Answer:
The period of oscillation is 3.14 sec.
Explanation:
Given that,
Spring constant = 20 N/m
Mass = 5.0 kg
We need to calculate the time period
Using formula of time period
[tex]T =2\pi\sqrt{\dfrac{m}{k}}[/tex]
m= mass
k = spring constant
[tex]T =2\times3.14\sqrt{\dfrac{5.0}{20}}[/tex]
[tex]T =3.14\ sec[/tex]
Hence, The period of oscillation is 3.14 sec.
Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an electric field of N/C at point P. Determine the work required of an external agent to move a 3.0-μC charge along the x axis from infinity to point P without any net change in the kinetic energy of the particle.
Answer:
0.018 J
Explanation:
The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by
[tex]W = q \Delta V[/tex]
where
[tex]q=3.0 \mu C = 3.0 \cdot 10^{-6} C[/tex] is the magnitude of the charge
[tex]\Delta V = 6.0 kV = 6000 V[/tex] is the potential difference between point P and infinity
Substituting into the equation, we find
[tex]W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J[/tex]
The work required to move a 3.0 μC charge from infinity to point P with an electric potential of 6.0 kV is 0.018 Joules. This is calculated using the formula W = qV. Substituting the given values into the formula yields the answer.
The work required to move a charge in an electric potential is given by the formula:
W = qV
where W is work, q is the charge, and V is the electric potential. Here, the electric potential V at point P is 6.0 kV (or 6000 V), and the charge q is 3.0 μC (or 3.0 × 10⁻⁶ C).
Substitute the values into the formula:
W = (3.0 × 10⁻⁶ C) × (6000 V)
Which simplifies to:
W = 0.018 J
Therefore, the work required of an external agent to move the 3.0 μC charge from infinity to point P is 0.018 Joules.
A transverse wave is traveling on a string stretched along the horizontal x-axis. The equation for the vertical displacement y of the string is given by y = (1.8 x 10-3)cos[π(13x - 60t)], where all quantities are in SI units. What is the maximum speed of a particle of the string
Answer:
The maximum speed of a particle of the string is 0.34 m/s
Explanation:
The equation for the vertical displacement y of the string is given by y :
[tex]y=1.8\times 10^{-3}cos[\pi(13x-60t)][/tex].............(1)
The general equation of transverse wave is given by :
[tex]y=A(kx-\omega t)[/tex].............(2)
On comparing equation (1) and (2) we get,
[tex]\omega=60\pi[/tex]
Speed of a particle of the string is maximum when displacement is equal to zero. Maximum speed is given by :
[tex]v_{max}=A\omega[/tex]
Where, A = amplitude of wave
[tex]\omega=60\pi[/tex]
So, [tex]v_{max}=1.8\times 10^{-3}\times 60\pi[/tex]
[tex]v_{max}=0.34\ m/s[/tex]
Hence, this is the required solution.
The maximum speed of any particle on the string can be determined by the product of the amplitude and the angular frequency of the wave. In this case, the maximum speed is found to be 0.108 m/s.
Explanation:The equation given represents the vertical displacement of a transverse wave on a string. We're trying to find the maximum speed of a particle of the string. This refers to the highest speed at which any string particle will be moving as the wave passes by.
From the given wave equation, the particle on the string will be moving in a simple harmonic motion. The maximum speed would be at the equilibrium position with a value equal to the product of the amplitude (A) and the angular frequency (ω).
In your case, the given equation is: y = (1.8x10-3)cos[π(13x - 60t)]. In this equation, the angular frequency (ω) is equal to 60 and the amplitude (A) is equal to 1.8 x 10-3.
Therefore, the maximum speed (vmax) of any particle on the string would be given by: vmax = Aω = (1.8 x 10-3) * 60 = 0.108 m/s
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A 1400kg automobile moving at a maximum speed of 23m/s on a level circular track of readius of 95m. What is the coefficient of friction?
Answer:
The coefficient of friction is 0.56
Explanation:
It is given that,
Mass of the automobile, m = 1400 kg
Speed of the automobile, v = 23 m/s
Radius of the track, r = 95 m
The automobile is moving in a circular track. The centripetal force is given by :
[tex]F_c=\dfrac{mv^2}{r}[/tex]............(1)
Frictional force is given by :
[tex]F_f=\mu mg[/tex]...................(2)
[tex]\mu[/tex] = coefficient of friction
g = acceleration due to gravity
From equation (1) and (2), we get :
[tex]\dfrac{mv^2}{r}=\mu mg[/tex]
[tex]\mu=\dfrac{v^2}{rg}[/tex]
[tex]\mu=\dfrac{(23\ m/s)^2}{95\ m\times 9.8\ m/s^2}[/tex]
[tex]\mu=0.56[/tex]
So, the coefficient of friction is 0.56. Hence, this is the required solution.
