Answer:
The balloon falls to the ground before the professor gets there. The student is DEFINITELY in for some TROUBLE!
Explanation:
The balloon picks up speed due to gravity and we can calculate the time taken for it to fall to the ground as follows:
Gravity (g) = 9.81 m/s^2
Height or distance (s) = 11 meters
Initial Speed (u) = 0 m/s
[tex]s = u*t + 0.5 * (a*t^2)[/tex]
[tex]11 = 0*t + 0.5 (9.81*t^2)[/tex]
[tex]t= 1.4975 s[/tex]
So we can see that the balloon takes 1.4975 seconds to fall to the ground, and since the professor takes 2 seconds to get to that place, the balloon hits the ground right before the professor gets there.
24 A uniform electric field of magnitude 1.1×104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the electric flux through the sheet?
Answer:
[tex]44,000 Nm^2/C[/tex]
Explanation:
The electric flux through a certain surface is given by (for a uniform field):
[tex]\Phi = EA cos \theta[/tex]
where:
E is the magnitude of the electric field
A is the area of the surface
[tex]\theta[/tex] is the angle between the direction of the field and of the normal to the surface
In this problem, we have:
[tex]E=1.1\cdot 10^4 N/C[/tex] is the electric field
L = 2.0 m is the side of the sheet, so the area is
[tex]A=L^2=(2.0)^2=4.0 m^2[/tex]
[tex]\theta=0^{\circ}[/tex], since the electric field is perpendicular to the surface
Therefore, the electric flux is
[tex]\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C[/tex]
The electric flux through the sheet will be "44,000 Nm²/C".
Electric field:An electric field seems to be an area of space that surrounds an electrically charged particle as well as object whereby an electric charge will indeed experience attraction.
According to the question,
Magnitude of electric field, E = 1.1 × 10⁴ N/C
Length, L = 2.0 m
We know,
The area will be:
→ A = L²
By substituting the value,
= (2.0)²
= 4.0 m²
hence,
The electric flux will be:
→ [tex]\Phi[/tex] = EA Cosθ
By substituting the values,
= [tex](1.1.10^4)(4.0)(Cos 0^{\circ})[/tex]
= 44,000 Nm²/C
Thus the above answer is appropriate.
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An Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. Assume that the rope and pulley are massless and that there is no friction in the pulley. If the masses have the values m 1 = 17.7 kg m1=17.7 kg and m 2 = 11.1 kg, m2=11.1 kg, find the magnitude of their acceleration a a and the tension T T in the rope. Use g = 9.81 m/s 2 .
Explanation:
According to Newton's second law of motion,
[tex]m_{1}g - T = m_{1}a[/tex] ......... (1)
and, [tex]T - m_{2}g = m_{2}a[/tex] ......... (2)
When we add both equations, (1) and (2) then the expression obtained for "a" is as follows.
a = [tex]\frac{m_{1} - m_{2}}{m_{1} + m_{2}} \times g[/tex]
= [tex]\frac{17.7 - 11.1}{17.7 + 11.1} \times 9.8[/tex]
= [tex]\frac{6.6}{28.8} \times 9.8[/tex]
= 2.24 [tex]m/s^{2}[/tex]
Now, putting the value of "a" in equation (1) then we will calculate the tension as follows.
[tex]m_{1}g - T = m_{1}a[/tex]
[tex]17.7 \times 9.8 - T = 17.7 \times 2.24[/tex]
173.46 - T = 39.648
T = 133.812 N
Thus, we can conclude that the magnitude of their acceleration is 2.24 [tex]m/s^{2}[/tex] and the tension T is 133.812 N in the rope.
Answer:
Explanation:
m1 = 17.7 kg
m2 = 11.1 kg
Let a be the acceleration and T be the tension in the string.
use Newton's second law
m1 g - T = m1 x a ....(1)
T - m2 g = m2 x a ..... (2)
Adding both the equations
(m1 - m2) g = ( m1 + m2 ) x a
(17.7 - 11.1 ) x 9.8 = (17.7 + 11.1) x a
64.68 = 28.8 a
a = 2.25 m/s²
Put the value of a in equation (1)
17.7 x 9.8 - T = 17.7 x 2.25
173.46 - T = 39.825
T = 133.64 N
The maximum acceleration a pilot can stand without blacking out is about 7.0 g. In an endurance test for a jet plane's pilot, what is the maximum speed she can tolerate if she is spun in a horizontal circle of diameter 85 m?
The pilot's maximum endurance speed in a circle of 85 m diameter without blacking out can be calculated using the centripetal acceleration formula, taking into account the acceleration limit of 7 g, where 1 g equals 9.80 m/s².
Explanation:The question requires the application of concepts from physics, specifically the idea of centripetal force and acceleration when a body moves in a circular path.
Given that the maximum acceleration a pilot can endure without blacking out is 7 g, and knowing that 1 g is equivalent to 9.80 m/s², we can calculate the maximum speed under these conditions using the formula for centripetal acceleration, which is `a = v² / r`, where `a` is the acceleration, `v` is the speed, and `r` is the radius of the circle.
The diameter of the circle is given as 85 meters, which means the radius `r` will be half of that, 42.5 meters. The maximum acceleration of 7 g translates to 7 × 9.80 m/s², which is 68.6 m/s². We can rearrange the formula to solve for `v`, leading to `v = √(a × r)`.
