An observer on Starship B will measure the same time interval as that on Starship A, because they are both travelling at the same speed and in the same direction relative to Earth. Starship C's measurement would differ due to its opposite direction of travel relative to A and B.
Explanation:To determine which reference frame measures a time interval equal to the time interval measured by the clock aboard Starship A, we need to consider the principles of the theory of special relativity. Since Starship B is traveling at the same speed and in the same direction as Starship A relative to Earth, the time dilation effect will be the same for both starships. Therefore, an observer on Starship B will measure the same time interval on their clock as that measured on Starship A's clock.
In contrast, Starship C is moving at the same speed but in the opposite direction relative to Earth, and its relative velocity to Starships A and B is not zero. As a result, the time interval measured on Starship C will not match the time interval measured on Starship A's clock.
So, the time interval that equals the time interval measured by the clock aboard Starship A can be measured in the reference frame of Starship B, which is moving at the same speed and in the same direction as Starship A relative to Earth.
An alpha particle (a helium nucleus, consisting of two protons and two neutrons) has a radius of approximately 1.6 × 10-15 m. A certain heavy nucleus contains 79 protons in addition to all its neutrons and has a radius of approximately 5.8 × 10-15 m. An alpha particle is shot directly from a large distance at such a resting heavy nucleus.
What is the initial momentum of the alpha particle?
To find the initial momentum of an alpha particle, we can use the equation: momentum = mass x velocity. The mass of an alpha particle is approximately 4 atomic mass units (amu). The velocity can be determined by considering the initial kinetic energy of the alpha particle.
Explanation:The initial momentum of an alpha particle can be calculated using the equation:
momentum = mass x velocity
The mass of an alpha particle is approximately 4 atomic mass units (amu). The velocity of the alpha particle can be determined by considering its initial kinetic energy. Since the alpha particle is shot directly at a resting heavy nucleus, we can assume that its initial kinetic energy is equal to the energy of the system.
Using the equation:
kinetic energy = (1/2) x mass x velocity^2
we can solve for velocity. Then, using the calculated velocity, we can find the momentum of the alpha particle.
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In an electric circuit, resistance and current are ____
A. directly proportional
B. inversely proportional
C. have no effect on each other
In an electric circuit, resistance and current are ____
A. directly proportional
B. inversely proportional
C. have no effect on each other
Explanation:
A
A car has a momentum of 20,000 kg • m/s. What would the car’s momentum be if its velocity doubles? 10,000 kg • m/s 20,000 kg • m/s 40,000 kg • m/s 80,000 kg • m/s
Answer:40000kg•m/s
Explanation:
momentum=20000kg•m/s
Since they velocity is doubled
Momentum would also be doubled, therefore momentum would be 40000kg•m/s
Answer:
C: 40,000 kg · m/s
Explanation:
I got it right on edg :)
Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected by a tube that contains an ideal turbine (a mechanical device) and a valve. Initially, the valve is closed. Container A (volume VA = 2 m3) is filled with n = 1.2 moles of an ideal monatomic gas, and container B (volume VB = 3.5 m3) is empty (vacuum). When the valve is opened, gas flows between A and B, and the turbine is used to generate electricity. What is the maximum amount of work than can be done on the turbine as the gas reaches equilibrium? Remember: U = 1.5 nRT + const S = nR ln(V) + f(U,n)
Answer:
The maximum amount of work is [tex]W = 1563.289 \ J[/tex]
Explanation:
From the question we are told that
The temperature of the environment is [tex]T = 280\ K[/tex]
The volume of container A is [tex]V_A = 2 m^3[/tex]
Initially the number of moles is [tex]n = 1.2 \ moles[/tex]
The volume of container B is [tex]V_B = 3.5 \ m^3[/tex]
At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as
[tex]W = P_A V_A ln[ \frac{V_B}{V_A} ][/tex]
Now from the Ideal gas law
[tex]P_A V_A = nRT[/tex]
So substituting for [tex]P_A V_A[/tex] in the equation above
[tex]W = nRT ln [\frac{V_B}{V_A} ][/tex]
Where R is the gas constant with a values of [tex]R = 8.314 \ J/mol[/tex]
Substituting values we have that
[tex]W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ][/tex]
[tex]W = 1563.289 \ J[/tex]
The maximum amount of work done on the turbine can be found by calculating the change in internal energy of the gas as it reaches equilibrium. The work done on the turbine is given by the difference in pressure between the initial and final states of the gas, multiplied by the change in volume. In this case, the maximum work done on the turbine is (nRT/VA)(VB - VA).
Explanation:When the valve is opened and the gas flows from container A to container B, the gas will expand and reach equilibrium. Since the process is isothermal, the temperature remains constant at T = 280K. To find the maximum work done on the turbine, we need to calculate the change in internal energy of the gas using the equation ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the gas, and W is the work done on the gas. For an ideal monatomic gas, the internal energy is given by U=1.5nRT, where n is the number of moles, R is the gas constant, and T is the temperature. In this case, we have n = 1.2 moles, V = 2m3 for container A, and V = 3.5m3 for container B. Therefore, the total volume is V = VA + VB = 5.5m3. Using the ideal gas law PV = nRT, we can find the initial pressure of the gas in container A, which is P = nRT/VA. Since the gas expands to fill both containers at equilibrium, the final pressure is P = nRT/V. The work done on the turbine is given by W = P(VB - VA). Plugging in the values, we find that the maximum work done on the turbine is W = (nRT/VA)(VB - VA).
