292 kg.m²
A soap bubble is 103 nm thick and illuminated by white light incident perpendicular to its surface. What wavelength (in nm) and color of visible light is most constructively reflected, assuming the same index of refraction as water (nw = 1.33)?
Answer:
λ = 547.96 nm
Explanation:
given,
thickness of soap bubble = 103 nm
refractive index of thin film = 1.33
using formula of constructive interference
[tex]2 n t = (m + \dfrac{1}{2})\times \lambda[/tex]
t is thickness of the medium
n is refractive index
m = 0,1,2...
now,
[tex]2 n t = (m + 0.5)\times \lambda[/tex]
at m = 0
[tex]2\times 1.33 \times 103 =0.5\times \lambda[/tex]
λ = 547.96 nm
at m = 1
[tex]2 n t = (m + 0.5)\times \lambda[/tex]
[tex]2\times 1.33 \times 103 =1.5\times \lambda[/tex]
λ = 182.65 nm
the only visible light enhanced by the thin film is of wavelength
λ = 547.96 nm
A compound die is used to blank and punch a large washer out of 6061ST aluminum alloy sheet stock 3.2 mm thick. The outside diameter of the washer is 25.0 mm, and the inside diameter is 12.0 mm. Determine the punch and die sizes for (a) the blanking operation, and (b) the punching operation.
Answer:
24.616 mm
12.384 mm
Explanation:
[tex]D_b[/tex] = Blanking die diameter = 25 mm
[tex]D_h[/tex] = Punch diameter = 12 mm
a = Allowance = 0.06 for aluminium
t = Thickness = 3.2 mm
Clearance
[tex]c=at\\\Rightarrow c=0.06\times 3.2\\\Rightarrow c=0.192[/tex]
Diameter of punch is
[tex]d_p=D_b-2c\\\Rightarrow d_p=25-2\times 0.192\\\Rightarrow d_p=24.616\ mm[/tex]
Diameter of blanking punch is 24.616 m
For punching operation
[tex]d_d=D_h+2c\\\Rightarrow d_d=12+2\times 0.192\\\Rightarrow d_d=12.384\ mm[/tex]
The diameter of punching die is 12.384 mm
For the blanking operation, the punch diameter should be 12.0 mm and the die diameter should be 25.0 mm. For the punching operation, the punch diameter should be 12.0 mm and the die diameter should be slightly larger than 12.0 mm, typically by a clearance of 0.05 mm to 0.15 mm per side.
(a) Blanking Operation:
The blanking operation involves cutting the outside profile of the washer from the aluminum sheet stock. The punch and die sizes for this operation are determined by the outside and inside dimensions of the washer.
- The die diameter for the blanking operation should match the outside diameter of the washer, which is 25.0 mm. This is the size of the hole in the die through which the punch will pass to cut the washer's outer edge.
- The punch diameter for the blanking operation should match the inside diameter of the washer, which is 12.0 mm. This is the size of the punch that will push through the sheet stock to create the inner hole of the washer.
(b) Punching Operation:
The punching operation involves cutting the inside hole of the washer after the blanking operation has been completed. The punch and die sizes for this operation are determined by the inside diameter of the washer and the required clearance.
- The punch diameter for the punching operation should also be 12.0 mm, the same as the inside diameter of the washer. This ensures that the punch cuts the material exactly at the desired inner diameter.
- The die diameter for the punching operation should be slightly larger than the punch diameter to allow for clearance. Clearance is necessary to prevent the punch and die from binding together and to allow for the removal of the slug (the piece of material that is punched out). The typical clearance ranges from 0.05 mm to 0.15 mm per side. Therefore, the die diameter would be 12.0 mm plus two times the clearance. If we take the middle value of the clearance range, which is 0.10 mm per side, the die diameter would be 12.0 mm + 2 * 0.10 mm = 12.20 mm.
Three polarizing filters are stacked, with the polarizing axis ofthe second and third filters at angles of 22.2^\circ and 68.0^\circ, respectively, to that of thefirst. If unpolarized light is incident on the stack, the light hasan intensity of 75.5 W/cm^2 after it passes through thestack.
a) If the incident intensity is kept constant, what is theintensity of the light after it has passed through the stack if thesecond polarizer is removed?