Substituting the values provided, we get `v = √(68.6 m/s² × 42.5 m)
v = √2923.175 m²/s² ≈ 54.1 m/s
Therefore, the maximum speed the pilot can tolerate while spinning in the horizontal circle is approximately 54.1 m/s, which is equivalent to about 192 km/h.
Two objects are made of the same material, but they have different masses and temperatures.
Part A
If the objects are brought into thermal contact, which one will have the greater temperature change?
a. The one with the lesser mass.
b. The one with the lower initial temperature.
c. The one with the higher initial temperature.
d. The one with the higher specific heat.
e. The one with the greater mass.
f. Not enough information
c. The object with the higher initial temperature will have a greater temperature change when brought into thermal contact with another object.
Explanation:The object with the higher initial temperature will have a greater temperature change when brought into thermal contact with another object. This is because energy transfers from the object with the higher temperature to the object with the lower temperature until they reach thermal equilibrium.
Mass and specific heat do not directly affect the temperature change. The mass only affects the amount of energy transferred, while the specific heat determines how much energy is needed to raise the temperature.
Therefore, the correct answer is c. The one with the higher initial temperature.
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Many spacecraft have visited Mars over the years. Mars is smaller than the earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8 m
Incomplete question.The complete one is here
Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8m/s2. What is the orbital period of a spacecraft in a low orbit near the surface of Mars?
Answer:
[tex]T=5900s=99min[/tex]
Explanation:
Given
[tex]r_{satelite}=r_{mars}=3.37*10^{6}m\\ g_{mars}=3.8m/s^{2}\\[/tex]
To find
orbital period of a spacecraft T
Solution
An the initial calculating is computing the angular velocity of satellite :
[tex]w=\frac{2\pi }{T}\\ w=\frac{2\pi }{110min}(1min/60s)\\ w=9.52*10^{-4}rad/s[/tex]
Computing T
[tex]T=\frac{2\pi }{w}\\ as\\w=\sqrt{\frac{a}{r} }\\ So\\T=\frac{2\pi }{\sqrt{\frac{a}{r} }} \\T=\frac{2\pi }{\sqrt{\frac{3.8m/s^{2} }{3.37*10^{6} m} }}\\T=5900s=99min[/tex]
The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce math]3.15 \times 10^{4} J[/math] of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.
Answer:
The average current that this cell phone draws when turned on is 0.451 A.
Explanation:
Given;
voltage of the phone, V = 3.7 V
electrical energy of the phone battery, E = 3.15 x 10⁴ J
duration of battery energy, t = 5.25 h
The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;
[tex]P = \frac{E}{t}[/tex]
where;
P is power in watts
E is energy in Joules
t is time in seconds
[tex]P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W[/tex]
The average current that this cell phone draws when turned on:
P = IV
[tex]I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A[/tex]
Therefore, the average current that this cell phone draws when turned on is 0.451 A.
The average current that the cell phone draws when turned on is approximately [tex]\(0.450 \, \text{A}\)[/tex].
To find the average current drawn by the cell phone, we need to use the relationship between power, voltage, and current. Let's go step by step.
Step 1: Calculate the power consumed by the cell phone
We already have:
- The total energy provided by the battery: [tex]\( \Delta W = 3.15 \times 10^4 \, \text{J} \)[/tex]
- The total time of operation: [tex]\( \Delta t = 5.25 \, \text{hours} \)[/tex]
First, convert the operation time from hours to seconds:
[tex]\[\Delta t = 5.25 \, \text{hours} \times 3600 \, \text{seconds/hour} = 5.25 \times 3600 = 18900 \, \text{s}\][/tex]
Now, calculate the power P using the energy and time:
[tex]\[P = \frac{\Delta W}{\Delta t} = \frac{3.15 \times 10^4 \, \text{J}}{18900 \, \text{s}} \approx 1.666 \, \text{W}\][/tex]
Step 2: Use the power to find the average current
We know that power P, voltage V, and current I are related by the equation:
[tex]\[P = VI\][/tex]
We can solve for the current I
[tex]\[I = \frac{P}{V}\][/tex]
Given the battery voltage [tex]\(V = 3.70 \, \text{V}\)[/tex] and the power [tex]\(P \approx 1.666 \, \text{W}\)[/tex], we can calculate the current:
[tex]\[I = \frac{1.666 \, \text{W}}{3.70 \, \text{V}} \approx 0.450 \, \text{A}\][/tex]
The complete question is this:
The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce 3.15 x 104 J of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.
Ok, so what I did so far was convert time into seconds and found Power:
t = 18900 s
P = ΔW/Δt =student submitted image, transcription available below= 1.6666 W
I think you have to use the problem : P = VabI = I2R = εI - I2R
Rosalind mentions to Aliyeh the teacher's talking to Jeremy earlier about "escape speed." Aliyeh had never heard of it, either. Jeremy takes another hit of coffee and says, "Maybe that's what happens if we were to take a spaceship to Mars. We'd have to 'escape' from Earth. Yeah, maybe that's it. What speed does it take to escape from the surface of Earth?