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An eighteen gauge copper wire has a nominal diameter of 1.02mm. This wire carries a constant current of 1.67A to a 200w lamp. The density of free electrons is 8.5 x 1028 electrons per cubic metre. Find the magnitude of:
i. The current density ii. The drift velocity
The current density of the wire is approximately[tex]2.04 x 10^6 A/m²[/tex]elocity of the electrons in the wire is approximately [tex]1.51 x 10^-3 m/s.[/tex]
Explanation:To find the magnitude of the current density, we need to calculate the cross-sectional area of the wire and divide the current by that area. The current density (J) is given by J = I/A, where I is the current and A is the cross-sectional area. Since the wire is cylindrical, we can use the formula for the area of a circle, A = πr², where r is the radius of the wire.
Given that the diameter of the wire is [tex]1.02mm[/tex]radius can be calculated as [tex]r = (1.02mm)/2 = 0.51mm = 0.00051m[/tex]
Using this value, the cross-sectional area is [tex]A = π(0.00051m)² = 8.18 x 10^-7 m².[/tex]
Dividing the current of[tex]1.67A[/tex]e cross-sectional area, we get the current density [tex]J = (1.67A)/(8.18 x 10^-7 m²) ≈ 2.04 x 10^6 A/m².[/tex]
To find the drift velocity, we can use the formula v_d = J/(nq), where n is the density of free electrons and q is the charge of an electron. Given that the density of free electrons is [tex]8.5 x 10^28 electrons/m³[/tex]
the charge of an electron is [tex]q = 1.60 x 10^-19 C[/tex]
we can substitute these values into the formula: [tex]v_d = (2.04 x 10^6 A/m²) / (8.5 x 10^28 electrons/m³ * 1.60 x 10^-19 C/electron) ≈ 1.51 x 10^-3 m/s.[/tex]
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The current density in the wire is approximately 2.04 × 10^6 A/m^2, and the drift velocity of the electrons is approximately 1.51 × 10^-4 m/s.
Sure, let's break down the calculation step by step:
Given data:
Diameter of the wire, d = 1.02 mm = 1.02 × 10^-3 m
Current, I = 1.67 A
Power of the lamp, P = 200 W
Density of free electrons, n = 8.5 × 10^28 electrons/m^3
Calculate the cross-sectional area of the wire:
Radius, r = d/2 = 1.02 × 10^-3 m / 2 = 0.51 × 10^-3 m
Cross-sectional area, A = π * r^2
A = π * (0.51 × 10^-3 m)^2 ≈ 8.19 × 10^-7 m^2
Calculate the current density (J):
Current density is the current per unit area.
J = I / A
J = 1.67 A / 8.19 × 10^-7 m^2 ≈ 2.04 × 10^6 A/m^2
Calculate the charge carrier density (n):
Given density of free electrons, n = 8.5 × 10^28 electrons/m^3
Calculate the drift velocity (v_d):
Drift velocity is given by the formula: J = n * e * v_d
Where e is the charge of an electron (approximately 1.6 × 10^-19 C)
Rearranging for drift velocity: v_d = J / (n * e)
Substituting values: v_d = (2.04 × 10^6 A/m^2) / (8.5 × 10^28 electrons/m^3 * 1.6 × 10^-19 C)
v_d ≈ 1.51 × 10^-4 m/s
A 1.9 kg ball drops vertically onto a floor, hitting with a speed of 25 m/s. It rebounds with an initial speed of 9.0 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.0158 s, what is the magnitude of the average force on the floor from the ball?
Answer:
(a) 64.6 kg/ms
(b) 4088.61 N
Explanation:
(a)
I = mΔv.................. Equation 1
Where I = impulse acting on the ball, Δv = change in velocity.
But,
Δv = v-u.............. Equation 2
Where v = final velocity, u = initial velocity.
Substitute equation 2 into equation 1
I = m(v-u).................. Equation 3
Assuming: upwards to be positive
Given: m = 1.9 kg, v = 9 m/s, u = -25 m/s (downward)
Substitute into equation 3
I = 1.9[9-(25)]
I = 1.9(9+25)
I = 1.9(34)
I = 64.6 kgm/s.
(b)
F = I/t................... Equation 4
Where F = Average force on the floor from the ball, t = time of contact of the floor with the ball.
Given: I = 64.6 kgm/s, t = 0.0158 s
Substitute into equation 4
F = 64.6/0.0158
F = 4088.61 N
Answer:
A) Impulse = 64.6 N.s
B) Force = 4088.6 N
Explanation:
We are given;
Mass of ball;m = 1.9 kg
Initial velocity of ball;u = -25 m/s (negative value because ball was drop downward)
Final velocity of ball;v = 9 m/s
From Newton's second law of motion, we know that;
Impulse = Change in momentum
Thus;
Impulse = final momentum - initial momentum = mv - mu
Thus;
Impulse = m(v - u)
Plugging in the relevant values, we have;
Impulse = 1.9(9 - (-25)
Impulse = 1.9(9 + 25)
Impulse = 1.9(34)
Impulse = 64.6 N.s
B) As said earlier,
impulse = 64.6 N.s
Now, we are looking for the magnitude of the average force.
Now, Impulse is also expressed as Force x time.
Thus,
Force x time = 64.6 N.s
We are given time as 0.0158 s
Thus, making Force the subject of the formula, we now have;
Force = 64.6/0.0158
Force = 4088.6 N
Polly is pushing a box across the floor with a force of 30 N. The force of gravity is -8 N, and the normal force is 8 N. Which
value could describe the force of friction if Polly could not move the box?
-30 N
8N
8N
30 N
Mark this and retum
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Answer:
Force of friction is (-30 N).