To find the coefficient of friction, calculate the maximum speed at which the car can negotiate the curve without slipping using the formula v = √(μrg). Plug in the values and solve for μ to get the coefficient of friction. The approximate value is 0.24.
Explanation:To find the coefficient of friction, we will first calculate the maximum speed at which the car can negotiate the curve without slipping. The maximum speed is given by the formula:
v = √(μrg)
Where v is the maximum speed, μ is the coefficient of friction, r is the radius of the curve, and g is the acceleration due to gravity.
Plugging in the values, we get:
23 = √(μ * 1400 * 9.8 * 95)
Squaring both sides of the equation, we have:
529 = μ * 1400 * 9.8 * 95
Simplifying the equation, we find:
μ = 529 / (1400 * 9.8 * 95)
Calculating the value, we get:
μ ≈ 0.24
Therefore, the coefficient of friction is approximately 0.24.
Four blocks of weights are required using which any body whose weight is between 1kg and 40 kg can be weighed. Find the four weights
Answer:
The weights are 1 kg, 3kg, 9kg and 27kg.
Explanation:
The weights are 1 kg, 3kg, 9kg and 27kg.
1+3+9+27= 40
27+9+3= 39
27+9+3-1=38
27+9+1=37
27+9=36
27+9-1=35
27+9+1-3=34
27+9-3=33
27+9-3-1=32
27+3+1=31
27+3=30
27+3-1=29
27+1=28
27
27-1=26
27+1-3=25
27-3=24
27-3-1=23
27+3+1-9=22
27+3-9=21
27+3-9-1=20
Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.
Heavy water usually refers to water that : (A) has been frozen, and so is more dense (B) has had its hydrogen removed (C) is radioactive (D) contains deuterium
Answer:
Option (D)
Explanation:
The chemical formula for normal water is H2O and the chemical formula for heavy water is D2O.
Where D is deuterium which is the isotope of hydrogen.
There are three isotopes of hydrogen.
1H1 it is called protium.
1H2 it is called deuterium.
1H3 it is called tritium.
Heavy water usually refers to water that contains deuterium.
How many 90-W, 120-V light bulbs can be connected to a 20-A, 120-V circuit without tripping the circuit breaker? (Note: This description of a light bulb gives the power it dissipates when connected to the stated potential difference; that is, a 25-W, 120-V light bulb dissipates 25 W when connected to a 120-V line.) 26
Final answer:
To connect 90-W, 120-V light bulbs to a 20-A, 120-V circuit without tripping the breaker, you can connect a maximum of 26 bulbs.
Explanation:
To determine how many 90-W, 120-V light bulbs can be connected to a 20-A, 120-V circuit without tripping the circuit breaker, we need to calculate the total power consumption of the light bulbs and compare it to the circuit's maximum current capacity.
Using the formula: Power (P) = Voltage (V) x Current (I), we can rearrange it to solve for current: I = P/V.
For each bulb, I = 90 W / 120 V = 0.75 A.
Given the maximum current capacity of the circuit is 20 A, divide the total current capacity by the current per bulb:
Total bulbs = 20 A / 0.75 A = 26.67 bulbs.
Since we can't have a fraction of a bulb, the maximum number of 90-W, 120-V light bulbs that can be connected to the circuit without tripping the breaker is 26 bulbs.
Final answer:
By calculating the current drawn by each 90-W, 120-V light bulb, we find that about 26 bulbs can be connected to a 20-A, 120-V circuit without exceeding the current limit and tripping the circuit breaker.
Explanation:
The question involves calculating the number of 90-W, 120-V light bulbs that can be connected to a 20-A, 120-V circuit. First, it is important to understand that the power rating of a light bulb (like 90 W) indicates the power consumed when it operates at its specified voltage (120 V in this case).
The power, voltage, and current are related by the formula Power (P) = Voltage (V) × Current (I).
Therefore, the current for one 90-W bulb can be found by rearranging the formula:
I = P/V, which gives I = 90 W / 120 V = 0.75 A.
Since the circuit can handle up to 20 A, you divide the total allowable current by the current per bulb: 20 A / 0.75 A/bulb = approximately 26.67.
This means you can connect 26 bulbs without tripping the circuit breaker, since you cannot connect a fraction of a bulb.
You throw a baseball directly upward at time ????=0 at an initial speed of 14.9 m/s. What is the maximum height the ball reaches above where it leaves your hand? Ignore air resistance and take ????=9.80 m/s2.