Answer:
11.2 Km/Sec
Explanation:
Escape velocity is the velocity that one has to achieve to go out of the gravitational effect of a body. Here they are talking about escaping from the Earth's gravity. To travel to Mars, first we will have to go far enough so that the impact of Earth's gravity is negligible.
The escape velocity ([tex]v_{e}[/tex]) can be calculate using the given formula:
[tex]v_{e} = \sqrt{\frac{2GM}{r}}[/tex]
where,
G = Gravitational Constant
M = Mass of the object from which we need to escape (Here it is the mass of the Earth)
r = distance from the center of that object (In this case it is equal to the radius of Earth)
For Earth, this comes out to be 11.2 KM/Sec
A charge q = 2.00 μC is placed at the origin in a region where there is already a uniform electric field \vec{E} = (100 N/C) \hat{i}E → = ( 100 N / C ) i ^. Calculate the flux of the net electric field through a Gaussian sphere of radius R = 10.0 \text{ cm}R = 10.0 cm centered at the origin. (ε0 = 8.85 × 10-12 C2/N ∙ m2)
Answer:
Φ = 2.26 10⁶ N m² / C
Explanation:
The electric flow is
Ф = E .ds = [tex]q_{int}[/tex] / ε₀
Since we have two components for this flow. The uniform electric field and the flux created by the point charge.
The flux of the uniform electric field is zero since the flux entering is equal to the flux leaving the Gaussian surface
The flow created by the point load at the origin is
Φ = q_{int} /ε₀
Let's calculate
Φ = 2.00 10⁻⁶ /8.85 10¹²
Φ = 2.26 10⁶ N m² / C
A constant force of 12 N in the +x direction acts on a 4 kg block as it moves from the origin to the point (6 − 8) m. How much work is done on the block by this force during this displacement?
Answer:
72 Nm²
Explanation:
work = F . d
work = (12i ,0j) . (6i ,-8j )
work = 72 J
The 12N force is a vector acting in the direction of the positive x axis, so its vector notation is 12i + 0j.
It's not exactly clear what the point's location is but I interpret it to be x= m*6, y= -m*8 (I hope the "m" wasn't a typo). The direction vector from the origin is then m*6i -m*8j
When a force vector F acts in a nonparallel direction d, the work is given by:
W = |F|*|d|*cos(theta) where theta is the angle between the vectors.
alternatively you can use the dot product of the two vectors to get:
W = (12N)*(m*6) + (0N)*(-m*8) = 72 Nm²
Final answer:
The work done on the block by the constant force of 12 N in the +x direction is 24 J.
Explanation:
The work done on the block can be calculated by multiplying the force applied to the block by the displacement of the block in the direction of the force. In this case, the force applied is 12 N in the +x direction and the displacement is (6 - 8) m = -2 m. Since the force and displacement are in opposite directions, the work done will be negative. The formula for work is given by:
Work = Force * Displacement * cos(theta)
where theta is the angle between the force and the displacement.
In this case, the angle between the force and the displacement is 180 degrees, so cos(theta) = -1. Substituting the values into the formula:
Work = 12 N * -2 m * (-1) = 24 J
A charge −1.3 × 10−5 C is fixed on the x-axis at 7 m, and a charge 1 × 10−5 C is fixed on the y-axis at 4 m. Calculate the magnitude of the resultant electric field E~ at the origin. Answer in units of N/C.
Answer:
6104 N/C.
Explanation:
Given:
k = 8.99 × 10^9 Nm2/C^2
Qx = 1.3 × 10^-5 C
rx = 7 m
Qy = 1 × 10−5 C
ry = 4 m
E = F/Q
= kQ/r^2
Ex = (8.99 × 10^9 × 1.3 × 10^−5) ÷ 7^2
= 2385.1 N/C.
Ey = (8.99 × 10^9 × 1.0 × 10^−5) ÷ 4^2
= 5618.75 N/C
Eo = sqrt(Ex^2 + Ey^2)
= sqrt(3.157 × 10^7 + 5.69 × 10^6)
= 6104 N/C.
The magnitude of the electric field at the origin due to the charges fixed at (7 m, 0) and (0, 4 m) is approximately 6107.83 N/C. This was calculated by determining the electric fields from each charge and then using vector addition to find the resultant electric field. The calculation involved using the formula E = k |q| / r² for each charge, followed by determining the resultant using Pythagoras' theorem.
Let's determine the electric field at the origin due to two fixed charges. The first charge is -1.3 × 10⁻⁵ C located at (7 m, 0), and the second charge is 1 × 10⁻⁵C located at (0, 4 m).
The electric field due to a point charge q at a distance r is given by E = k |q| / r², where k = 8.99 × 10⁹ N·m²/C² is the Coulomb constant.
Step-by-Step Calculation
Calculate the distance from each charge to the origin:
Charge 1 at (7 m, 0): r₁ = 7 m
Charge 2 at (0, 4 m): r₂ = 4 m
Compute the electric field due to each charge:
For q₁ = -1.3 × 10⁻⁵ C:
E₁ = k |q₁| / r₁² = (8.99 × 10⁹ N·m²/C²) (1.3 × 10⁻⁵ C) / (7 m)² ≈ 2381.57 N/C along the negative x-direction.
For q₂ = 1 × 10⁻⁵ C:
E₂ = k |q₂| / r₂² = (8.99 × 10⁹ N·m²/C²) (1 × 10⁻⁵ C) / (4 m)² ≈ 5621.88 N/C along the positive y-direction.