Explanation:
The force applied on the box across the floor is 30 N.
The force of gravity is (-8 N) and the the normal force is 8 N.
It is based on Newton's third law of motion. Newton's third law of motion states that the force acting on object 1 to object 2 is equal in magnitude of the force from object 2 to 1 but in opposite direction.
Here there is force of 30 N is applied in horizontal direction. The frictional force act in opposite direction. So, the force of friction is -30 N so that box across the floor.
A 1.20 kg copper rod rests on two horizontal rails 0.78 m apart and carries a current of 45 A from one rail to the other. The coefficient of static friction between rod and rails is 0.61. What is the magnitude of the smallest magnetic field that puts the rod on the verge of sliding
Answer:
B = 0.204T
Explanation:
To find the value of the magnetic force you use the following formula:
[tex]F_B=ILBsin\theta[/tex]
I: current of the copper rod = 45A
B: magnitude of the magnetic field
L: 0.78m
you assume that magnetic field B and current I are perpendicular between them.
The magnetic force must be, at least, equal to the friction force, that is:
[tex]F_{f}=F_{B}\\\\\mu N=\mu Mg=ILB\\\\B=\frac{\mu Mg}{IL}[/tex]
M: mass of the rod = 1.20kg
μ: coefficient of static friction = 0.61
g: gravitational acceleration constant = 9.8m/s^2
By replacing the values of the parameters you obtain:
[tex]B=\frac{(0.61)(1.20kg)(9.8m/s^2)}{(45A)(0.78m)}=0.204T[/tex]
Answer:
The magnitude of the smallest magnetic field is [tex]B = 0.1744 \ T[/tex]
Explanation:
From the question we are told that
The mass of the copper is [tex]m = 1.20 kg[/tex]
The distance of separation for the rails is [tex]d = 0.78 \ m[/tex]
The current is [tex]I = 45 A[/tex]
The coefficient of static friction is [tex]\mu = 0.61[/tex]
The force acting along the vertical axis is mathematically represented as
[tex]F = mg - F_y[/tex]
Where [tex]F_y[/tex] is the force acting on copper rod due to the magnetic field generated this is mathematically represented as
[tex]F_y = I * d * B_1[/tex]
The magnetic field here is acting towards the west because according to right hand rule magnetic field acting toward the west generate a force acting in the vertical axis
So the equation becomes
[tex]F = mg - I * d * B_1[/tex]
Here [tex]B_1 = B sin \theta[/tex]
[tex]F = mg - I * d * Bsin(\theta )[/tex]
The in the horizontal axis is mathematically represented as
[tex]F_H = ma + F_x[/tex]
Since the rod is about to move it acceleration is zero
Now [tex]F_x[/tex] is the force acting in the horizontal direction due to the magnetic field acting downward this is because a according to right hand rule magnetic field acting downward generate a force acting in the horizontal positive horizontal direction. this mathematically represented as
[tex]F_H = 0 + I * d * B_2[/tex]
So the equation becomes
[tex]F_H = I * d * B_2[/tex]
Here [tex]B_2 = B cos \theta[/tex]
[tex]F_H = I * d * Bcos (\theta)[/tex]
Now the frictional force acting on this rod is mathematically represented as
[tex]F_F = \mu * F[/tex]
[tex]F_F = \mu * (mg -( I * d * Bsin(\theta )))[/tex]
Now when the rod is at the verge of movement
[tex]F_H = F_F[/tex]
So [tex]I * d * Bcos (\theta) = \mu * (mg -( I * d * Bsin(\theta )))[/tex]
=> [tex]B = \frac{\mu mg }{I * d (cos \theta + \mu (sin \theta ))}[/tex]
Now [tex]\theta[/tex] is the is the angle of the magnetic field makes with the vertical and the horizontal and this can be mathematically evaluated as
[tex]\theta = tan^{-1} (\mu )[/tex]
Substituting value
[tex]\theta = tan^{-1} ( 0.61 )[/tex]
[tex]\theta = 31.38^o[/tex]
Substituting values into the equation for B
[tex]B = \frac{0.61 (1.20) (9.8)}{(45) (0.78) (cos (31.38) + 0.61 (sin (31.38)) )}[/tex]
[tex]B = 0.1744 \ T[/tex]
Your mental health is a reflection of your...
O
Thoughts
Emotions
Behaviors
O
All of the above
Answer:all of the above
Explanation:
because thoughts relate to your emotions and your behaviors reflect to your emotions
Mental health is a reflection of one's thoughts, emotions, and behaviors. Positive thinking, managed emotions and constructive behaviors contribute to a healthy mental state.
Explanation:Your mental health is indeed a reflection of your thoughts, emotions and behaviors. Accumulative thoughts and feelings can shape our mental well-being to a great extent. Similarly, our behaviors, influenced by these thoughts and feelings, can either nurture or deteriorate our mental health. A balance in all three – positive thinking, managing emotions and constructive behaviors lead to a healthy mental state.
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Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, straight wire. You may want to review (Pages 926 - 929) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Magnetic field of a single wire. Part A If you were unfortunate enough to be 4.9 m away from such a lightning bolt, how large a magnetic field would you experience? Express your answer using two significant figures. B1 = nothing T Request Answer Part B How does this field (B1) compare to one (B2) you would experience by being 4.9 cm from a long, straight household current of 10 A? Express your answer using two significant figures.
Given Information:
Current due to Lightning bolt = I₁ = 20 kA
Current of household = I₂ = 10 A
Distance = r₁ = 4.9 m
Distance = r₂ = 4.9 cm = 0.049 m
Required Information:
a) Magnetic field due Lightning bolt = B₁ = ?
b) Magnetic field due to household current = B₂ = ?