Answer:
Maximum height, h = 11.32 meters
Explanation:
It is given that,
The baseball is thrown directly upward at time, t = 0
Initial speed of the baseball, u = 14.9 m/s
Ignoring the resistance in this case and using a = g = 9.8 m/s²
We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :
[tex]v^2-u^2=2ah[/tex]
At maximum height, v = 0
and a = -g = -9.8 m/s²
[tex]h=\dfrac{v^2-u^2}{2a}[/tex]
[tex]h=\dfrac{0-(14.9\ m/s)^2}{2\times -9.8\ m/s^2}[/tex]
h = 11.32 meters
Hence, the maximum height of the baseball is 11.32 meters.
A level pipe contains a nonviscous, incompressible fluid with a density 1200 kg/m3 that is flowing steadily. At one position within the pipe, the pressure is 300 kPa and the speed of the flow is 20.0 m/s. At another position, the pressure is 200 kPa. What is the speed of the flow at this second position?
Answer:
V2 = 23.8m/s
Explanation:
as we know That,
p1/(r*g) + (v1^2)/2*g = p2/(r*g) + (v2^2)/2*g
p1=300kPa=300000Pa
p2=200kPa=200000Pa
r=1200kG/m^3
v1 = 20 m/s
250+200 = 166.66+(v2^2)/2
v2=(450-166.66)*2
v2^2 =566.68
v2=23.8m/s
Your answer is : V2 = 23.8m/s
The speed of the flow at this second position is 23.8 m/s.
The given parameters:
Density of the fluid, ρ = 1200 kg/m³ Pressure of the pipe, P₁ = 300 kPaSpeed of the flow, v₁ = 20.0 m/sPressure at second position, P₂ = 200 kPaThe speed of the flow at this second position is calculated by applying Bernoulli's equation as follows;
[tex]P_1 + \frac{1}{2} \rho v_1^2= P_2 + \frac{1}{2} \rho v_2 ^2\\\\(P_1 - P_2) + \frac{1}{2} \rho v_1^2 = \frac{1}{2} \rho v_2 ^2\\\\2(P_1-P_2) + \rho v_1^2 = \rho v_2^2\\\\\frac{2(P_1-P_2) }{\rho} + v_1^2 = v_2^2\\\\\sqrt{\frac{2(P_1-P_2) }{\rho} + v_1^2} = v_2\\\\\sqrt{\frac{2(300,000-200,000) }{1200} + 20^2}= v_2 \\\\\sqrt{566.67} = v_2\\\\23.8\ m/s = \ v_2[/tex]
Thus, the speed of the flow at this second position is 23.8 m/s.
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|| Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down the runway. A smaller plane with the same acceleration has a takeoff speed of 40 m/s. Starting from rest, after what distance will this smaller plane reach its takeoff speed?
The distance at which the smaller plane will reach its takeoff speed, we can use the equation of motion:
[tex]\[ v^2 = u^2 + 2as \][/tex]
v is the final velocity (takeoff speed) of the smaller plane (40 m/s)
u is the initial velocity (rest) of the smaller plane (0 m/s)
a is the acceleration of both planes (which is the same)
s is the distance covered by the smaller plane before reaching its takeoff speed (what we want to find)
[tex]\[ s = \frac{{v^2 - u^2}}{{2a}} \][/tex]
[tex]\[ s = \frac{{(40 \, \text{m/s})^2 - (0 \, \text{m/s})^2}}{{2a}} \][/tex]
[tex]\[ a = \frac{{v - u}}{{t}} = \frac{{(80 \, \text{m/s}) - (0 \, \text{m/s})}}{{30 \, \text{s}}} \][/tex]
[tex]\[ s = \frac{{(40 \, \text{m/s})^2}}{{2 \times \left(\frac{{80 \, \text{m/s}}}{30 \, \text{s}}\right)}} \]\[ s = \frac{{1600 \, \text{m}^2/\text{s}^2}}{{\frac{{160 \, \text{m/s}}}{30 \, \text{s}}}} \]\[ s = \frac{{1600 \, \text{m}^2/\text{s}^2 \times 30 \, \text{s}}}{{160 \, \text{m/s}}} \]\[ s = \frac{{48,000 \, \text{m}^2/\text{s}}}{{160 \, \text{m/s}}} \]\[ s = 300 \, \text{m} \][/tex]
Therefore, the smaller plane will reach its takeoff speed after traveling a distance of 300 meters.
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By using the equation of motion v² = u² + 2as, we can calculate the acceleration of the first plane based on its initial velocity, final velocity, and distance covered. Then, using the same acceleration and applying it to the second plane's parameters, we can find the distance it covers before reaching its takeoff speed.
Explanation:To solve this problem, we'll make use of the equation of motion usually represented as v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered.
For the first plane, we have v = 80 m/s, u = 0 m/s (because the plane starts from rest), and s = 1200 m. Plugging these into our equation, we can solve for a.