Determine the resultant electric field at the origin using vector addition. Since the fields are perpendicular:
Resultant E = √(E₁² + E₂²)
≈ √((2381.57 N/C)² + (5621.88 N/C)²)
≈ √(5662179.6649 + 31643689.7344)
≈ √(37305869.3993)
Resultant E ≈ 6107.83 N/C
The magnitude of the resultant electric field at the origin is therefore approximately 6107.83 N/C.
4. Will a light bulb glow more brightly when it is connected to a battery as shownbelow, when it is connected to an ammeter on the left or to a voltmeter on the right?
Answer:
The diagram is in the attachment
Explanation:
An ammeter is use to know the current flowing in a circuit,
A voltmeter is use to know the potential difference across an element.
The ideal voltmeter and the ideal ammeter has zero internal resistance, so as to drop as little voltage as possible as current flows through it.
1. Let analyse the first circuit i.e the ammeter connection
The ammeter is connected rightly and all the current coming from the battery will flow into the bulb and the bulb will glow bright using only the current from the battery and the ammeter work in the circuit is only to measure the current from the battery.
Now let analyse the second circuit, the voltmeter connection.
This is a wrong connection and if this is done it will act has high resistance to the current flow. The connecting of voltmeter in series is equivalent to connecting a very high resistance in series with the circuit. By this only small insignificant amount of current flow through the circuit and nearly results in an open circuit.
Conclusion,
The first connection of ammeter bulb will glow brightly because it uses up all the current from the battery but for the voltmeter connection the current has been reduced due to the high resistance of voltmeter and thus reduces current.
A 6.93 µF capacitor charged to 94 V and a 1.76 µF capacitor charged to 81 V are connected to each other, with the two positive plates connected and the two negative plates connected. What is the final potential difference across the 1.76 µF capacitor? Answer in units of V.
Answer:
225.56V
Explanation:
Q=cv
6.93 µF capacitor was charged to 94 V, therefore Q₁ =6.93 *10^-6 x 94 =651.42µC
1.76 µF capacitor charged to 81 V, therefore Q₂ =1.76 *10^-6 x 81 =142.56µC
When the capacitor are brought together, the charges on them move to maintain equilibrum, so therefore, each capacitor has 0.5(142.56µC + 651.42µC )= 396.99µC
The final potential difference across the 1.76 µF capacitor = Q/c = 396.99µC/1.76 µF = 225.56V
To find the final potential difference across the 1.76 µF capacitor, use charge conservation by equating the total charge before and after connection. After calculating the charges on each capacitor, rearrange the equation to solve for the final potential difference across the 1.76 µF capacitor.
Explanation:In order to find the final potential difference across the 1.76 µF capacitor, we can use the concept of charge conservation.
First, let's calculate the total charge on the capacitors before they are connected. The charge on a capacitor is given by Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference across the capacitor.
For the 6.93 µF capacitor, the charge is Q1 = (6.93 µF)(94 V) = 650.42 µC. For the 1.76 µF capacitor, the charge is Q2 = (1.76 µF)(81 V) = 142.56 µC.
After connecting the capacitors, the total charge remains the same. So, we have Q1 + Q2 = Qf, where Qf is the final charge on the capacitors. Rearranging the equation, we can find the final potential difference across the 1.76 µF capacitor:
Vf = (Qf - Q1) / C2 = (142.56 µC - 650.42 µC) / (1.76 µF).
Calculating this expression will give us the final potential difference across the 1.76 µF capacitor.
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For the tread on your car tires, which is greater: the tangential acceleration when going from rest to highway speed as quickly as possible or the centripetal acceleration at highway speed? For the tread on your car tires, which is greater: the tangential acceleration when going from rest to highway speed as quickly as possible or the centripetal acceleration at highway speed? The centripetal acceleration at highway speed is greater. The tangential acceleration when going from rest to highway speed as quickly as possible is greater. Under different realistic conditions the answer can be either acceleration.
Answer:
The centripetal acceleration at highway speed is greater.
Explanation:
We assume the motion of the car is uniformly accelerated. Let the highway speed be v.
By the equation of motion,
[tex]v=u+at[/tex]
[tex]a=\dfrac{v-u}{t}[/tex]
u is the initial velocity, a is acceleration and t is time
Because the car starts from rest, u = 0.
[tex]a_T=\dfrac{v}{t}[/tex]
This is the tangential acceleration of the thread of the tire.
The centripetal acceleration is given by
[tex]a_C=\dfrac{v^2}{r}[/tex]
r is the radius of the tire.
Comparing both accelerations and applying commonly expected values to r and t, the centripetal acceleration is seen to be greater. The radius of a tyre is, on the average, less than 0.4 m. Then the centripetal acceleration is about
[tex]a_C=\dfrac{v^2}{0.3}=2.5v^2[/tex]
The tangential acceleration can only be greater in the near impossible condition that the time to attain the speed is on the order of microseconds.
A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the tube has magnitude f. You attach the upper end of a lightweight vertical spring of force constant k to the cap at the top of the tube, and attach the lower end of the spring to the top of the cylinder. Initially the cylinder is at rest and the spring is relaxed. You then release the cylinder. What vertical distance will the cylinder descend before it comes momentarily to rest? Express your answer in terms of the variables m, f, and constants g, k.