Answer:
a) Magnetic field due Lightning bolt = B₁ = 8.16×10⁻⁴ T
b) Magnetic field due to household current = B₂ = 4.08×10⁻⁵ T
B₁ = 20B₂
Explanation:
The magnetic field produced in a long straight wire carrying a current (I) at distance r is given by
B = μ₀I/2πr
Where μ₀ is the permeability of free space and its value is 4π×10⁻⁷
a) The magnetic field produced due to the lightning bolt current is
B₁ = μ₀I₁/2πr₁
B₁ = (4π×10⁻⁷*20,000)/2π*4.9
B₁ = 0.000816
B₁ = 8.16×10⁻⁴ T
Therefore, the strength of magnetic field due to the lightning bolt current is 8.16×10⁻⁴ T
b) The magnetic field produced due to the household current is
B₂ = μ₀I₂/2πr₂
B₂ = (4π×10⁻⁷*10)/2π*0.049
B₂ = 0.00004081
B₂ = 4.08×10⁻⁵ T
Therefore, the strength of magnetic field due to the household current is 4.08×10⁻⁵ T
The ratio of magnetic field produced by the lightning bolt current to the magnetic field produced by the household current is
B₁/B₂ = 8.16×10⁻⁴/4.08×10⁻⁵
B₁/B₂ = 20
B₁ = 20B₂
Which means that the magnetic field produced by the lightning bolt current is 20 time greater than the magnetic field produced by the household current.
a) The magnetic field due Lightning bolt will be 8.16×10⁻⁴ T
b)The magnetic field due to household current will be 4.08×10⁻⁵ T
What is a magnetic field?
It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained it is the field felt around a moving electric charge.
The number of magnetic flux lines on a unit area passing perpendicular to the given line direction is known as induced magnetic field strength .it is denoted by B.
a) The magnetic field due Lightning bolt will be 8.16×10⁻⁴ T
The magnetic field produced in a long straight wire carrying a current (I) at distance r is given by
B = μ₀I/2πr
Where μ₀ is the permeability of free space and its value is 4π×10⁻⁷
[tex]\rm B_1= \frac{\mu_0i_1}{2 \pi r} \\\\ \rm B_1= \frac{4 \pi \times 10^{-7} \times 20,00}{2 \times 3.14 4.9} \\\\ \rm B_1= 8.16 \times 10^{-4} \ T[/tex]
b) Magnetic field due to household current will be 4.08×10⁻⁵ T
[tex]\rm B_2= \frac{\mu_0i_2}{2 \pi r} \\\\ \rm B_2= \frac{4 \pi \times 10^{-7} \times 10}{2 \times 3.14 0.049} \\\\ \rm B_2= 4.08\times 10^{-5} \ T[/tex]
Hence the magnetic field due to household current will be 4.08×10⁻⁵ T
The magnetic field created by lightning bolt current divided by the magnetic field produced by home electricity equals
[tex]\rm \frac{B_1}{B_2} = \rm \frac{8.16 \times 10^{-4}}{4.08 c\times 10^{-5}} \\\\ \rm \frac{B_1}{B_2} = 20 \\\\[/tex]
That is, the magnetic field created by a lightning bolt current is 20 times stronger than the magnetic field produced by a home current.
Hence the magnetic field due to the Lightning bolt will be 8.16×10⁻⁴and the magnetic field due to household current will be 4.08×10⁻⁵ T.
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What is the structure's center of gravity? (b) Calculate the structure's moment of inertia. (c) What is the angular acceleration experienced by the masses? (d) Determine the structures angular velocity after 10 s. (e) After 30 s, the external force is removed. What is the total energy of the system?om/homework-help/questions-and-answers/far-space-5-kg-ball-10-kg-ball-connected-10-m-long-rigid-massless-rod-steady-force-10-n-ap-q50426001
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on, the initial current decreases until the filament reaches its steady-state temperature. The temperature coefficient of resistivity of the filament is 4 times 10-3 K-1. The final current through the filament is one- eighth the initial current. What is the change in temperature of the filament
Answer:
The change in temperature is [tex]\Delta T = 1795 K[/tex]
Explanation:
From the question we are told that
The temperature coefficient is [tex]\alpha = 4 * 10^{-3 }\ k^{-1 }[/tex]
The resistance of the filament is mathematically represented as
[tex]R = R_o [1 + \alpha \Delta T][/tex]
Where [tex]R_o[/tex] is the initial resistance
Making the change in temperature the subject of the formula
[tex]\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ][/tex]
Now from ohm law
[tex]I = \frac{V}{R}[/tex]
This implies that current varies inversely with current so
[tex]\frac{R}{R_o} = \frac{I_o}{I}[/tex]
Substituting this we have
[tex]\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ][/tex]
From the question we are told that
[tex]I = \frac{I_o}{8}[/tex]
Substituting this we have
[tex]\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ][/tex]
=> [tex]\Delta T = \frac{1}{3.9 * 10^{-3}} (8 -1 )[/tex]
[tex]\Delta T = 1795 K[/tex]
The lamp filament's temperature rise can be calculated using the equations for resistance and the current. Given the temperature coefficient of resistivity and the ratio of initial and final currents, the change in temperature is calculated to be 875K.
