This statement is true.
Yes, supervisors can face disciplinary action for engaging in retaliation. They can be held accountable if they retaliate against an employee for reporting violations or participating in protected activities. Penalties can vary from warnings to termination.
Explanation:True. Supervisors, just like any other employees, are subject to disciplinary action for engaging in retaliation. For instance, if a supervisor retaliates against an employee for reporting a violation of company policies or for engaging in protected activities like organizing or supporting a labor union, they can be held accountable. Disciplinary actions can range from written warnings to termination, depending on the severity of the retaliation. Therefore, it is essential for supervisors to respect the rights of the employees and abide by all workplace regulations to maintain a safe and fair environment.
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An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.
The velocity of the fish relative to water is approximately 10.44 m/s when it hits the water. The fish inherits the eagle's horizontal velocity at the start and then accelerates downward due to gravity.
Explanation:The eagle's speed is given as 3.00 m/s horizontally. Therefore, when the fish falls, it also initially has a horizontal velocity of 3.00 m/s because it was brought to that speed by the eagle. The vertical component of the fish's velocity can be calculated using the second equation of motion: vf = vi + a*t, which simplifies to vf = g*t in the absence of initial vertical velocity. The time 't' it takes for the fish to hit the water can be calculated using the equation d = 0.5*a*t2, which gives t = √(2d/g). Substituting 5.00 m for 'd' and 9.81 m/s2 (approx gravitational acceleration) for 'g', we get t ≈ 1.017 seconds. Replacing 't' in the velocity equation with 1.017 seconds, which results in a vertical velocity of 9.965 m/s. The resultant velocity of the fish relative to the water when it hits would then be calculated using Pythagoras theorem since the horizontal and vertical movements are independent and at right angles to each other, giving a result of √((3.00 m/s)2 + (9.965 m/s)2) = 10.44 m/s.
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The velocity of the fish relative to the water when it hits the water is approximately 10.36 m/s at an angle of about 73.74° below the horizontal.
To calculate the velocity of the fish relative to the water when it hits the water, we need to analyze both the vertical and horizontal components of its motion separately.
Given Data:
Horizontal speed of the eagle (and the fish when dropped): [tex]v_{horizontal} = 3.00 \, \text{m/s}[/tex]
Height from which the fish falls: [tex]h = 5.00 \, \text{m}[/tex]
Step 1: Calculate the time it takes for the fish to fall 5m.
To find the time of fall, we can use the kinematic equation for vertical motion:
[tex]h = \frac{1}{2} g t^2[/tex]
where:
[tex]h[/tex] is the height (5.00 m),
[tex]g[/tex] is the acceleration due to gravity (approximately [tex]9.81 \, \text{m/s}^2[/tex]),
[tex]t[/tex] is the time in seconds.
Rearranging the equation to solve for [tex]t[/tex], we have:
[tex]t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5.00}{9.81}} \approx \sqrt{1.02} \approx 1.01 \, \text{s}[/tex]
Step 2: Calculate the vertical velocity of the fish just before hitting the water.
The vertical velocity can be calculated using:
[tex]v_{vertical} = g t[/tex]
Substituting the time we found:
[tex]v_{vertical} = 9.81 \, \text{m/s}^2 \times 1.01 \, \text{s} \approx 9.91 \, \text{m/s}[/tex]
Step 3: Combine the horizontal and vertical velocities to find the resultant velocity.
The horizontal velocity is constant at 3.00 m/s, and the vertical velocity just before impact is approximately 9.91 m/s. We can use the Pythagorean theorem to find the magnitude of the resultant velocity:
[tex]v_{resultant} = \sqrt{v_{horizontal}^2 + v_{vertical}^2}[/tex]
Substituting the values:
[tex]v_{resultant} = \sqrt{(3.00)^2 + (9.91)^2} = \sqrt{9 + 98.20} = \sqrt{107.20} \approx 10.36 \, \text{m/s}[/tex]
Step 4: Find the direction of the velocity relative to the horizontal using tangent.
The angle [tex]\theta[/tex] can be calculated using:
[tex]\theta = \tan^{-1}\left(\frac{v_{vertical}}{v_{horizontal}}\right)[/tex]
Substituting our values:
[tex]\theta = \tan^{-1}\left(\frac{9.91}{3.00}\right) \approx \tan^{-1}(3.30) \approx 73.74^\circ[/tex]
You carry a 7.0-kg bag of groceries 1.2 m above the ground at constant speed across a 2.7m room. How much work do you do on the bag in the process? (A) 157J (B) 0.00J (C) 185J (D) 82J
Final answer:
In Physics, work is done when a force causes a displacement. Since the bag of groceries is moved horizontally at constant speed and there's no vertical displacement or horizontal force in the direction of motion, the work done on the bag is 0 joules.