The cylinder will descend a distance of f/k before coming momentarily to rest.
Explanation:
When the cylinder is released, it will initially accelerate downwards due to the force of gravity. As it accelerates, the spring attached to it will begin to stretch, exerting an upward force on the cylinder. The cylinder will continue to descend until the force exerted by the spring is equal in magnitude to the force of kinetic friction between the cylinder and the walls of the tube. At this point, the net force on the cylinder will be zero, and it will come momentarily to rest.
To find the distance the cylinder descends before coming to rest, we can equate the force exerted by the spring with the force of kinetic friction:
kx = f
Where k is the force constant of the spring and x is the distance the cylinder has descended. Solving for x gives:
x = f/k
Therefore, the vertical distance the cylinder descends before it comes momentarily to rest is given by x = f/k.
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The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that four of your neighbors are all mowing their lawns using identical "Super-Silent" mowers. When they are each 20 m from your open bedroom window, what is the intensity level of the sound in your bedroom? You can neglect any absorption, reflection, or interference of the sound. The lowest detectable intensity is 1.0 × 10 -12 W/m 2.
Answer: The intensity level of sound in the bedroom is 80dB
Explanation:
Intensity of lawn mower at r=1m is 100dB
Beta1= 10dBlog(I1/Io)
100dB= 10dB log(I1/Io)
10^10= I1/Io
I1= Io(10^10)
10^12)×(10^10)= I1
I1=10^-2w/m^2
Intensity of lawn mower at r=20m
I2/I1=(r1/r2)^2 =(1/20)^2
I2= I1(1/400)
I2=2.5×10^-3W_m^2
Intensity of 4 lown mowers at 20m fro. Window
= 10dBlog(4I2/Io)
= 10^-4/10^-12
=80dB
A water tank is in the shape of a right circular cone with height 18 ft and radius 12 ft at the top. If it is filled with water to a depth of 15 ft, find the work done in pumping all of the water over the top of the tank.
Answer:
210,600πft-lb
Explanation:
Force is a function F(x) of position x then in moving from
x = a to x= b
Work done = [tex]\int\limits^b_a {Fx} \, dx[/tex]
Consider a water tank conical in shape
we will make small horizontal section of the water at depth h and thickness dh and also assume radius at depth h is w
we will have ,
[tex]\frac{w}{12} = \frac{(18-h)}{18} \\w = \frac{2}{3} (18-h)a[/tex]
weight of slice under construction
weight = volume × density × gravitational constant
[tex]weight = \pi \times w^2 \times dh \times 62.4\\= (62.4\pi w^2dh)lb[/tex]
Now we can find work done
[tex]W = \int\limits \, dw\\[/tex]
[tex]\int\limits^{18}_{3} {(62.4\pi } \, dx w^2dh)h\\= 62\pi \frac{4}{9} \int\limits^{18}_{3} {(18-h)^2hdh} \,[/tex]
= [tex]62.4\pi \times\frac{4}{9} (\frac{324}{2} h^2-\frac{36}{3} + \frac{h^4}{4})^1^8_3[/tex]
= 210,600πft-lb
Suppose a student recorded the following data during Step 10 of this experiment,
A 6.2 deg 4.7 deg 3.2 deg 2.4 deg 1.8 deg
Time 2 s 4 s 6 s 8 s 10 s
The student was careful to start taking data when the angular position of the pendulum was zero. Unfortunately he does not have a computer available to fit the data as you will in Step 12, but he knows that the amplitude of oscillations decays exponentially according to A= A0e-(Bt)+ C, where C = 1. Use the data obtained by the student to plot a straight line graph of the appropriate variables on the graph paper, and use it to find values for the parameters A0 and B.
Answer:
B = -0.215 s⁻¹, A₀ = 1.537º
Explanation:
To be able to make this graph we must linearize the data, the best procedure is to calculate the logarithm of the values
A = A₀ [tex]e^{-Bt}[/tex] + C
The constant creates a uniform displacement of the graph if we start the graph at the angle of 1, the constant disappears, this is done by subtracting 1 from each angle, so the equation is
(A-1) = A₀ e^{-Bt}
We do the logarithm
Log (A-1) = log Ao –Bt log e
Make this graph because paper commercially comes in logarithm 10, if we use graph paper if we can calculate directly in base logarithm e, let's perform this calculation
Ln (A-1) = Ln A₀ - Bt
To graph the points we subtract 1 from each angle and calculate the logarithm, the data to be plotted are
θ’= ln (θ -1)
θ(º) t(s) θ'(º)
6.2 2 1.6487
4.7 4 1.3083
3.2 6 0.7885
2.4 8 0.3365
1.8 10 -0.2231
We can use a graph paper and graph on the axis and the primary angle (θ') and on the x-axis time, mark the points and this graph is a straight line, we see that the point for greater time has a linearity deviation , so we will use the first three for the calculations
To find the line described by the equation
y -y₀ = m (x -x₀)
m = (y₂ -y₁) / (x₂ -x₁)
Where m is the slope of the graph e (x₀, y₀) is any point, let's start as the first point of the series
(x₀, y₀) = (2, 1.65)
(x₂, y₂) = (2, 1.65)
(x₁, y₁) = (6, 0.789)
We use the slope equation
m = (1.65 - 0.789) / (2-6)
m = -0.215
The equation is
y - 1.65 = -0.215 (x- 2)
y = -0.215 x +0.4301
We buy the two equations and see that the slope is the constant B
B = -0.215 s⁻¹
The independent term is
b = ln A₀
A₀ = [tex]e^{b}[/tex]
A₀ = [tex]e^{0.43}[/tex]
A₀ = 1.537º
A curve that has a radius of 100 m is banked at an angle of θ = 10.4 ∘ . If a 1200 kg car navigates the curve at 65 km / h without skidding, what is the minimum coefficient of static friction μ s between the pavement and the tires?