Answer:
I₂ = 25.4 W
Explanation:
Polarization problems can be solved with the malus law
I = I₀ cos² θ
Let's apply this formula to find the intendant intensity (Gone)
Second and third polarizer, at an angle between them is
θ₂ = 68.0-22.2 = 45.8º
I = I₂ cos² θ₂
I₂ = I / cos₂ θ₂
I₂ = 75.5 / cos² 45.8
I₂ = 155.3 W
We repeat for First and second polarizer
I₂ = I₁ cos² θ₁
I₁ = I₂ / cos² θ₁
I₁ = 155.3 / cos² 22.2
I₁ = 181.2 W
Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized
I₁ = I₀ / 2
I₀ = 2 I₁
I₀ = 2 181.2
I₀ = 362.4 W
Now we remove the second polarizer the intensity that reaches the third polarizer is
I₁ = 181.2 W
The intensity at the exit is
I₂ = I₁ cos² θ₂
I₂ = 181.2 cos² 68.0
I₂ = 25.4 W
Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of useful mechanical power to the water. The free surface of the upper reservoir is 45 m higher than the surface of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine the irreversible head loss of the system and the lost mechanical power during this process. Take the density of water to be 1000 kg/m3.
According to the information presented, it is necessary to take into account the concepts related to mass flow, specific potential energy and the power that will determine the total work done in the system.
By definition we know that the change in mass flow is given by
[tex]\dot{m} = \rho AV[/tex]
[tex]\dot {m} = \rho Q[/tex]
Remember that the Discharge is defined as Q = AV, where A is the Area and V is the speed.
Substituting with the values we have we know that the mass flow is defined by
[tex]\dot{m} = 1000*0.03[/tex]
[tex]\dot{m} = 30kg/s[/tex]
To calculate the power we need to obtain the specific potential energy, which is given by
[tex]\Delta pe = gh[/tex]
[tex]\Delta pe = 9.8*45[/tex]
[tex]\Delta pe = 441m^2/s^2[/tex]
So the power needed to deliver the water into the storage tank would be
[tex]\dot {E} = \dot{m}\Delta pe[/tex]
[tex]\dot {E} = 30*441[/tex]
[tex]\dot{E} = 13230W = 13.23kW[/tex]
Finally the mechanical power that is converted to thermal energy due to friction effects is:
[tex]\dot{W}_f = \dot{W}_s - \dot{E}[/tex]
[tex]\dot{W}_f 20-13.23[/tex]
[tex]\dot{W} = 6.77kW[/tex]
Therefore the mechanical power due to friction effect is 6.77kW
The irreversible head loss in this pump system is 23 meters and the lost mechanical power during the pumping process is 6.8 kW.
Explanation:In this system, the pump supplies energy to the water and drives it up to the higher reservoir. Out of this energy provided by the pump, part of it is used for lifting the water while some is lost as head loss due to friction and other losses in the system. Using the principle of conservation of energy and the Bernoulli's equation, one can calculate the irreversible head loss and the mechanical power loss.
Given the mechanical power provided by the pump, P = 20 kW, the height difference h = 45 m, the flow rate Q = 0.03 m3/s, and the density of water ρ = 1000 kg/m3, the useful power for lifting the water equals ρghQ=1000x 9.81x 45 x 0.03=13.2 kW. Thus, the lost power due to irreversible head loss equals the total power minus the useful power, or 20 kW - 13.2 kW = 6.8 kW. The corresponding head loss equals the lost power divided by the power used for lifting the water, or h_loss = (6.8 kW)/(ρgQ) = 23 m.
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Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and diameter 25 cm. As this cylinder is turned with a crank, the rope raises the bucket. The mass of a bucket of water is 20 kg. Someone cranks the bucket up and then lets go of the crank, and the bucket of water falls down to the bottom of the well. Without friction or air resistance, what is the angular acceleration of the 50-kg cylinder?
To find the angular acceleration of the cylinder, calculate the moment of inertia of the cylinder, determine the torque due to the falling bucket, and then use Newton's second law for rotation to find the angular acceleration.
Explanation:The subject involves calculating the angular acceleration of a cylinder when the bucket of water tied to a rope around the cylinder is let go. We assume no friction or air resistance in this case.
First, we need to calculate the moment of inertia (I) of the solid cylinder, which can be calculated using the formula I = 0.5 * m * r^2. Here, m is the mass of the cylinder (50 kg) and r is the radius (half of the diameter, 0.125 m). After calculating I, use Newton's second law for rotation τ = I * α, where τ is torque and α is the angular acceleration.
The torque τ can be calculated as the product of the force exerted by the falling bucket (F) and the radius r of the cylinder. The force F is the weight of the falling bucket, m * g, where m is the mass of the bucket (20 kg) and g is acceleration due to gravity (about 9.8 m/s^2). Once you have calculated τ, you can solve for α by rearranging the equation to α = τ/I.
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The angular acceleration of the 50-kg cylinder is found to be approximately 62.7 rad/s².