Once we have the acceleration, we can then apply it to the second plane to find the distance it covers (s) before reaching its takeoff speed of 40 m/s. Given a = acceleration calculated above and v = 40 m/s, u = 0 m/s (since the second plane also starts from rest), we can rearrange our equation of motion to solve for s.
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Carlos attempts to get a refrigerator magnet to stick on the S end of a bar magnet. He is unable to do so and states that the bar magnet is broken. Evaluate his argument: is this correct or incorrect? Explain.
Answer:
It is incorrect.
Explanation:
Refrigerator magnets are actually multiple layers of magnets, and they are stacked using opposite polarities, so the south end of the magnet might be under a north, so if he puts the magnet on its side it will probably attract.
The argument that Carlos attempts to get a refrigerator magnet to stick on the S end of a bar magnet. He is unable to do so and states that the bar magnet is broken is incorrect.
What is magnet?A substance which is negative charged at one end and positive charged at another end creating the magnetic field around it.
A bar is a magnet which can attract the particle which are charged. It has two poles North and South.
Refrigerator magnets are having multiple layers of magnets. They are stacked using opposite polarities. So, south pole of the magnet will definitely attract a north even if it is broken.
Thus, the argument is incorrect.
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A cardiac defibrillator stores 1275 J of energy when it is charged to 5.6kV What is the capacitance? O 11.3 pF 96.1 pF O 813 F
Answer: [tex]813.13(10)^{-7}F[/tex]
Explanation:
The answer is not among the given options. However, this is a good example of the relation between the energy stored in a capacitor [tex]W[/tex]and its capacitance [tex]C[/tex], which is given by the following equation:
[tex]W=\frac{1}{2}CV^{2}[/tex] (1)
Where:
[tex]W=1275J[/tex]
[tex]V=5.6kV=5.6(10)^{3}V[/tex] is the voltage
[tex]C[/tex] is the capacitance in Farads, the value we want to find
Isolating [tex]C[/tex] from (1):
[tex]C=\frac{2W}{V^{2}}[/tex] (2)
[tex]C=\frac{2(1275J)}{(5.6(10)^{3}V)^{2}}[/tex] (3)
Finally:
[tex]C=0.000081313F=813.13(10)^{-7}F[/tex] This is the capacitance of the cardiac defibrillator
Two forces of magnitudes 4.0 Newtons and 3.0 Newtons pull on a box. The forces make an angle of 40° with each other. What is the magnitude of a third force that must be applied to keep the box in equilibrium?
Answer:
6.6 N
Explanation:
Let's take the direction of the force of 4.0 N as positive x-direction. This means that the force of 3.0 N is at 40 degrees above it. So the components of the two forces along the x- and y-directions are:
[tex]F_{1x} = 4.0 N\\F_{1y} = 0[/tex]
[tex]F_{2x} = 3.0 N cos 40^{\circ}=2.3 N\\F_{2y} = 3.0 N sin 40^{\circ} = 1.9 N[/tex]
So the resultant has components
[tex]F_x = F_{1x}+F_{2x}=4.0 N +2.3 N = 6.3 N\\F_y = F_{1y} + F_{2y} = 0 + 1.9 N = 1.9 N[/tex]
So the magnitude of the resultant is
[tex]F=\sqrt{F_x^2 +F_y^2}=\sqrt{(6.3)^2+(1.9)^2}=6.6 N[/tex]
And in order for the body to be balanced, the third force must be equal and opposite (in direction) to this force: so, the magnitude of the third force must be 6.6 N.
The magnitude of the resultant force when a 40 N force acts in the x-direction and a 30 N force acts in the y-direction is calculated using the Pythagorean theorem. Since these forces are perpendicular, the resultant force is the hypotenuse of a right triangle formed by the two forces, which equals 50 N.
Given a force of 40 N in the positive x-direction and a force of 30 N in the positive y-direction, these forces are perpendicular to each other. Hence, the resultant force is the hypotenuse of a right-angled triangle where the sides are the two forces.
To find the magnitude of the resultant force (R), we use the formula:
R = sqrt((40 N)^2 + (30 N)^2)
Calculate the R value (with calculator):
R = sqrt((1600) + (900))
R = sqrt(2500 N^2)
R = 50 N
Therefore, the magnitude of the resultant force is 50 N.
A cord, 6 m long and fixed at both ends, supports a standing wave with a total of 4 nodes. What is the wavelength of the standing wave?
Answer:
The wavelength of the standing wave is 3 m.
Explanation:
Given that,
Length = 6 m
Nodes = 4
We need to calculate the wave length
Using formula of nodes
[tex]L=\dfrac{n}{2}\lamda[/tex]
[tex]\lambda=\dfrac{2L}{n}[/tex]
Where, l = length
n = number of nodes
[tex]\lambda[/tex] = wavelength
Put the value into the formula
[tex]\lambda=\dfrac{2\times6}{4}[/tex]
[tex]\lambda=3\ m[/tex]
Hence, The wavelength of the standing wave is 3 m.