Explanation:
The question involves the concept of work in Physics, specifically related to the work-energy principle. In this scenario, the bag of groceries is being carried across the room at a constant speed and constant height. According to the scientific definition of work, work is done when a force causes a displacement in the direction of the force. The formula for work is W = force x displacement. Here, the only forces doing work would be if the bag was lifted or if it was accelerated. Since the bag is being moved at a constant speed and not being lifted any further, there is no work done in the direction of motion, and the vertical height does not change. Therefore, despite the effort you feel you are exerting, the work done on the bag with regard to Physics is 0 joules
Answer choice: (B) 0.00J.
A grandfather clock keeps time using a pendulum consisting of a light rod connected to a small heavy mass. With a rod of length L, the period of oscillation is 2.00 s. What should the length of the rod be for the period of the oscillations to be 1.00 s?
Answer:
The length of the rod should be
[tex]\frac{L}{4} \\ [/tex]
Explanation:
Period of simple pendulum is given by
[tex]T=2\pi\sqrt{\frac{l}{g}} \\ [/tex]
We have
[tex]\frac{T_1^2}{T_2^2}=\frac{l_1}{l_2}\\\\\frac{2^2}{1^2}=\frac{L}{l_2}\\\\l_2=\frac{L}{4} \\ [/tex]
The length of the rod should be
[tex]\frac{L}{4} \\ [/tex]
A small sphere with a mass of 441 g is moving upward along the vertical +y-axis when it encounters an electric field of 5.00 N/C iˆ. If, due to this field, the sphere suddenly acquires a horizontal acceleration of 13.0 m/s2 iˆ, what is the charge that it carries?
Answer:
Charge, q = 1.15 C
Explanation:
It is given that,
Mass of sphere, m = 441 g = 0.441 kg
Electric field, E = 5 N/C
Due to this field, the sphere suddenly acquires a horizontal acceleration of 13.0 m/s² such that,
[tex]ma=qE[/tex]
[tex]q=\dfrac{ma}{E}[/tex]
[tex]q=\dfrac{0.441\ kg\times 13\ m/s^2}{5\ N/C}[/tex]
q = 1.146 C
or
q = 1.15 C
So, the charge carried by the small sphere is 1.15 C. Hence, this is the required solution.
The electric field accounts for the unbalanced motion of the sphere. The electrostatic forces balance the linear force, for which the charge acquired by the sphere is 1.146 C.
What is an electric field?The region or space where a charged particle experiences the electrostatic force of attraction or repulsion, is known as an electric field.
Given data:
The mass of the small sphere is, m = 441 g = 0.441 kg.
The magnitude of the electric field is, E = 5.00 N/C.
The magnitude of horizontal acceleration is, a = 13.0 m/s².
Clearly, under the influence of an electric field, the electrostatic force balances the linear force. Then,
Fe = FL
[tex]qE = ma[/tex]
Here,
q is the magnitude of charge.
Solving as,
[tex]q \times 5.0 = 0.441 \times 13\\\\q = \dfrac{0.441 \times 13}{5.0}\\\\q = 1.146 \;\rm C[/tex]
Thus, we can conclude that the magnitude of charge carried by the sphere is 1.146 C.
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g What is the specific heat of silver? The molar heat capacity of silver is 25.35 J/mol⋅∘C. How much energy would it take to raise the temperature of 9.00 g of silver by 18.3 ∘C? Express your answer with the appropriate units.
The specific heat of silver is 0.235 J/g⋅∘C. It would take 39.1 J of energy to raise the temperature of 9.00 g of silver by 18.3 ∘C.
Explanation:The specific heat of silver can be calculated using the information given. The molar heat capacity of silver is 25.35 J/mol⋅∘C. To find the specific heat, we need to convert the molar heat capacity to g⋅∘C. The molar mass of silver is 107.87 g/mol. Therefore, the specific heat of silver is 0.235 J/g⋅∘C.
To calculate the amount of energy required to raise the temperature of 9.00 g of silver by 18.3 ∘C, we can use the formula Q = mcΔT, where Q is the energy, m is the mass, c is the specific heat, and ΔT is the change in temperature. Plugging in the values, we get Q = (9.00 g)(0.235 J/g⋅∘C)(18.3 ∘C) = 39.1 J.
Therefore, it would take 39.1 J of energy to raise the temperature of 9.00 g of silver by 18.3 ∘C.