Explanation:The problem is solved by using the formula for the change in resistance due to temperature, R = R0(1 + αΔT), where R is the final resistance, R0 is the initial resistance, α is the temperature coefficient of resistivity, and ΔT is the change in temperature. We can relate the initial and final currents (I and If) using Ohm’s law: V = I*R = If Rf, where V is the constant voltage. Given that If = I/8, and substituting from R = R0(1 + αΔT), we get that ΔT = [(8 - 1)/(α*7)] = 875K.
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A moving van collides with a sports car in a high-speed head-on collision. During the impact, the truck exerts a force Ftruck on the car and the car exerts a force Fcar on the truck. Which of the following statements about these forces is true?A. The force exerted by the truth on the car is the same size as the force exerted by the car on the truck: Ftruck = Fcar B. Ftruck < Fcar C. Ftruck > Fcar
Answer:
A. The force exerted by the truth on the car is the same size as the force exerted by the car on the truck: Ftruck = Fcar
Explanation:
Both vehicles will experience equal both opposite forces one on the other.
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that Group of answer choices the moment of inertia of the system increases and the angular speed increases. the moment of inertia of the system increases and the angular speed decreases. the moment of inertia of the system decreases and the angular speed increases. the moment of inertia of the system decreases and the angular speed decreases.
Answer:
The moment of inertia of the system decreases and the angular speed increases.
Explanation:
This very concept might not seem to be interesting at first, but in combination with the law of the conservation of angular momentum, it can be used to describe many fascinating physical phenomena and predict motion in a wide range of situations.
In other words, the moment of inertia for an object describes its resistance to angular acceleration, accounting for the distribution of mass around its axis of rotation.
Therefore, in the course of this action, it is said that the moment of inertia of the system decreases and the angular speed increases.
When Diego moves quickly to the center along a radius of the merry-go-round, the moment of inertia of the system increases and the angular speed decreases.
Explanation:When Diego moves quickly to the center along a radius of the merry-go-round, the moment of inertia of the system increases and the angular speed decreases.
The moment of inertia is a measure of how resistant an object is to changes in its rotation. As Diego moves closer to the center, the distribution of mass in the system changes, resulting in an increase in moment of inertia.
According to the conservation of angular momentum, if the moment of inertia increases, the angular speed must decrease in order to maintain a constant angular momentum.
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Jeff's body contains about 5.31 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells and the rest is plasma. The density of blood cells is approximately 1125 kg/m3, and about 1% of the cells are white blood cells, the rest being red blood cells. The red blood cells are about 7.50µm across (modeled as spheres). What is the mass of the blood mblood in Jeff's body?
Answer:
The mass of the blood is 5.6286 kg
Explanation:
Given:
V = volume of blood = 5.31 L = 0.00531 m³
ρ = density = 1060 kg/m³
Question: What is the mass of the blood mblood in Jeff's body, m = ?
To calculate the mass of the blood you just have to multiply the density of the blood by the volume occupied by it:
[tex]m=\rho *V=1060*0.00531=5.6286kg[/tex]
While visiting the planet Mars, Moe leaps straight up off the surface of the Martian ground with an initial velocity of 105 m/s to catch a sleeping juicy Martian cricket. It took Moe 56 seconds to reach the bug. The gravitational acceleration on Mars is 3.75 2 m / s . Find how far the cricket was off the ground when it became Moe’s lunch. Must use integration to derive your results.
Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
A uniform hollow spherical ball of mass 17 kg and radius 50.0 cm rolls without slipping up a ramp at an angle of 19 degrees above the horizontal. The speed of the rolling ball at the base of the ramp is 5.0 m/s. While the ball is rolling up the ramp, what are the magnitude and direction of the acceleration of its center of mass, and what is the magnitude and direction of the friction force acting on the ball from the surface of the ramp
Answer:
The acceleration of its center of mass is
The frictional force is [tex]f = 21.65 \ N[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 17 kg[/tex]
The radius of the ball is [tex]r = 50cm = \frac{50}{100} = 0.5m[/tex]
the angle with the horizontal is [tex]\theta = 19 ^ o[/tex]
The of the ball at the base is [tex]v = 5.0 \ m/s[/tex]
This setup is shown on the first uploaded image
looking at the diagram we see that the force acting on the ball can be mathematically evaluated as
[tex]mg sin \theta -f = ma[/tex]
Where f is the frictional force
The torque on the ball is mathematically represented as
[tex]\tau = f * r[/tex]
This torque can also be mathematically represented as
[tex]\tau = I \alpha[/tex]
where I is the moment of inertia of the ball which is mathematically represented as
[tex]I = \frac{2}{3} m r^2[/tex]
While [tex]\alpha[/tex] is the angular acceleration which is mathematically represented as
[tex]\alpha = \frac{a}{r}[/tex]
So [tex]\tau = \frac{2}{3} m r^2 * \frac{a}{r}[/tex]
Equating the both formula for torque
[tex]f * r = \frac{2}{3} m r^2 * \frac{a}{r }[/tex]
=> [tex]f = \frac{2}{3} ma[/tex]
Substituting this for f in the above equation
[tex]mg sin \theta = ma + \frac{2}{3} ma[/tex]
[tex]g sin \theta = \frac{5}{3} a[/tex]
[tex]a = \frac{3}{5} * g * sin \theta \alpha[/tex]
Substituting values
[tex]a = 1.91 m/s^2[/tex]
Now substituting into the equation frictional force equation
[tex]f = \frac{2}{3} * 17 * 1.91[/tex]
[tex]f = 21.65 \ N[/tex]
Answer:
[tex]a=-1.92 m/s^{2}[/tex]
[tex]F_{f}=-21.76 N[/tex]
Explanation:
We can use the definition of the torque:
[tex]\tau=I\alpha[/tex]
When:
I is the inertia of a uniform hallow sphere [tex]I = (2/3)mR^{2}[/tex]α is the angular acceleration (α = a/R)Now, torque is the product of the friction force times the radius.