Answer:
Should be....
λ = 2(6)/3 = 4 m
Explanation:
λ = 2L/n where n is the antinode
n = 3 instead of 4 (4 is where the wave crosses zero)
A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held by cables so that it is initially compressed 50 mm. Knowing that the velocity of the package is 2 m/s when it is 8 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest.
Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.
We use the principles of Kinetic and Potential Energy to establish an energy conservation equation, balancing the initial kinetic energy of the box against the potential energy stored in the spring at rest. Solving this equation gives the maximum additional deformation of the spring.
Explanation:From the problem, we know that the mass of the package (m) is 50 kg, the initial speed (v) is 2 m/s, the spring's constant (k) is 30 kN/m or 30000 N/m, and the initial compression of the spring (xi) is 50 mm or 0.05 m. We are tasked to determine the maximum additional deformation (xf) of the spring in bringing the package to rest.
Your problem involves Kinetic Energy and Potential Energy concepts. If we consider the instant when the package begins to hit the spring and the moment when it comes to rest, the energy conservation law can be applied, stating that the initial kinetic energy of the box equals the work done by the spring (which is the potential energy stored in it).
So, the kinetic energy of the package when it begins to hit the spring equals 1/2 * m * v^2, and the potential energy of the spring when the package comes to rest equals 1/2 * k * (xi+xf)^2.
Therefore, equating both expressions and resolving the quadratic equation for xf, provides us with the maximum additional deformation of the spring.
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An ultracentrifuge spins biological samples at very high speeds in order to separate their constituent microscopic components. One such instrument spins at 120,000 rpm, and the samples experience a centripetal acceleration of 6.6×10^6 m/s^2. What is the radius of the circle around which the samples travel?
The radius of the circle around which samples travel in the mentioned ultracentrifuge is approximately 4.18 cm. This was calculated using the centripetal acceleration and the rotational speed of the ultracentrifuge, demonstrating how principles of physics apply to scientific and medical practices.
Explanation:An ultracentrifuge utilizes very high speeds and centripetal acceleration to separate minute biological components in a sample. Given that the ultracentrifuge mentioned spins at 120,000 rotations per minute (rpm) and the centripetal acceleration recorded is 6.6×10⁶ m/s², we need to determine the radius of the circle around which these samples travel. This can be formulated using the physics concept of uniform circular motion.
To solve this, we will use the formula for centripetal acceleration, which is a = ω²r, where 'a' is centripetal acceleration, 'ω' is angular velocity, and 'r' is the radius of the circle. First, we need to convert the rotational speed from rpm to radians per second: 1 rpm is equal to 2π rad/60 s, so 120,000 rpm equals 2π * 120,000 / 60 = 12,566 rad/s. Then, we substitute these values into the formula and solve for the radius, giving us r = a/ω² ≈ 0.0418 meters or 4.18 cm.
The samples would therefore travel around a circular path with a radius of approximately 4.18 cm. This is a great example of how the principles of physics are used in scientific and medical practices, like the operation of an ultracentrifuge.
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The radius of the circle around which the samples travel in the ultracentrifuge is approximately 13.2 millimeters.
Calculating the Radius of Circular Motion in an Ultracentrifuge
To determine the radius of the circle around which the samples travel in an ultracentrifuge, use the formula for centripetal acceleration: a = ω²r, where a is the centripetal acceleration, ω is the angular velocity in radians per second, and r is the radius.
First, convert the given rotational speed from revolutions per minute (rpm) to radians per second:
Given: 120,000 rpm
Convert rpm to revolutions per second: 120,000 rpm ÷ 60 seconds/min = 2,000 revolutions per second
Convert revolutions per second to radians per second (since there are 2π radians in one revolution): 2,000 rev/s × 2π radians/rev = 4,000π radians/second
Now that we have the angular velocity, we can use the centripetal acceleration formula. Rearrange the formula to solve for radius r:
r = a / ω²
Substitute the given values (where a = 6.6×106 m/s² and ω = 4,000π rad/s):
r = 6.6×106 m/s² / (4,000π rad/s)²
Calculate the radius:
Square the angular velocity: (4,000π rad/s)² = 16,000,000π² rad²/s²
Divide the acceleration by the squared angular velocity: r = 6.6×[tex]10^6[/tex] m/s² / 16,000,000π² rad²/s² = 0.131 / π² m
Simplify using π ≈ 3.14159: r ≈ 0.131 / (3.14159)² ≈ 0.0132 m or 13.2 mm
Thus, the radius of the circle around which the samples travel in the ultracentrifuge is approximately 13.2 millimeters.