To find the angular acceleration of the 50-kg cylinder when the bucket of water falls down to the bottom of the well, we will use the concept of rotational dynamics. The moment of inertia (I) of a solid cylinder is given by:
I = (1/2) [tex]\times[/tex] M[tex]\times[/tex] R²
where M is the mass and R is the radius. Given the mass M is 50 kg and the diameter is 25 cm, so the radius R is 0.125 m:
I = (1/2) * 50 kg[tex]\times[/tex] (0.125 m)² = 0.390625 kg m²
The torque (τ) caused by the falling bucket is:
τ = Force[tex]\times[/tex] RadiusThe force exerted by the bucket of water is its weight, F = m [tex]\times[/tex] g, where m is the mass of the bucket and water (20 kg) and g is the acceleration due to gravity (9.8 m/s²):
F = 20 kg [tex]\times[/tex] 9.8 m/s² = 196 NSo,
τ = 196 N [tex]\times[/tex] 0.125 m = 24.5 N mUsing Newton's second law for rotation:
τ = I[tex]\times[/tex] αwhere α is the angular acceleration:
24.5 N m = 0.390625 kg m²[tex]\times[/tex] αSolving for α:
α = 24.5 N m / 0.390625 kg m² ≈ 62.7 rad/s²Therefore, the angular acceleration of the 50-kg cylinder is approximately 62.7 rad/s².
5.00 g of molten NaCl is at its melting temperature. The salt is added to 25.00 mL water at 25.0 °C in a coffee cup calorimeter. The salt becomes solid and the temperature of the water increases to 47.9 °C. a) Calculate the amount of heat transferred.
Answer:
heat absorbed by water = 2.39 KJ
Explanation:
given,
mass of molten NaCl = 5 g
mass of water = 25 g
Temperature of water = 25.0 °C
Temperature increased to = 47.9 °C
heat absorbed by water = m S ΔT
S is specific heat in J/gm °C
Specific heat of water = 4.184 J/gm °C
m is mass of water
heat absorbed by water = 25 x 4.184 x (47.9-25)
= 25 x 4.184 x 22.9
= 2395.34 J
heat absorbed by water = 2.39 KJ
The amount of heat transferred is 2.39 KJ.
What is Heat?
This is the form of energy which is transferred from one body to another as the result of a difference in temperature.
Parameters
mass of molten NaCl = 5 g
mass of water = 25 g
Temperature of water = 25.0 °C
Final Temperature of water = 47.9 °C
Specific heat of water = 4.184 J/gm °C
Heat = m S ΔT
Heat absorbed by water = 25 x 4.184 x (47.9 -25)
= 25 x 4.184 x 22.9 = 2395.34 J
= 2.39 KJ.
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The size (radius) of an oxygen molecule is about 2.0 ×10−10 m. Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures (T= 300K ). Assume that deviatons would be noticeable when volume of the gas per molecule equals the volume of the molecule itself.Express your answer using one significant figure.P = ? Pa
Answer:
[tex]1\times 10^{8}\ Pa[/tex]
Explanation:
k = Boltzmann constant = [tex]1.38\times 10^{-23}\ J/K[/tex]
r = Radius of gas molecule = [tex]2\times 10^{-10}\ m[/tex]
t = Temperature = 300 K
P = Pressure
Volume of gas per molecule is given by
[tex]V=\frac{4}{3}\pi r^3\\\Rightarrow V=\frac{4}{3}\pi (2\times 10^{-10})^3\\\Rightarrow V=3.35103\times 10^{-29}\ m^3[/tex]
From the ideal gas law we have
[tex]PV=kt\\\Rightarrow P=\frac{kt}{V}\\\Rightarrow P=\frac{1.38\times 10^{-23}\times 300}{3.35103\times 10^{-29}}\\\Rightarrow P=123544104.35\ Pa=1\times 10^{8}\ Pa[/tex]
The pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior is [tex]1\times 10^{8}\ Pa[/tex]
The pressure at which the finite volume of oxygen molecules causes deviations from ideal-gas behavior can be estimated using the ideal gas law and considering the volume of a molecule. At ordinary temperatures, the approximate pressure is 2.04 × 10^11 Pa.
Explanation:To estimate the pressure at which the finite volume of oxygen molecules causes noticeable deviations from ideal-gas behavior, we need to equate the volume of gas per molecule to the volume of a molecule. The volume of the gas per molecule is equivalent to the volume of a sphere with a radius of 2.0 × 10-10 m, which is approximately 4.19 × 10-30 m3. Using the ideal gas law, we can calculate the pressure using the formula PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Assuming 1 mole of oxygen gas (which contains 6.022 × 1023 molecules), the number of moles can be calculated by dividing the volume of the gas per molecule by Avogadro's number. The approximate pressure at which deviations from ideal-gas behavior become noticeable is around 2.04 × 1011 Pa (Pascals).