The minimum coefficient of static friction between the pavement and the tires is 0.156.
Minimum coefficient of static friction
The minimum coefficient of static friction is calculated by applying Newton's second law of motion in determining the net force.
[tex]F_c = Wsin(\theta) + \mu_s W cos(\theta)\\\\\frac{mv^2}{r} = mg sin(\theta) + \mu_s mg cos(\theta)\\\\\mu_s mg cos(\theta)\ = \frac{mv^2}{r} - mg sin(\theta) \\\\\mu _ s = \frac{mv^2 \ - \ mgr sin(\theta)}{mg rcos(\theta)}[/tex]
where;
m is the mass = 1200 kgv is the speed = 65 km/h = 18.1 m/sr is the radius = 100 mg is gravityθ = 10.4 ∘[tex]\mu _ s = \frac{1200 \times 18.1^2 \ - \ 1200 \times 9.8 \times 100 sin(10.4)}{1200 \times 9.8 \times 100 \times cos(10.4)}\\\\\mu_s = 0.156[/tex]
Thus, the minimum coefficient of static friction between the pavement and the tires is 0.156.
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For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser pointer (λ=680nm) through the slit onto a screen 5.5 m away, a diffraction pattern appears. The bright band in the center of the pattern is 7.8 cm wide. What is the width of the slit?
Answer:
a = 4.8 10⁻⁵ m
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
How the pattern is observed on a distant screen
tan θ = y / L = sin θ / cos θ
Since the angle is very small in these experiments ’we can approximate the tangent function
tan θ = sin θ = y / L
We substitute
a y / L = m λ
The first minimum occurs for m = 1
a = λ L / y
a = 680 10⁻⁹ 5.5 / 0.078
a = 4.8 10⁻⁵ m
How fast do they need to push the mass at the beginning (now at a height equal to the top of the loop-the-loop) to get the mass around the loop-the-loop without falling off the track
Answer:
Check attachment for complete questions, the question is not complete
Explanation:
Check attachment for solution
Complete Question
The complete Question is shown on the first and second uploaded image
Answer:
The speed at which they need to push the mass is v = 13.1 m/s
Explanation:
In order to solve this problem we need to consider conservation of energy when the block is at the top of the inclined plane and also when it is on top of the loop
Now Applying the law of conservation of energy
[tex]mg (2R) + \frac{1}{2} mv^2 = \frac{1}{2} mv_{top}^2 + mg(2R)[/tex]
where [tex]mg (2R)[/tex] is potential energy and [tex]\frac{1}{2} mv^2[/tex] is kinetic energy
and [tex]v_{top}[/tex] is the velocity at the top inclined plane and the top of the loop
Now considering the formula
[tex]\frac{1}{2} mv^2 = \frac{1}{2} mv_{top}^2[/tex]
[tex]v^2 = v_{top}^2[/tex]
[tex]v = v_{top}[/tex]
Now to obtain [tex]v_{top}[/tex]
Looking at the question we can say that the centripetal force that made the block move around loop without leaving the track is q=equivalent to the centripetal force so we have
[tex]mg = \frac{mv_{top}^2}{R}[/tex]
The m would cancel out each other then cross- multiplying
[tex]gR = v^2_{top}[/tex]
[tex]v_{top} = \sqrt{gR}[/tex]
[tex]= \sqrt{(9.8 m/s^2)(17.4\ m)}[/tex]
[tex]= 13.05 m/s[/tex]
[tex]\approx 13.1 m/s[/tex]
A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards th
The angular velocity of the merry-go-round increases as a child walks towards the center, due to the conservation of angular momentum and a decrease in the system's moment of inertia.
Explanation:Angular Momentum and Angular Velocity on a Merry-Go-RoundWhen a child walks from the outer edge of a rotating merry-go-round towards the center, the system's angular momentum remains constant due to the conservation of angular momentum. However, the moment of inertia of the system decreases as the child moves closer to the center. Since angular momentum is the product of the moment of inertia and angular velocity (L = I∙ω), and it is conserved, the angular velocity of the merry-go-round must increase to compensate for the decrease in moment of inertia.
This concept can be compared to a figure skater pulling in their arms to spin faster. As the skater's arms come closer to the body, their moment of inertia decreases, causing an increase in their spinning speed, or angular velocity, with no external torques acting on the system.
The same principle applies to the merry-go-round: as the child moves inward, the system's angular velocity increases to conserve angular momentum. This can be observed in playgrounds and is a practical example of conservation laws in rotational motion.