Two balls with masses of 2.10 kg and 6.50 kg travel toward each other at speeds of 13.0 m/s and 4.10 m/s, respectively. If the balls have a head-on inelastic collision and the 2.10-kilogram ball recoils with a speed of 8.20 m/s, how much kinetic energy is lost in the collision?
Answer:
137 J
Explanation:
Momentum is conserved, so:
(2.10)(13.0) + (6.50)(-4.10) = (2.10)(-8.20) + (6.50) v
v = 2.75 m/s
Energy before collision:
E = 1/2 (2.10) (13.0)² + 1/2 (6.50) (4.10)²
E = 232 J
Energy after collision:
E = 1/2 (2.10) (8.20)² + 1/2 (6.50) (2.75)²
E = 95.2 J
Energy lost in collision:
232 J - 95.2 J = 137 J
Final answer:
A total of 161.488 Joules of kinetic energy is lost in this collision.
Explanation:
The subject question involves a head-on inelastic collision between two balls of different masses traveling towards each other at different speeds and asks for the amount of kinetic energy lost in the collision. To find the kinetic energy lost, we first calculate the initial and final kinetic energies and then take their difference.
Initial kinetic energy (Eki) is the sum of the kinetic energies of the two balls before the collision:
Kinetic energy of the 2.10 kg ball: [tex](1/2) * 2.10 kg * (13.0 m/s)^2[/tex]
Kinetic energy of the 6.50 kg ball: [tex](1/2) * 6.50 kg * (4.1 m/s)^2[/tex]
Final kinetic energy (Ekf) involves only the kinetic energy of the 2.10 kg ball, as the final velocity of the 6.50 kg ball is not given:
Kinetic energy of the 2.10 kg ball: (1/2) * 2.10 kg * [tex](8.20 m/s)^2[/tex]
Kinetic energy lost during the collision is calculated as:
Elost = Eki - Ekf
After calculation:
Initial kinetic energy, Eki = (1/2) * 2.10 kg * [tex](13.0 m/s)^2[/tex] + (1/2) * 6.50 kg * [tex](4.1 m/s)^2[/tex]
Final kinetic energy, Ekf = (1/2) * 2.10 kg * [tex](8.20 m/s)^2[/tex]
Kinetic energy lost, Elost = Eki - Ekf
Plugging the numbers in:
Eki = 0.5 * 2.10 kg * [tex](13.0 m/s)^2[/tex] + 0.5 * 6.50 kg * [tex](4.1 m/s)^2[/tex]
= (0.5 * 2.10 * 169) + (0.5 * 6.50 * 16.81)
= 177.45 J + 54.64 J
= 232.09 J
Ekf = 0.5 * 2.10 kg * [tex](8.20 m/s)^2[/tex]
= (0.5 * 2.10 * 67.24)
= 70.602 J
Therefore, kinetic energy lost, Elost = 232.09 J - 70.602 J = 161.488 J
A total of 161.488 Joules of kinetic energy is lost in this collision.
The critical angle for diamond (n-2.42) surrounded by air is approximately 24 35 48 66
Answer:
sin(C) = 1/n = 1/2.42
C = 24.4 deg
The critical angle for the diamond-air interface is 24 degrees.
What is the critical angle for the diamond-air interface?If the ray of light in the diamond medium strikes the separating surface between the diamond and air making an angle more than 24 degrees from the normal to the point of incidence then that ray of light will undergo a total internal reflection and comes back to the diamond medium. This is the reason behind the sparkles of diamonds.Hence, the critical angle for the diamond-air interface is 24 degrees.To learn more about the critical angle for the diamond-air interface, refer to:https://brainly.com/question/1304781
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A beam of white light from a flashlight passes through a red piece of plastic. a. What is the color of the light that emerges from the plastic? b. Is the emerging light as intense as, more intense than, or less intense than the white light? Explain. c. The light then passes through a blue piece of plastic. Describe the color and intensity of the light that emerges.
The light emerging from a red filter is red and less intense than the original white light. When passed through blue plastic, it would be very dim or blocked, as red light cannot pass through a blue filter.
When a beam of white light passes through a red piece of plastic, the color of the light that emerges is red. This color is seen because the red plastic acts as a filter, absorbing other colors and only allowing red light to pass through. The emerging light is less intense than the white light because some of the light energy has been absorbed by the plastic.
If the light that has already passed through a red filter is then passed through a blue piece of plastic, the emerging light would likely be very dim or completely blocked. This is because the red light, lacking blue components, will be mostly absorbed by the blue plastic, which only allows blue light to pass through.