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1, the rocket engine is shut down and the sled moves with constant velocity v for a time t2. If the total distance traveled by the sled is 18,350 ft and the total time is 90 s, find the following. (a) the times t1 and t2 t1 = s t2 = s (b) the velocity v
Answer:
a) t₁ = 4.76 s, t₂ = 85.2 s
b) v = 209 ft/s
Explanation:
Constant acceleration equations:
x = x₀ + v₀ t + ½ at²
v = at + v₀
where x is final position,
x₀ is initial position,
v₀ is initial velocity,
a is acceleration,
and t is time.
When the engine is on and the sled is accelerating:
x₀ = 0 ft
v₀ = 0 ft/s
a = 44 ft/s²
t = t₁
So:
x = 22 t₁²
v = 44 t₁
When the engine is off and the sled is coasting:
x = 18350 ft
x₀ = 22 t₁²
v₀ = 44 t₁
a = 0 ft/s²
t = t₂
So:
18350 = 22 t₁² + (44 t₁) t₂
Given that t₁ + t₂ = 90:
18350 = 22 t₁² + (44 t₁) (90 − t₁)
Now we can solve for t₁:
18350 = 22 t₁² + 3960 t₁ − 44 t₁²
18350 = 3960 t₁ − 22 t₁²
9175 = 1980 t₁ − 11 t₁²
11 t₁² − 1980 t₁ + 9175 = 0
Using quadratic formula:
t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22
t₁ = 4.76, 175
Since t₁ can't be greater than 90, t₁ = 4.76 s.
Therefore, t₂ = 85.2 s.
And v = 44 t₁ = 209 ft/s.
A force of 6 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching it from its natural length to 8 in. beyond its natural length?
Explanation and answer:
Given:
6 lb is needed to stretch 4 inches beyond natural length.
Need work done to stretch same string from natural length to 8 inches.
Solution:
string stiffness, K
= Force / stretched distance
= 6 lb / 4 inches
= 1.5 lb/inch
Work done on a string of stiffness K
= (Kx^2)/2 lb-in
= 1.5 lb/in *(8 in)^2)/2
= 48 lb-in.
Final answer:
To find the work done in stretching a spring, one calculates the spring constant using Hooke's Law and then applies the work-energy principle. For a spring stretched 8 inches beyond its natural length, with a spring constant derived as 1.5 lb/in, the work done is 48 lb-in.
Explanation:
To calculate the work done in stretching a spring from its natural length to 8 inches beyond it, we use Hooke's Law and the work-energy principle. Given that a force of 6 lb is required to hold the spring stretched 4 inches beyond its natural length, we first find the spring constant, k, using the formula F = kx, where F is the force applied and x is the displacement from the spring's natural length.
So, k = F / x = 6 lb / 4 in = 1.5 lb/in. The work done, W, in stretching the spring from its natural length to 8 inches beyond can be calculated using the formula W = 1/2 k x², where x is the total displacement from the spring's natural length. In this case, x = 8 in, so W = 1/2 * 1.5 lb/in * (8 in)² = 48 lb-in.
A 28-kg particle exerts a gravitational force of 8.3 x 10^-9 N on a particle of mass m, which is 3.2 m away. What is m? A) 140 kg B) 8.5 x 10^-10 kg C) None of the choices are correct D) 46 kg E)1300 kg
Answer:
Mass of another particle, m = 46 kg
Explanation:
it is given that,
Mass of first particle, m₁ = 28 kg
Gravitational force, [tex]F=8.3\times 10^{-9}\ N[/tex]
Distance between the particles, d = 3.2 m
We need to find the mass m of another particle. It is given by the formula as follows :
[tex]F=G\dfrac{m_1m}{d^2}[/tex]
[tex]m=\dfrac{Fd^2}{Gm_1}[/tex]
[tex]m=\dfrac{8.3\times 10^{-9}\ N\times (3.2\ m)^2}{6.67\times 10^{-11}\times 28\ kg}[/tex]
m = 45.5 kg
or
m = 46 kg
So, the correct option is (d) "46 kg". Hence, this is the required solution.
The correct answer is B) 8.5 x 10⁻¹⁰ kg is the mass.
[tex]\[ F = G \frac{M m}{r^2} \][/tex]
Where G is the gravitational constant, approximately equal to [tex]\( 6.674 \times 10^{-11} \) N(m/kg)^2[/tex].