[tex]F_{f}*R=\frac{2}{3}mR^{2}*\frac{a}{R}[/tex]
[tex]F_{f}=\frac{2}{3}ma[/tex] (1)
Now, let's analyze the force acting over the sphere using the Newton's second law.
[tex]F=ma[/tex]
[tex]-mgsin(\theta)-F_{f}=ma[/tex] (2)
Let's put F(f) of the equation (1) into the equation (2):
[tex]-mgsin(\theta)-\frac{2}{3}ma=ma[/tex]
[tex]a=-\frac{3}{5}gsin(\theta)[/tex]
[tex]a=-\frac{3}{5}*9.81*sin(19)[/tex]
[tex]a=-1.92 m/s^{2}[/tex]
Hence: [tex]F_{f}=\frac{2}{3}ma=\frac{2}{3}*17*(-1.92)[/tex]
[tex]F_{f}=-21.76 N[/tex]
I hope it helps you!
A 2 kg object is subjected to three forces that give it an acceleration −→a = =(8.00m/s2 )ˆi + (6.00m/s2 )ˆj. If two of the three forces, are −→F1 = (30.0N)ˆi + (16.0N)ˆj and −→F2 = =(12.0N)ˆi + (8.00N)ˆj, find the third force.
Answer:
The third force = [tex]-(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})[/tex]
Explanation:
Here we note that, the formula for the resultant force is as follows;
[tex]\Sigma F = \Sigma F_x + \Sigma F_y[/tex]
Also
∑F = m×a
Where:
[tex]F_x[/tex] = Force components in the x direction
[tex]F_y[/tex] = Force components in the y direction
∑F = Resultant force vector
m = Mass of the object
a = Acceleration vector ob the object =
[tex]a = 8.00 \, m/s^2 \, \hat{i} + 6.00 \, m/s^2 \, \hat{j}[/tex]
[tex]F_1 = 30.0 \, N \, \hat{i} + 16.0 \, N \, \hat{j}[/tex]
[tex]F_2 = 12.0 \, N \, \hat{i} + 8.00 \, N \, \hat{j}[/tex]
Therefore, since ∑F = m×a, we have;
[tex]\Sigma F = 2 kg \times (8.00 \, m/s^2 \, \hat{i} + 6.00 \, m/s^2 \, \hat{j})[/tex]
[tex]\Sigma F = (16.00 \, m/s^2 \, \hat{i} + 12.00 \, m/s^2 \, \hat{j})[/tex]
Hence from [tex]\Sigma F = \Sigma F_x + \Sigma F_y[/tex], we have;
[tex]F_{x1} + F_{x2} + F_{x3} = 16[/tex]
That is 30 + 12 + [tex]F_{x3}[/tex] = 16
∴ [tex]F_{x3}[/tex] = 16 - (30 + 12) = -26
Similarly,
[tex]F_{y1} + F_{y2} + F_{y3} = 12[/tex]
Therefore, [tex]F_{y3}[/tex] = 12 - (16 + 8) = -12
Hence, [tex]F_3 = -26.0 \, N \, \hat{i} - 12.00 \, N \, \hat{j} = -(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})[/tex]
The third force, [tex]F_3, = -(26.0 \, N \, \hat{i} + 12.00 \, N \, \hat{j})[/tex]
The third force applied on the object calculated using Newton's second law provides a magnitude of (-26.00 N î - 12.00 N ĵ).
Explanation:The subject of this problem is Newton's second law (F=ma), where F is the total force, m is the mass of the object, and a is acceleration. Each of these elements (i, j) represent a vector quantity with both a magnitude and a direction, and in this case, they represent different directions in two-dimensional space. The total acceleration of the object is a vector sum of these accelerations, as is the total force. Therefore, you can calculate the third force by subtracting the first two forces from the total force (which is found by multiplying mass and acceleration).
Total Force (F) = m*a = (2 kg)*(8.00m/s² î + 6.00m/s² ĵ) = 16.00 N î + 12.00 N ĵ.
We know two forces, F1 = 30.00 N î + 16.00 N ĵ and F2 = 12.00 N î + 8.00 N ĵ. Adding them, we get F1 + F2 = 42.00 N î + 24.00 N ĵ.
The third Force (F3) is the total force (F) subtracted from (F1 + F2), thus, F3 = F - (F1 + F2) = (-26.00 N î - 12.00 N ĵ).
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What is the wavelength of a sound wave traveling at 360 m/s and a
frequency of 150 Hz? *
1.0 m
2.4 m
оооо
Е Е ЕЕ
0.4 m
0.8 m
Answer:2.4m
Explanation:
Velocity=360m/s
Frequency=150Hz
Wavelength=velocity ➗ frequency
Wavelength=360 ➗ 150
Wavelength=2.4m
Final answer:
The wavelength of a sound wave traveling at 360 m/s and a frequency of 150 Hz is 2.4 m.
Explanation:
The wavelength of a sound wave can be calculated using the equation:
λ = v / ƒ
where λ is the wavelength, v is the speed of sound, and ƒ is the frequency of the sound wave. Given that the speed of sound is 360 m/s and the frequency is 150 Hz, we can substitute these values into the equation:
λ = 360 m/s / 150 Hz = 2.4 m
Therefore, the wavelength of the sound wave is 2.4 m.