A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on a proton is 1.60218×10−19 C and its mass is 1.67262 × 10−27 kg. What is the cyclotron frequency? Answer in units of rad/s.
Explanation:
It is given that,
Magnetic field, B = 0.15 T
Charge on a proton, [tex]q=1.60218\times 10^{-19}\ C[/tex]
Mass of a proton, [tex]m=1.67262 \times 10^{-27}\ kg[/tex]
The cyclotron frequency is given by :
[tex]f=\dfrac{qB}{2\pi m}[/tex]
[tex]f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}[/tex]
f = 2286785.40 Hz
or
[tex]\omega=14368296.44\ rad/s[/tex]
[tex]\omega=1.43\times 10^7 rad/s[/tex]
Hence, this is the required solution.
You are a passenger on a jetliner that is flying at constant velocity. You get up from your seat and walk toward the front of the plane. Because of this action, your forward momentum increases. What happens to the forward momentum of the plane itself?
Answer:
It feels like it has more weight
A 1.7cm diameter pipe widens to 4.7cm. Liquid flows through the first segment at a speed of 4.7m/s. What is the speed of the liquid in the second segment?
Answer:
Speed of the liquid in the second segment = 0.61 m/s
Explanation:
This discharge is constant.
That is
Q₁ = Q₂
A₁v₁ = A₂v₂
[tex]\frac{\pi d_1^2}{4}\times v_1=\frac{\pi d_2^2}{4}\times v_2\\\\\frac{\pi \times 1.7^2}{4}\times 4.7=\frac{\pi \times 4.7^2}{4}\times v_2\\\\v_2=0.61m/s[/tex]
Speed of the liquid in the second segment = 0.61 m/s
How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J c.4.09 x 10^5 J d.4.77 x 10^5 J
Answer:
Work done by the frictional force is [tex]3.41\times 10^5\ J[/tex]
Explanation:
It is given that,
Mass of the car, m = 1000 kg
Initial velocity of car, u = 26.1 m/s
Finally, it comes to rest, v = 0
We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.
[tex]W=k_f-k_i[/tex]
[tex]W=\dfrac{1}{2}m(v^2-u^2)[/tex]
[tex]W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)[/tex]
W = −340605 J
or
[tex]W=3.41\times 10^5\ J[/tex]
Hence, the correct option is (a).
The work done by frictional forces to slow a 1000 kg car from 26.1 m/s to rest is 3.41 x 10^5 J (a). This value is derived from the car's initial kinetic energy, which is lost due to friction.
Explanation:The question is asking how much work the frictional forces need to do to bring a 1000 kg car to rest, from an initial speed of 26.1 m/s. In such a scenario, these forces would be working against the car's kinetic energy.
The car's initial kinetic energy can be calculated using the formula 1/2 m v^2 (m = mass, v = speed). So the kinetic energy = 1/2 * 1000 kg * (26.1 m/s)^2 = 3.41 x 10^5 J. Since work done by friction is equal to the change in kinetic energy, and the car is brought to rest, the total work done by frictional forces is 3.41 x 10^5 J. Therefore, the correct answer is (a) 3.41 x 10^5 J.
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A cylindrical container with 25 ft in diameter is filled to a depth of 22 ft with gasoline (S.G.-0.68). Find the weight of the gasoline and the number of gallons that exist in the tank.
Answer:
Weight of gasoline = 2039.93 kN
Volume = 80783.80 gallon
Explanation:
Volume = Base area x Depth
[tex]\texttt{Base area = }\frac{\pi d^2}{4}=\frac{\pi \times 25^2}{4}=490.88ft^2[/tex]
Depth = 22 ft
Volume = 490.88 x 22 = 10799.22 ft³ = 10799.22 x 0.0283168 = 305.8 m³
Density of gasoline = 680 kg/m³
Mass of gasoline = 305.8 x 680 = 207943.65 kg
Weight of gasoline = 207943.65 x 9.81 = 2039.93 kN
We have 1 m³ = 264.172 gallon
305.8 m³ = 305.8 x 264.172 = 80783.80 gallon
You throw a baseball directly upward at time t = 0 at an initial speed of 12.3 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9.80 m/s2.
Explanation:
At the maximum height, the ball's velocity is 0.
v² = v₀² + 2a(x - x₀)
(0 m/s)² = (12.3 m/s)² + 2(-9.80 m/s²)(x - 0 m)
x = 7.72 m
The ball reaches a maximum height of 7.72 m.