Given that the force F is [tex]\( 8.3 \times 10^{-9} \)[/tex] N, the mass M is 28kg, and the distance ris 3.2 m, we can rearrange the equation to solve for m:
[tex]\[ m = \frac{F r^2}{G M} \][/tex]
Plugging in the given values:
[tex]\[ m = \frac{(8.3 \times 10^{-9} \text{ N}) \times (3.2 \text{ m})^2}{6.674 \times 10^{-11} \text{ N(m/kg)}^2 \times 28 \text{ kg}} \] \[ m = \frac{(8.3 \times 10^{-9}) \times (3.2)^2}{6.674 \times 10^{-11} \times 28} \] \[ m = \frac{(8.3 \times 10^{-9}) \times 10.24}{1.86872 \times 10^{-9}} \][/tex]
[tex]\[ m = \frac{8.5056 \times 10^{-8}}{1.86872 \times 10^{-9}} \] \[ m \approx 4.55 \times 10^{-1} \] \[ m \approx 8.5 \times 10^{-10} \text{ kg} \][/tex]
Therefore, the mass m of the second particle is approximately [tex]\( 8.5 \times 10^{-10} \)[/tex] kg, which corresponds to option B.
Which one has larger kinetic energy: a 500-kg object moving at 40 m/s or a 1000-kg object moving at 20 m/s? a. The 500-kg object b. The 1000-kg object c. Both have the same kinetic energy.
Final answer:
Using the kinetic energy formula, the 500-kg object moving at 40 m/s has a kinetic energy of 400,000 J, while the 1000-kg object moving at 20 m/s has a kinetic energy of 200,000 J. Therefore, the 500-kg object has larger kinetic energy.
Explanation:
To determine which object has larger kinetic energy, we can use the formula Ek = 1/2 m v2, where Ek is the kinetic energy, m is the mass of the object, and v is the velocity of the object. Plugging in the values for both objects:
For the 500-kg object moving at 40 m/s: Ek = 1/2 (500 kg) (40 m/s)2 = 1/2 (500) (1600) = 400,000 JFor the 1000-kg object moving at 20 m/s: Ek = 1/2 (1000 kg) (20 m/s)2 = 1/2 (1000) (400) = 200,000 JComparing the results, the 500-kg object has larger kinetic energy than the 1000-kg object. Therefore, the correct answer is:
a. The 500-kg object
A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car's motion, the eastward component of the car's velocity is given by υx(t)=(0.980m/s3)t2.
What is the acceleration of the car when υx = 14.5 m/s ?
Answer:
7.54 m/s²
Explanation:
uₓ(t) = (0.980 m/s³) t²
Acceleration is the derivative of velocity with respect to time.
aₓ(t) = 2 (0.980 m/s³) t
aₓ(t) = (1.96 m/s³) t
When uₓ = 14.5 m/s, the time is:
14.5 m/s = (0.980 m/s³) t²
t = 3.85 s
Plugging into acceleration equation:
aₓ = (1.96 m/s³) (3.85 s)
aₓ = 7.54 m/s²
This question is dealing with velocity, acceleration and time of motion.
Acceleration is; aₓ = 7.54 m/s²
We are told that the eastward component of the car's velocity is;
uₓ(t) = (0.980 m/s³) t²
Now, from calculus differentiation in maths, we know that with respect to time, the derivative of velocity is equal to the acceleration.
Thus;
aₓ(t) = du/dt = 2t(0.980 m/s³)
aₓ(t) = 1.96t m/s³
We w ant to find the acceleration of the car when velocity is; uₓ = 14.5 m/s. Let us find the time first and then plug the value into the acceleration equation.
Thus;
14.5 m/s = (0.980 m/s³) t²
14.5/0.98 = t²
t = 3.85 s
Putting 3.85 for t in the acceleration equation to get;
aₓ = (1.96 m/s³) (3.85 s)
aₓ = 7.54 m/s²
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in ideal diesel cycle the working substance is (A) air (B) diesel (C) misture of air and diesel (D) any combustible gas
Answer:
(D) any combustible gas
Explanation:
In ideal diesel cycle the working substance is any combustible gas.
Final answer:
The working substance in the ideal diesel cycle, during the initial intake and compression process, is air. Diesel fuel is injected only after the air has been adiabatically compressed and heated to a sufficient temperature for ignition.
Explanation:
In the ideal diesel cycle, the working substance during the intake stroke is air alone. This air is then compressed adiabatically to a high temperature before diesel fuel is injected. It is important to note that the diesel fuel itself is not part of the initial intake but is rather injected after the air has been compressed, which subsequently ignites due to the high temperature from compression.
The compression stroke significantly raises the air's temperature by compressing it adiabatically from state A to state B. Then during the power stroke, the diesel fuel added to the hot, high pressure air causes ignition and continues the cycle. Therefore, the correct answer to the question is (A) air, as it is the only substance involved in the complete cycle from the intake to the compression phase before fuel injection.
After two half lives, the fraction of nuclei remaining is : (A) 1/2 (B) 1/9 (C) 1/8 (D) 1/4
Answer:
D
Explanation:
After each half life, half the original amount remains.
So after two half lives, (1/2)^2 remains, or 1/4.