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As the child walks towards the center of the rotating merry-go-round, the system's angular momentum remains constant due to the conservation of angular momentum. The moment of inertia decreases, causing the angular velocity to increase.
A child is standing on the edge of a merry-go-round that is rotating with frequency f.
As the child walks towards the center of the merry-go-round, the angular momentum of the system (child plus merry-go-round) remains constant.
This is due to the conservation of angular momentum, which states that if no external torque acts on the system, the total angular momentum remains unchanged.
Angular momentum (L) is given by the formula:
L = I * ωwhere I is the moment of inertia and ω is the angular velocity.
As the child moves towards the center, the moment of inertia (I) decreases because it depends on the distance from the axis of rotation. To keep angular momentum constant, the angular velocity (ω) must increase.Suppose the initial angular momentum is Lo = Io * ωo. As the child walks inward, the new moment of inertia In becomes smaller, and thus the new angular velocity ωn must be greater to satisfy Lo = In * ωn.complete question:
A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards the center of the rotating merry-go-round?
Launch the simulation, then answer the question Which statement is not correct regarding the deformation of a circular shaft in torsion?
Cross sections remain flat.
Longitudinal lines remain straight.
Circular sections remain circular.
Radial lines on the sections remain straight.
Answer:
correct answer is (b) Longitudinal lines remain straight
Explanation:
solution
As we know that Deformation of circular shaft in the torsion is associate with twisting of shaft more than an specify with the yielding limit.
so when any angle of twist is obtain in the torsion and that is beyond the specified safety limit of shaft
than that shaft will be fail.
but it does not regain its original shape and it will cause permanent deformation
so that we can say longitudinal lines which is twist, they will not regain to original back position as straight
but they will remain in curved shape.
so here incorrect statement is b Longitudinal lines remain straight
Though torsion theory assumes that cross sections of a torsionally loaded shaft remain flat, in reality, under heavy loading, they can warp and so this statement is not entirely accurate.
Explanation:When a circular shaft deforms under torsion, certain assumptions are made about its deformation according to the torsion theory. These assumptions include:
1) cross sections remain flat and perpendicular to the axis of the shaft,
2) longitudinal lines remain straight,
3) circular sections remain circular, and
4) radial lines on the sections remain straight. However, the statement that is not entirely accurate is that cross sections remain flat. In reality, under severe torsional loading, the cross sections might warp and not remain entirely flat.
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Suppose you want to operate an ideal refrigerator with a cold temperature of − 13.5 °C , and you would like it to have a coefficient of performance of at least 7.75. What is the maximum hot reservoir temperature for such a refrigerator?
Answer:
Th = 50°C
Explanation:
See the attachment below.
K = 7.75
Tc = 13.5°C = 273+13.5 = 286.5K
Answer:
292.984 K or 19.98 °C
Explanation:
Using,
η = Tc/(Th-Tc)................ Equation 1
Where η = coefficient of performance, Tc = cold temperature of the refrigerator, Th = hot temperature of the refrigerator.
Make Th the subject of the equation
η (Th-Tc) = Tc
η Th-η Tc = Tc
Th = (Tc+ηTc)/η ....................... Equation 2
Given: Tc = -13.5°C+273 = 259.5 °C, η = 7.75
Substitute into equation 2
Th = (259.5+7.75×259.5)/7.75
Th = (259.5+2011.125)/7.75
Th = 2270.625/7.75
Th = 292.984 °C or 19.98 °C
Hence the maximum hot reservoir temperature = 292.984 °C or 19.98 °C
xperiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest?
Answer:
a) b) d)
Explanation:
The question is incomplete. The Complete question might be
In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.
a)2 N; 2 N
b) 200 N; 200 N
c) 200 N; 201 N
d) 2 N; 2 N; 4 N
e) 2 N; 2 N; 2 N
f) 2 N; 2 N; 3 N
g) 2 N; 2 N; 5 N
h ) 200 N; 200 N; 5 N
For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces
a) 2+(-2)=0 here minus sign is to show the opposing firection of force
b) 200+(-200)=0
c) 200+(-201)[tex]\neq[/tex]0
d) 2+2+(-4)=0
e) 2+2+(-2)[tex]\neq[/tex]0
f) 2+2+(-3) [tex]\neq[/tex]0; 2+(-2)+3[tex]\neq[/tex]0
g) 2+2+(-5)[tex]\neq[/tex]0; 2+(-2)+5[tex]\neq[/tex]0
h)200 + 200 +(-5)[tex]\neq[/tex]0; 200+(-200)+5[tex]\neq[/tex]0
A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire
Answer:
Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T
Explanation:
Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :
[tex]B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2} } }[/tex] ......(1)
Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.
In this problem,
Current, I = 0.7 A
Length of wire, L = 0.62 m
Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m
Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m
Substitute these values in equation (1).
[tex]B=\frac{4\pi\times10^{-7}\times 0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2} } }[/tex]
B = 6.99 x 10⁻⁶ T
A 60-N box rests on a rough horizontal surface with a coefficient of static friction of 0.5. A horizontal force of 23 N acts on the box but the box is observed to be at rest. What is the value of the static friction force
Answer:23 N
Explanation:
Given
Weight of box [tex]W=60\ N[/tex]
Coefficient of static friction is [tex]\mu _s=0.5[/tex]
Applied force [tex]F=23\ N[/tex]
When Force is applied box is observed to be at rest i.e. static friction is overcoming the applied force.