The times where the ball passes through half that height is:
x = x₀ + v₀ t + ½ at²
(7.72 m / 2) = (0 m) + (12.3 m/s) t + ½ (-9.8 m/s²) t²
3.86 = 12.3 t - 4.9 t²
4.9 t² - 12.3 t + 3.86 = 0
Using quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 12.3 ± √(12.3² - 4(4.9)(3.86)) ] / 9.8
t = 0.368, 2.14
The ball reaches half the maximum height after 0.368 seconds and after 2.14 seconds.
The maximum height the ball reaches above where it leaves your hand is 7.72 m
The time taken for the ball to pass half of its maximum height moving upwards is 2.14s and 0.37 s when moving downwards.
The given parameters;
initial velocity, u = 12.3 m/s
acceleration due to gravity, g = 9.8 m/s²
The maximum height the ball reaches above where it leaves your hand is calculated as;
[tex]v_f^2 = v_0^2 - 2gh\\\\at \ maximum \ height \ final \ velocity \ v_f = 0\\\\2gh = v_0 ^2\\\\h = \frac{v_0^2}{2g} \\\\h = \frac{(12.3)^2}{2(9.8)} \\\\h = 7.72 \ m[/tex]
The time for the ball to reach half of the maximum height is calculated as;
[tex]half \ of \ the \ maximum \ height = \frac{7.72}{2} = 3.86 \ m[/tex]
[tex]h = v_0t - \frac{1}{2} gt^2\\\\3.86 = 12.3t - 0.5\times 9.8t^2\\\\3.86 = 12.3t - 4.9t^2\\\\4.9t^2- 12.3t + 3.86=0\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 4.9, \ \ b = -12.3, \ \ c = 3.86\\\\t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-b \ \ +/- \ \ \sqrt{(-12.3)^2 - 4(4.9\times 3.86)} }{2(4.9)} \\\\t = 2.14 \ s \ \ or \ \ 0.37 \ s[/tex]
The time taken for the ball to pass half of its maximum height moving upwards is 2.14s and 0.37 s when moving downwards.
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. A hot-air balloon is drifting straight downward with a constant speed of 2.40 m/s. When the balloon is 7.63 m above the ground, the balloonist decides to drop one of the ballast sandbags to the ground below. How much time elapses before the sandbag hits the ground?
Answer:
1.03 seconds
Explanation:
x = x₀ + v₀ t + ½ at²
0 = 7.63 + (-2.40) t + ½ (-9.8) t²
0 = 7.63 - 2.40 t - 4.9 t²
Solve with quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 2.40 ± √(2.40² - 4(-4.9)(7.63)) ] / -9.8
t = -1.52, 1.03
Since t can't be negative here, the sandbag hits the ground after 1.03 seconds.
The sandbag is dropped from the hot air balloon with the balloon's existing speed. Using the second equation of motion and the known values for initial velocity, acceleration due to gravity, and distance, we can solve for the time taken for the sandbag to hit the ground.
Explanation:In the given question, a hot-Air balloon is drifting straight downward with a constant speed of 2.40 m/s. When the balloon is 7.63 m above the ground, a ballast sandbag is dropped.
Given the bag is dropped from rest relative to the balloon (i.e., the initial velocity (u) of the sandbag is -2.4 m/s, negative because the direction is downward), and the acceleration due to gravity (a) is -9.8 m/s² (negative due to the downward direction), we can use the second equation of motion (s = ut + 0.5at²) where s is the distance covered, which is 7.63 m, to find time (t).
Setting this up, -7.63 = -2.4t + 0.5(-9.8)t². Solving this quadratic equation for time should give us the answer. The result will be the time taken by the sandbag to hit the ground after it is dropped from the hot air balloon.
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Starting from rest and height of 7 m, a 3kg object slides down a 30degrees incline, reaching the bottom with a speed of 10m/s . what is the work done by friction? please elaborate.
Answer:
The work done by the friction is 55.8 J.
Explanation:
Given that,
Height h = 7 m
Mass of the object = 3 kg
Angle = 30°
Speed = 10 m/s
The initial potential energy of the object is converted in to the kinetic energy at the bottom of the incline and the work done by the friction
[tex]mgh=\dfrac{1}{2}mv^2+W[/tex]
Where, m = mass of the object
v = final velocity
g = acceleration due to gravity
h = height
Put the value in the equation
[tex]3\times9.8\times7=\dfrac{1}{2}\times3\times(10)^2+W[/tex]
[tex]W = (205.8-150)\ J[/tex]
[tex]W=55.8\ J[/tex]
Hence, The work done by the friction is 55.8 J.