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A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave is 0.4. What is the largest ratio of tensile forces between the two ends of the rope before the rope starts to slide over the sheave?
Answer:3.51
Explanation:
Given
Coefficient of Friction [tex]\mu =0.4 [/tex]
Consider a small element at an angle \theta having an angle of [tex]d\theta [/tex]
Normal Force[tex]=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}[/tex]
[tex]N=T\cdot d\theta [/tex]
Friction [tex]f=\mu \times Normal\ Reaction[/tex]
[tex]f=\mu \cdot N[/tex]
and [tex]T+dT-T=f=\mu Td\theta [/tex]
[tex]dT=\mu Td\theta [/tex]
[tex]\frac{dT}{T}=\mu d\theta [/tex]
[tex]\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta [/tex]
[tex]\frac{T_2}{T_1}=e^{\mu \pi}[/tex]
[tex]\frac{T_2}{T_1}=e^{0.4\times \pi }[/tex]
[tex]\frac{T_2}{T_1}==e^{1.256}[/tex]
[tex]\frac{T_2}{T_1}=3.51[/tex]
What is the rate of change of flux through a circle with radius r1 inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis? Express your answer in terms of the variables R, B, r1, and appropriate constants.
Answer:
[tex]\frac{d\phi}{dt}=\pi r^2\frac{dB}{dt}[/tex]
Explanation:
the rate of the magnetic flux is given by,
[tex]\frac{d\phi}{dt}= \frac{d(BA)}{dt}[/tex]
where B= magnetic strength
A= area of cross section
r= radius of circle
\phi= flux
=[tex]A\frac{dB}{dt}[/tex]
=[tex]\pi r^2\frac{dB}{dt}[/tex]
An electromagnetic wave is propagating in the positive x direction. At a given moment in time, the magnetic field at the origin points in the positive y direction. In what direction does the electric field at the origin point at that same moment?a. Positive xb. Negative xc. Positive yd. Negative ye. Positive zf. Negative z
To solve this exercise it is necessary to take into account the concepts related to the magnetic field and its vector representation through the cross product or vector product.
According to the definition the direction of the electromagnetic wave propagation is given by
[tex]\hat{n} = \vec{E} \times \vec{B}[/tex]
Where,
E = Electric Field
B = Magnetic Field
According to the information provided, the direction of propagation of the electromagnetic wave is on the X axis, which for practical purposes we will denote as [tex]\hat {i}[/tex], on the other hand it is also indicated that the magnetic field is in the Y direction, that for practical purposes we will denote it as [tex]\hat {j}[/tex]. In this way using the previous equation we would have to,
[tex]\hat{n} = \vec{E} \times \vec{B}[/tex]
[tex]\hat{i} = \vec{E} \times \hat{j}[/tex]
The cross product identity is
[tex]\hat{i}=-\hat{k} \times \hat{j}[/tex]
From the equation we can notice that the electric field would be given by,
[tex]\vec{E} = -\hat{k}[/tex]
Therefore the direction of electric field is negative z-axis.
A high-speed drill rotating counterclockwise takes 2.5 s to speed up to 2400 rpm. (a) What is the drill’s angular acceleration?
(b) How many revolutions does it make as it reaches top speed?
Answer:
(a) Angular acceleration will be [tex]100.53rad/sec^2[/tex]
(b) 50 revolution
Explanation:
We have given time t = 2.5 sec
Initial speed of the drill [tex]\omega _0=0rad/sec[/tex]
Speed after 2.5 sec [tex]=2400rpm=\frac{2\pi \times 2400}{60}=251.327rad/sec[/tex]
From first equation of motion we know that
[tex]\omega =\omega _0+\alpha t[/tex]
[tex]251.327 =0+\alpha \times 2.5[/tex]
[tex]\alpha =100.53rad/sec^2[/tex]
(b) From second equation of motion we know that
[tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2[/tex]
[tex]\Theta =0\times 2.5+\frac{1}{2}\times 100.53\times 2.5^2=314.16rad=\frac{314.16}{2\pi }=50revolution[/tex]
A small car and a heavy pickup truck are both out of gas. The truck has twice the mass of the car. After you push both the car and the truck for the same amount of time with the same force, what can you say about the momentum and kinetic energy (KE) of the car and the truck? Ignore friction.
a. They have the same momentum, but the car has more kinetic energy than the truck.
b. They have the same momentum and the same KE.
c. They have the same kinetic energy, but the truck has more momentum than the car.
d. The car has more momentum and more KE than the truck.
e. The truck has more momentum and more KE than the car.