Thus Static friction [tex]F_s[/tex]=applied force
[tex]F_s=23\ N[/tex]
Although it maximum value can go up to [tex](F_s)_{max}=\mu _sN[/tex]
[tex]F_s=0.5\times 60[/tex]
[tex]F_s=30\ N[/tex]
Compare waves on a pond and electromagnetic waves. 1. A wave on a pond is a mechanical wave which doesn’t require a medium to travel. 2. A wave on a pond is an electromagnetic wave which doesn’t require a medium to travel. 3. A wave on a pond is an electromagnetic wave which requires a medium to travel 4. A wave on a pond is a mechanical wave which requires a medium to travel.
Answer:
The correct answer is the number 4. A wave on a pond is a mechanical wave which requires a medium to travel.
Explanation:
Mechanical waves are those that need a material medium to propagate. The waves of the sea and the waves that we produce on a guitar string, the sound, are examples of mechanical waves. Electromagnetic waves are energetic pulses capable of propagating in a vacuum. This way, a wave on a pond is a mechanical wave which requires a medium to travel.
Final answer:
A wave on a pond is a mechanical wave requiring water to propagate, while electromagnetic waves, such as light, don't require a medium and can travel through a vacuum. Therefore, statement 4 is correct. This reflects fundamental differences in how mechanical and electromagnetic waves propagate and their need for a medium.
Explanation:
To compare waves on a pond with electromagnetic waves, we should understand that a wave on a pond is a mechanical wave which requires a medium to travel (option 4). This is because mechanical waves are disturbances that propagate through a material medium and the pond water serves as this medium. On the other hand, electromagnetic waves do not need a medium; they can travel through a vacuum because they consist of oscillating electric and magnetic fields.
So, the correct statement would be that a wave on a pond is a mechanical wave which requires a medium to travel. Therefore, the waves on a pond are similar to sound waves since both are mechanical and need a medium. In contrast, light waves are an example of electromagnetic waves, which means they can travel through the vacuum of space without any medium.
Furthermore, all electromagnetic waves move at the same speed in a vacuum, which is known as the speed of light. This is a fundamental difference between mechanical and electromagnetic waves since the speed of mechanical waves depends on the medium in which they are traveling. The discovery that electromagnetic waves do not require a medium led to the abandonment of the aether theory, which was invented to provide a medium for light and other electromagnetic wave propagation.
Find the intensity in decibels [i(db)] for each value of i. normal conversation: i = 106i0 i(db) = power saw a 3 feet: i = 1011i0 i(db) = jet engine at 100 feet: i = 1018i0 i(db) =
Answer:
Normal Conversation: i=106i0
i(dB)=60
Power saw a 3 feet: i=1011i0
i(dB)=110
Jet engine at 100 feet: i=1018i0
i(dB)=180
Explanation:
if these are the same as edge, then these are the answers! :)
Normal Conversation: i=106i0
The intensity in decibels is 60,110,180, respectively.
What is sound intensity in decibels?The intensity of a legitimate is the power of the sound in Watts divided by means of the place the sound covers in square meters. The loudness of a valid relates the intensity of any given sound to the intensity at the brink of hearing. its miles are measured in decibels (dB). the brink of human hearing has a depth of approximately.
⇒ normal conversation: i =60(dB)
⇒power saw 3 feet: i =110(dB)
⇒jet engine at 100 feet: i = 180(dB)
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A 6-kg block slides down an incline with a 5-meter vertical drop over an 8-meter horizontal distance. If the block starts from rest and friction is negligible, then what is its kinetic energy at the bottom?
To solve this problem we will apply the concepts related to energy conservation. We know that potential energy is transformed into kinetic and vice versa energy. Since the energy accumulated in the upper part is conserved as potential energy, when the object is thrown all that energy will be converted into kinetic energy. Therefore we will have the following relation,
[tex]KE = PE[/tex]
[tex]KE = mgh[/tex]
Here,
m = mass
g = Gravitational acceleration
h = Height
Replacing,
[tex]KE = (6)(9.8)(5)[/tex]
[tex]KE = 294J[/tex]
Therefore the kinetic energy at the bottom is 294J
A pig enjoys sliding down a ramp. The farmer who owns the pig discovers that if he greases the pig, there is no fiction and the pig enjoys the slide more (happy pig, better bacon). The time required for the pig to reach the bottom of the slide with friction is twice the time without friction. Assumed the pig starts from rest. Derive an expression for the coefficient of friction. (
Answer:
Explanation:
Acceleration without friction on an inclined plane = g sinθ
Acceleration with friction on inclined plane = g sinθ - μ g cosθ
s = 1/2 a t²
For motion on friction-less surface
s = 1/2 g sinθ t₁²
For motion on frictional surface
s = 1/2( g sinθ - μ g cosθ) t₂²
t₂ = 2t₁
( sinθ - μ cosθ) t₂² = sinθ t₁²
( sinθ - μ cosθ) 4t₁² = sinθ t₁²
( sinθ - μ cosθ) 4 = sinθ
4sinθ - sinθ = 4μ cosθ
3sinθ = 4μ cosθ
3 / 4 tanθ =μ
μ = .75 tanθ