Two simple pendulums are in two different places. The length of the second pendulum is 0.4 times the length of the first pendulum, and the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum. Determine the comparison of the frequency of the first pendulum to the second pendulum. a) 2/3. b) 3/2 (5 marks) c) 4/9. d) 9/4
Answer:
[tex]\sqrt{\frac{4}{9}}[/tex]
Explanation:
The frequency of a simple pendulum is given by:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}[/tex]
where
g is the acceleration of gravity
L is the length of the pendulum
Calling [tex]L_1[/tex] the length of the first pendulum and [tex]g_1[/tex] the acceleration of gravity at the location of the first pendulum, the frequency of the first pendulum is
[tex]f_1=\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}[/tex]
The length of the second pendulum is 0.4 times the length of the first pendulum, so
[tex]L_2 = 0.4 L_1[/tex]
while the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum, so
[tex]g_2 = 0.9 g_1[/tex]
So the frequency of the second pendulum is
[tex]f_2=\frac{1}{2\pi}\sqrt{\frac{g_2}{L_2}}=\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}[/tex]
Therefore the ratio between the two frequencies is
[tex]\frac{f_1}{f_2}=\frac{\frac{1}{2\pi}\sqrt{\frac{g_1}{L_1}}}{\frac{1}{2\pi} \sqrt{\frac{0.9 g_1}{0.4 L_1}}}=\sqrt{\frac{0.4}{0.9}}=\sqrt{\frac{4}{9}}[/tex]
Final answer:
The frequency of the first pendulum compared to the second pendulum is in the ratio of 2/3, since the length of the second pendulum is 0.4 times the first and gravity on the second is 0.9 times the first, impacting their respective periods and inversely their frequencies.
Explanation:
The frequency of a simple pendulum is dependent on its length and the acceleration due to gravity it experiences. The formula to find the frequency (f) of a pendulum is given by:
f = 1 / T
Where T is the period of the pendulum, which can be calculated by:
T = 2π√(L/g)
Let's denote L1 and g1 as the length and the acceleration due to gravity for the first pendulum, and L2 and g2 for the second pendulum. Since the length of the second pendulum is 0.4 times that of the first, we have L2 = 0.4L1. And because the acceleration of gravity is 0.9 times for the second pendulum, g2 = 0.9g1.
Using these proportions:
T1 = 2π√(L1/g1)
T2 = 2π√(L2/g2) = 2π√((0.4L1)/(0.9g1))
We can simplify T2 by putting it in terms of T1:
T2 = T1√(0.4/0.9)
Now, since frequency is the inverse of period:
f1 = 1/T1
f2 = 1/T2
So the ratio of their frequencies (f1/f2) will be the inverse of the ratio of their periods (T2/T1):
f1/f2 = T2/T1
Plugging in the ratio:
f1/f2 = √(0.4/0.9) = √(4/9) = 2/3
Therefore, the correct answer is (a) 2/3.
A charge q = 2.00 μC is placed at the origin in a region where there is already a uniform electric field E⃗ = (100 N/C) iˆ . Calculate the flux of the net electric field through a Gaussian sphere of radius R = 10.0 cm centered at the origin. (ε 0 = 8.85 × 10-12 C2/N · m2)
Answer:
The flux of the net electric field is [tex]2.26\times10^{5}\ Nm^2/C[/tex]
Explanation:
Given that,
Charge [tex]q= 2.00 \mu C[/tex]
Electric field E = 100 N/C i
Radius R = 10.0 cm
We need to find the flux. It can be calculate using Gauss's law
The flux of the net electric field
[tex]\phi=\dfrac{q}{\epsilon_{0}}[/tex]
[tex]\phi=\dfrac{2.00\times10^{-6}}{8.85\times10^{-12}}[/tex]
[tex]\phi=2.26\times10^{5}\ Nm^2/C[/tex]
Hence, The flux of the net electric field is [tex]2.26\times10^{5}\ Nm^2/C[/tex]
An engine is designed to obtain energy from the temperature gradient of the ocean. What is the thermodynamic efficiency of such an engine if the temperature of the surface of the water is 59°F (15°C) and the temperature well below the surface is 41°F (5°C)
Answer:
0.035 (3.5 %)
Explanation:
The thermodynamic efficiency is given by:
[tex]\eta = 1 - \frac{T_C}{T_H}[/tex]
where
[tex]T_C[/tex] is the cold temperature
[tex]T_H[/tex] is the hot temperature
In this problem we have
[tex]T_C = 5 ^{\circ}C+ 273 = 278 KT_H = 15^{\circ}C+273 = 288 K[/tex]
So the efficiency is
[tex]\eta = 1 - \frac{278 K}{288 K}=0.035[/tex]
The maximum thermodynamic efficiency of an engine operating with a hot reservoir at 288 K and a cold reservoir at 278 K, according to the Carnot efficiency formula, is approximately 3.47%.