Answer:
They have the same momentum, but the car has more kinetic energy than the truck.
Explanation:
Let [tex]m_c[/tex] is the mass of the car and [tex]m_t[/tex] is the mass of truck such that,
[tex]m_t=2m_c[/tex]
You push both the car and the truck for the same amount of time with the same force. Momentum of an object is given by :
[tex]p=mv=F\times t[/tex]
Since, force and time are same, so they have same momentum. The kinetic energy of an object is given by :
[tex]K=\dfrac{1}{2}mv^2[/tex]
Since, the mass of truck is more, it will have maximum kinetic energy. So, the correct option is (a) "They have the same momentum, but the car has more kinetic energy than the truck".
The car has more momentum and more KE than the truck because car mass is lower so it moves fast as compared to truck.
Relationship between mass, momentum and kinetic energyMomentum is the product of its mass and velocity. When we compare car with truck , car has a lower mass so when the mass is lower the velocity will be higher so car has more momentum.
We also know that kinetic energy also depends on mass so less mass have more kinetic energy.
Learn more about kinetic energy here: https://brainly.com/question/20658056
A. Situation A: A 20-kg block is pressed against a relaxed spring on a frictionless surface. The spring is compressed by ∆x = 10 cm from its equilibrium position x = x 0 e . The block is released and the spring uncompresses. The speed of the block at x = x 0 e is va,f = 5.0 m/s.
B. Situation B: The same 20-kg block is pressed against two relaxed springs that are attached in series on a frictionless surface. (These two springs are both identical to the spring in Situation A). The entire spring assembly is compressed by ∆x = 10 cm from its equilibrium position x = x 0 e . The block is released and the spring assembly uncompresses. The speed of the block at x = x 0 e is vb,f Question:
What is the value of vb,f ?
(A) 3.5 m/s
(B) 2.5 m/s/s
(C) 5.0 m/s
(D) 6.4 m/s
(E) 7.0 m/s.
Final answer:
In Situation B, the speed of the block at x = x0e is 3.5 m/s.
Explanation:
In Situation A, a 20-kg block is pressed against a single spring and released. The speed of the block at x = x0e is va,f = 5.0 m/s. In Situation B, the same 20-kg block is pressed against two identical springs in series. The entire spring assembly is compressed by Δx = 10 cm. The question asks for the speed of the block at x = x0e, which we can find using the principle of conservation of mechanical energy.
The speed of the block in Situation B, vb,f, can be found as follows:
Calculate the potential energy stored in the spring assembly when it is compressed by Δx = 10 cm. This is done using the formula U = (1/2)k(Δx)^2, where k is the spring constant.Use the principle of conservation of mechanical energy to relate the potential energy of the spring assembly to the kinetic energy of the block at x = x0e. This can be written as (1/2)mvb,f^2 = U.Solve for vb,f to find the speed of the block at x = x0e.Solving this equation gives us vb,f = 3.5 m/s.
A double-slit experiment is performed using a helium-neon laser, with wavelength of 632.8 nm. The slit distance is 0.5 mm, and the distance to the screen is 1.8 m.
What is the width of the central bright fringe?
Answer:
Width of central bright fringe will be 0.00228 m
Explanation:
We have given wavelength of light [tex]\lambda =632.8nm=632.8\times 10^{-9}m[/tex]
Distance between the slits [tex]d=0.5mm=0.5\times 10^{-3}m[/tex]
Distance between slit and screen D = 1.8 m
We have to find the width
We know that width is given by
[tex]width=\frac{\lambda D}{d}=\frac{632.8\times 10^{-9}\times 1.8}{0.5\times 10^{-3}}=0.00228m[/tex]
So width of central bright fringe will be 0.00228 m
Applying the given values to the formula for the width of the central bright fringe in a double-slit interference pattern yields a result of approximately 4.556 × 10^(-3) m.
The width of the central bright fringe in a double-slit interference pattern can be calculated using the formula:
w = λL / d
where:
w is the width of the central bright fringe,
λ is the wavelength of the light,
L is the distance from the slits to the screen, and
d is the distance between the slits.
In this case:
λ = 632.8 nm = 6.328 × 10^(-7) m,
L = 1.8 m, and
d = 0.5 mm = 5 × 10^(-4) m.
Plug these values into the formula:
w = (6.328 × 10^(-7) m × 1.8 m) / (5 × 10^(-4) m)
Calculating this expression gives:
w ≈ 4.556 × 10^(-3) m
So, the width of the central bright fringe is indeed approximately 4.556 × 10^(-3) m
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