Problem 16.71 A bowler projects an 8-in.-diameter ball weighing 12 lb along an alley with a forward velocity v0 of 15 ft/s and a backspin 0 of 9 rad/s. Knowing that the coefficient of kinetic friction between the ball and the alley is 0.10, determine (a) the time t1 at which the ball will start rolling without sliding, (b) the speed of the ball at time t1, (c) the distance the ball will have traveled at time t1.
Answer:
Find the attachments for complete solution
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0 m/s when the hand is 1.90 m above the ground. You may want to review (Pages 49 - 51) . For help with math skills, you may want to review: Quadratic Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. Part A How long is the ball in the air before it hits the ground
Answer:
t = 3.38 s
Explanation:
We have,
Initial speed of the ball that leaves the student's hand is 16 m/s
Initially, the hand is 1.90 m above the ground.
It is required to find the time for which the ball in the air before it hits the ground. We can use the equation of kinematics as :
[tex]y_f=y_i+ut+\dfrac{1}{2}at^2[/tex]
Here, [tex]y_f=-1.9\ m, y_i=0[/tex] and a=-g
The equation become:
[tex]-1.9=16t-\dfrac{1}{2}\times 9.8t^2[/tex]
After rearranging we get the above equation as :
[tex]4.9t^2-16t-1.9=0[/tex]
It is a quadratic equation, we need to find the value of t. On solving the above equation, we get :
t = -0.115 s and t = 3.38 s (ignore t = -0.115 s )
So, the ball is in air for 3.38 seconds before it hits the ground.
Select the three correct parts that make up the cell theory.
A
All living things are composed of cells.
B
All cells are microscopic.
C
Cells are the basic unit of structure and function in animals only.
D
All cells are produced from other cells.
E
Cells are the basic unit of structure and function in all living things.
F
Cells are made of organelles that perform functions necessary for the cell to do its job.
Plsss help me!! :/
Answer:
Options A, D and E....make up cell theory
Rudolf Virchow was the first to propose the concept of Omnis cellula-e cellula in relation to cell division.
What is the cell theory?It is a scientific theory that was proposed in the mid-nineteenth century.
Who proposed it?It was proposed by Theodore Schwann a British Zoologist and Matthias Schleiden a German botanist.
What are the three principles of the cell theory?The three principles are:
All living things are composed of cells.Cells are the basic unit of structure and function in all living things.All cells are produced from pre-existing cells.Hence options A, D, E are correct.
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The amount of light entering a camera or your eye is regulated by
a. an eyepiece
c. an aperture.
b. the cornea.
d. a set of compound lenses.
Answer:
c. an aperture
Explanation:
Aperture: It relates to the size of the opening, like a doorway, through which light moves into the eye, camera lens or a telescope. In human eye aperture is known as pupil, the black part in the center of the eye. The size of the pupil can increase or decrease depending upon the amount of light available. The same thing happens with a camera as well. The amount of light passing through the lens can be varied by varying the size of the aperture.
If the amount of work done on a book was 10 J and the force required to move the book was 2.5 N, what was the distance the book was moved?
Answer:
Explanation:4meters
Work=10J
Force=2.5N
Distance=work ➗ force
Distance=10 ➗ 2.5
Distance=4meter
Final answer:
Given that the work done on a book was 10 J and the force required to move the book was 2.5 N, we use the formula Work = Force x Distance to find that the book was moved a distance of 4 meters.
Explanation:
The question asks to calculate the distance a book was moved given that the work done on the book was 10 J and the force required to move the book was 2.5 N.
To find the distance, we can use the formula for work, which is:
Work (W) = Force (F) * Distance (d)
From the formula, we can solve for distance (d) by rearranging the equation:
Distance (d) = Work (W) / Force (F)
Plugging in the given values:
d = 10 J / 2.5 N
This calculation yields:
d = 4 m
Therefore, the book was moved a distance of 4 meters.
Weather balloons are high altitude balloons that are capable of reaching an altitude of 40 km (or 25 miles). They are usually filled with hydrogen or helium gas. One such balloon is filled with gas that is initially at a temperature of 300 K and at atmospheric pressure. This particular balloon can expand to a maximum diameter of 32 m when it reaches an altitude where the pressure and temperature are 0.028 atm and 190 K respectively. Determine the diameter of the balloon at lift off
Answer:
[tex]D_{1}=11.32 m[/tex]
Explanation:
We will need to use the ideal gas equation. The equation is given by:
[tex]PV=nRT[/tex]
P is the pressureV is the volumen is the amount of moleculesR is the ideal gas constantAs we have the same amount of molecules in the initial and final steps, therefore we can do this:
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex] (1)
- P(1) is the atmospheric pressure (P(1) = 1 atm) and P(2) is 0.028 atm
- T(1) is 300 K and T(2) is 190 K
- V(1) is the volume of the balloon in the first step, we can consider a spherical geometry so:
[tex]V_{1}=\frac{4}{3}\pi (\frac{D_{1}}{2})^{3}[/tex] (2)
[tex]V_{2}=\frac{4}{3}\pi (\frac{D_{2}}{2})^{3}[/tex] (3)
- D(2) = 32 m
So [tex]V_{2}=17157.3 m^{3}[/tex]
Let's solve the equation (1) for V(1)
[tex]V_{1}=\frac{T_{1}P_{2}V_{2}}{P_{1}T_{2}}[/tex]
[tex]V_{1}=\frac{300*0.028*17157.3}{1*190}[/tex]
[tex]V_{1}=758.53 m^{3}[/tex]
And using the equation (2) we can find D.