Explanation:The student has asked about the thermodynamic efficiency of an engine that uses the temperature gradient of the ocean. To calculate this, we can use the Carnot efficiency formula:
Carnot Efficiency Formula
η = 1 - Tc/Th, where η represents the efficiency, Tc is the cold temperature, and Th is the hot temperature. It's important to remember these temperatures need to be in Kelvin.
First, convert the given temperatures from degrees Fahrenheit to Kelvin:
59°F (15°C) = 288 K (Th)41°F (5°C) = 278 K (Tc)Now, apply the Carnot efficiency formula:
η = 1 - (278 K / 288 K) = 1 - 0.9653 = 0.0347 or 3.47%
Therefore, the maximum thermodynamic efficiency of this oceanic temperature gradient engine would be approximately 3.47%.
A particle moves with constant acceleration along a straight line starting from rest. The percentage increase in its displacement during the 4th second compared to its displacement in the 3rd second (1) 33 % . is (2) 40 % (4) 77 % / (3) 66 % A particle is thrown upward from groun
Answer:
Option 2 is the correct answer.
Explanation:
We have equation of motion s = ut + 0.5at²
Here u = 0 m/s
So, s = 0.5at²
Distance traveled in first second = 0.5 x a x 1² = 0.5 a
Distance traveled in second second = 0.5 x a x 2² - 0.5 x a x 1²= 1.5 a
Distance traveled in third second = 0.5 x a x 3² - 0.5 x a x 2²= 2.5 a
Distance traveled in fourth second = 0.5 x a x 4² - 0.5 x a x 3²= 3.5 a
The percentage increase in its displacement during the 4th second compared to its displacement in the 3rd second
[tex]=\frac{3.5a-2.5a}{2.5a}=0.4=40\%[/tex]
Option 2 is the correct answer.
An overhang hollow shaft carries a 900 mm diameter pulley, whose centre is 250 mm from the centre of the nearest bearing. The weight of the pulley is 600 N and the angle of lap is 180°. The pulley is driven by a motor vertically below it. If permissible tension in the belt is 2650 N and if coefficient of friction between the belt and pulley surface is 0.3, estimate, diameters of shaft, when the internal diameter is 0.6 of the external. Neglect centrifugal tension and assume permissible tensile and shear stresses in the shaft as 84 MPa and 68 MPa respectively.
Answer:yu
Explanation:
A person in a kayak starts paddling, and it accelerates from 0 to 0.680 m/s in a distance of 0.428 m. If the combined mass of the person and the kayak is 82.7 kg, what is the magnitude of the net force acting on the kayak?
To find the magnitude of the net force acting on the kayak, we will use the following kinematic equation derived from Newton's second law of motion:
v2 = u2 + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance over which the acceleration occurs.
We know that the kayak starts from rest, so u = 0.
The final velocity v is 0.680 m/s and the distance s is 0.428 m. Rearranging the equation for acceleration a, we get:
a = (v2 - u2) / (2s)
By plugging in the given values, we find that the acceleration a is:
a = (0.6802 - 02) / (2 imes 0.428) = 0.6802 / 0.856 = 0.541 m/s2
Now, using Newton's second law (Force = mass x acceleration), we find the net force F:
F = mass x a
F = 82.7 kg x 0.541 m/s2 = 44.7397 N
The magnitude of the net force acting on the kayak is approximately 44.74 N.
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a distance of 1.4 m in a time of 1.1 s. The readout on the display indicates that the average power you are producing is 84 W. What is the magnitude of the force that you exert on the handle?
Answer:
Magnitude of the force exerted on the handle = 66 N
Explanation:
Power is the ratio of work and time.
[tex]P=\frac{W}{t}\\\\W=Pt=84\times 1.1=92.4J[/tex]
We have work done by rower = 92.4 J
We also have
Work = Force x Displacement
92.4 = Force x 1.4
Force = 66 N
Magnitude of the force exerted on the handle = 66 N
A satellite orbits Earth at an altitude of 350 km. What is its orbital period? A) 1.1 min B) It can not be determined without the satellite's mass. C) 65 min D) 91 min E) 1440 min
The orbital period of a satellite at an altitude of 350 km above Earth's surface is approximately 1.93 hours or around 115.8 minutes. Given the options, the closest one is option D, with a period of 91 minutes.
Explanation:To answer the question about the satellite orbiting Earth at an altitude of 350 km, we can utilize some physics concepts related to orbital mechanics. Note that a significant factor in determining the orbital period of a satellite (time required for a satellite to complete one orbit) is not the satellite's mass but rather its altitude or, more accurately, the total distance from the center of the Earth, considering the Earth's radius plus the satellite's altitude.
In general, the closer a satellite is to Earth, the shorter its orbital period. The information given indicates that a satellite at an altitude of 350 km will likely have an orbital period around 1.93 hours, which is roughly equivalent to 115.8 minutes. Given that, none of the available options exactly match this result. But the value closest to it would be 91 minutes, which is represented by option D.
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The electric field between square plates of a parallel-plate capacitor has magnitude E. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The magnitude of the electric field between the plates is now equal to a)E b)E/4 c)E/2 d)4E e)2E
Answer:
the Answer is c: E/2
Explanation:
see attachment
The cylinder rotates about the fixed z-axis in the direction indicated. If the speed of point A is vA = 2.7 ft/sec and the magnitude of its acceleration is aA = 20.6 ft/sec2, determine the magnitudes of the angular velocity and angular acceleration of the cylinder. Is knowledge of the angle θ necessary?
To solve this problem, we need to utilize the relationship between linear and angular motion.
Firstly, we know the speed of point A is vA = 2.7 ft/sec. In a rotating cylinder, the linear speed v of a point at distance r from the center of rotation is given by v = ω * r, where ω is the angular speed. This would allow us to find ω by dividing the speed at point A by its distance from the center.
However, in this problem, we don't know the exact distance of point A from the center. So let's denote the unknown distance as r ft. Then, the angular speed ω = vA / r.
Secondly, it's given that the magnitude of the acceleration of the point A is aA = 20.6 ft/sec². Angular acceleration α is related to linear acceleration a by the factor of the radius, i.e., a = α * r.
We don't know r itself, but we know vA and aA, so we are able to compute the ratio of α to ω or α/ω = aA/vA, which equals to 20.6/2.7. This calculation enables us to conclude that the angular velocity and angular acceleration are proportional to each other, regardless of the values of radius r and angle θ
Taken together, from this reasoning, we can conclude that the ratio of angular to linear velocity/acceleration is approximately 7.63.
Finally, considering the issue of whether the knowledge of the angle θ is necessary, we can see that the angle θ does not appear in our ratio, and it does not affect the ratios of angular to linear velocities and accelerations. Therefore, the knowledge of the angle θ is not necessary for this calculation.
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Which of the following is NOT an example of retaliation?
Getting A Sudden Shift Change
Being Fired
Getting A Promotion
Getting A Bad Performance Review
Getting a promotion is NOT an example of retaliation.
C. Getting A Promotion
What is retaliation?the action of harming someone because they have harmed oneself; revenge.
An example of to retaliate is for a person to punch someone who has hit him. (intransitive) To do something harmful or negative to get revenge for some harm; to fight back or respond in kind to damage or affront. Johny insulted Peter to retaliate for Peter's acid remark earlier.
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If a planet has the same surface gravity as the earth (that is, the same value of g at the surface), what is its escape speed? (i) The same as the earth's: (ii) less than the earth'sii) greater than the earth's (iv) any of these are possible. I
Answer:
Explanation:
The escape velocity on a planet is directly proportional to the square root of acceleration due to gravity and the radius of the planet.
According to the question acceleration due to gravity that means g is same but there is no idea about the radius.
So if radius is change then escape velocity at that planet may be more or less depending on the radius.
If the radius is also same then the escape velocity is same.
A person is in a room whose walls are maintained at a temperature of 17 °C. The temperature of the person’s skin is 32 °C. The surface area of the person’s skin is 1.4 m2. Determine the netheat transfer rate leaving the person.
Answer:
sned me short notes of Physics of all branches (post graduation level) thanks
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Explanation:
You connect a voltmeter to an unknown resistor in a circuit in order to measure the potential difference across that resistor. How should you connect it to the resistor and what would the voltmeter's ideal internal resistance be? A. In series Fint B. In parallel; int- O C. In series; rint- 0 D. In parallel; rint →
Answer:
option (a)
Explanation:
To make a galvanometer into voltmeter, we have to connect a high resistance in series combination.
The voltmeter is connected in parallel combination with teh resistor to find the voltage drop across it.
An ideal voltmeter has very high resistance that means it has a resistance as infinity.
Answer:
A. In series Fint
Explanation:
If you connect it to the resistor, the voltmeter's ideal internal resistance be In series Fint.