[tex]D_{1}=2(\frac{3}{4}V_{1})^{1/3}[/tex]
[tex]D_{1}=2(\frac{3}{4\pi}*758.53)^{1/3}[/tex]
[tex]D_{1}=11.32 m[/tex]
I hope it helps you!
what is the only part of a motor through which a current does not flow
Answer:
An electric motor
Explanation:
Answer:
the only part it doesn't flow through is an insulator
Explanation:
A conductor is material through which current flows easily; an insulator is material through which current does not flow easily.
mymathlab A diving pool that is 8 m deep and full of water has a viewing window on one of its vertical walls. Find the force on a window that is a%E2%80%8B circle, with a radius of 2m, tangent to the bottom of the pool.
Answer:
492.6 kN
Explanation:
Since the window as a radius of 2 m, its diameter is thus 4 m. Since the diving pool is 8 m deep, the height of water from the top of the window to the bottom of the pool is 8 - 4 = 4 m. The actual pressure acting on the window is thus
P = ρgh were ρ = density of water = 1000 kg/m³ g = 9.8 m/s² and h = 4 m
P = 1000 kg/m³ × 9.8 m/s² × 4 m = 39200 N/m².
Since P = F/A were F = force and A = area,
F = PA were A = area of window = πr² = π2² = 4π m² = 12.57 m²
F = 39200 N/m² × 12.57 m² = 492601.73 N = 492.6 kN
The force on the window of a diving pool due to the water pressure can be calculated by multiplying the pressure at the depth of the pool by the window's surface area. The pressure is found using density, gravity and depth, and the area is calculated from the radius of the circular window. The force comes out to be about 987 KN.
Explanation:This problem deals with hydrostatic pressure in a diving pool. The force on the window can be calculated by multiplying the pressure at the depth of the window by the surface area of the window.
First, we calculate the pressure at the depth of the pool using the formula P = ρgh, where P is the pressure, ρ (rho) is the density of the water (typically 1000 kg/m³ for fresh water), g is acceleration due to gravity (9.81 m/s²), and h is the depth (8m). This yields P = 1000 kg/m³ * 9.81 m/s² * 8m = 78480 Pa (Pascal).
Second, we calculate the area of the circular window. The area A of a circle is given by πr², where r is the radius. In this case, the radius r is 2m, so A = π * (2m)² = 12.57 m².
Finally, we calculate the force F by multiplying the pressure P by the area A. This gives F = P * A = 78480 Pa * 12.57 m² = 986857.6 N or approximately 987 KN.
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A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a radius of 2.97 m and a rotational inertia of 358 kg·m2 about the axis of rotation. A 69.5 kg student walks slowly from the rim of the platform toward the center. If the angular speed of the system is 1.96 rad/s when the student starts at the rim, what is the angular speed when she is 1.06 m from the center?
Answer:
4.36 rad/s
Explanation:
Radius of platform r = 2.97 m
rotational inertia I = 358 kg·m^2
Initial angular speed w = 1.96 rad/s
Mass of student m = 69.5 kg
Rotational inertia of student at the rim = mr^2 = 69.5 x 2.97^2 = 613.05 kg.m^2
Therefore initial rotational momentum of system = w( Ip + Is)
= 1.96 x (358 + 613.05)
= 1903.258 kg.rad.m^2/s
When she walks to a radius of 1.06 m
I = mr^2 = 69.5 x 1.06^2 = 78.09 kg·m^2
Rotational momentuem of system = w(358 + 78.09) = 436.09w
Due to conservation of momentum, we equate both momenta
436.09w = 1903.258
w = 4.36 rad/s
A certain computer chip that has dimensions of 3.67 cm and 2.93 cm contains 3.5 million transistors. If the transistors are square, what must be their maximum dimensions (in terms of microns or µm)? (Note: Devices other than transistors are also on the chip, and there must be room for the interconnections among the circuit elements. Transistors smaller than 0.7 μm are now commonly and inexpensively fabricated.)
Answer:
1.56 × 10^-3 cm.
Explanation:
So, we are given the following parameters from the question above;
Length = 3.67 cm, breadth = 2.93 cm, and the number of embedded transistors = 3.5 million.
Step one: find the area of the computer chip.
Therefore, Area = Length × breadth.
Area = 3.67 cm × 2.93 cm.
Area of the computer chip = 10.7531 cm^2. = 10.75 cm^2.
Step two: find the area of one transistor
The area of one transistor is; (area of the computer chip) ÷ (number of embedded transistors).
Hence;
The area of one transistor= 10.7531/4.4 × 10^6.
The area of one transistor= 2.44 × 10^-6 cm^2.
=> Note that We have our transistors as square, therefore;
The maximum dimension = √ (2.44 × 10^-6) cm^2.
The maximum dimension= 1.56 × 10^-3 cm.
An oven mitt is made of a material that is a poor thermal conductor.
Which is the most likely reason that the material makes a good oven mitt?
A: it is made mostly of water
B: it lacks electrons
C: it is made of tightly packed particles
D: its electrons do not move freely
Answer:
D
Explanation:
The most likely reason that a material that is a poor thermal conductor makes a good oven mitt is because its electrons do not move freely.
Explanation:The most likely reason that a material that is a poor thermal conductor makes a good oven mitt is because its electrons do not move freely.
In a good thermal conductor, such as metal, the electrons are able to move freely and transfer heat quickly. However, in a poor thermal conductor, such as the material of an oven mitt, the electrons are not able to move freely, which means they cannot transfer heat easily and thus provide insulation.
Therefore, option D: its electrons do not move freely is the most likely reason that a material makes a good oven mitt.
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