(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic field is 0.301 T. Each has a speed of 7.92 × 10^5 m/s, but one ion is singly charged and the other is doubly charged. Find the radius of the circular path followed by the singly charged ion in the field. Answer in units of cm. (b) Find the radius of the circular path followed by the doubly charged ion in the field. (c) Find the distance of separation when they have moved through one-half of their circular path and strike a piece of photographic paper.
Answer:
Part a)
[tex]R_1 = 0.072 m[/tex]
Part b)
[tex]R_2 = 0.036 m[/tex]
Part c)
d = 0.072 m
Explanation:
Part a)
As we know that the radius of the charge particle in constant magnetic field is given by
[tex]R = \frac{mv}{qB}[/tex]
now for single ionized we have
[tex]R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}[/tex]
[tex]R_1 = 0.072 m[/tex]
Part b)
Similarly for doubly ionized ion we will have the same equation
[tex]R = \frac{mv}{qB}[/tex]
[tex]R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}[/tex]
[tex]R_2 = 0.036 m[/tex]
Part c)
The distance between the two particles are half of the loop will be given as
[tex]d = 2(R_1 - R_2)[/tex]
[tex]d = 2(0.072 - 0.036)[/tex]
[tex]d = 0.072 m[/tex]
Each plate of an air-filled parallel-plate air capacitor has an area of 0.0040 m2, and the separation of the plates is 0.080 mm. An electric field of 5.3 × 106 V/m is present between the plates. What is the energy density between the plates? (ε0 = 8.85 × 10-12 C2/N · m2)
Answer:
124.3 J/m^3
Explanation:
The energy density between the plates of a parallel-plate capacitor is given by
[tex]u=\frac{1}{2}\epsilon_0 E^2[/tex]
where
ε0 = 8.85 × 10-12 C2/N · m2 is the vacuum permittivity
E is the electric field strength
In this problem,
E = 5.3 × 106 V/m
So the energy density is
[tex]u=\frac{1}{2}(8.85\cdot 10^{-12}F/m) (5.3\cdot 10^6 V/m)^2=124.3 J/m^3[/tex]
The energy density between the plates of a capacitor can be calculated using the formula U = 0.5 * ε0 * E². By substituting the given values for the electric field (E) and the permittivity of free space (ε0) into the formula, we can ascertain the energy density.
Explanation:To find the energy density between the plates of the capacitor we can use the formula U = 0.5 * ε0 * E².
First, we need to find the electric field (E) which is given to us as 5.3 × 10^6 V/m. The permittivity of free space (ε0) is 8.85 × 10^-12 C^2/N * m^2.
Then we substitute the values into the formula to calculate the energy density:
U = 0.5 * ε0 * E² = 0.5 * 8.85 × 10^-12 C^2/N * m^2 * (5.3 × 10^6 V/m)²
So, the energy density between the plates of the capacitor is calculated by using the given values of the electric field and permittivity of free space.
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A banked road comer of radius 146 m is to be constructed. If the comer is designed for vehicles moving with a speed of 20.0 m/s, what should the banking angle be, in units of degrees?
Answer:
Angle of banking be 15.64 degrees.
Explanation:
It is given that,
Radius of banked road comer, r = 146 m
Velocity of vehicle, v = 20 m/s
Banking angle is defined as the angle at which a vehicle is in inclined position. Angle of banking is given by :
[tex]tan\ \theta=\dfrac{v^2}{rg}[/tex] , g = acceleration due to gravity
[tex]tan\ \theta=\dfrac{(20\ m/s)^2}{146\ m\times 9.8\ m/s^2}[/tex]
[tex]tan\ \theta=0.28[/tex]
[tex]\theta=tan^{-1}(0.28)[/tex]
[tex]\theta=15.64\ degrees[/tex]
Where, [tex]\theta[/tex] = angle of banking
Hence, this is the required solution.
(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of energy into a circular spot 2.00 mm in diameter in 4.00 s
Answer:
4.42 x 10⁷ W/m²
Explanation:
A = energy absorbed = 500 J
η = efficiency = 0.90
E = Total energy
Total energy is given as
E = A/η
E = 500/0.90
E = 555.55 J
t = time = 4.00 s
Power of the beam is given as
P = E /t
P = 555.55/4.00
P = 138.88 Watt
d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m
Area of the circular spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Intensity of the beam is given as
I = P /A
I = 138.88 / (3.14 x 10⁻⁶)
I = 4.42 x 10⁷ W/m²
The intensity of a laser beam : 4.42.10⁷ W/m²
Further explanationThe energy transferred by waves per unit area per unit time is called wave intensity
Because energy per unit time is Power, the intensity of the wave is equal to Power divided by area
[tex]\rm P=\dfrac{W}{t}[/tex]
P = power, watt
W = energy, J
t = time, s
For waves that spread in all directions, the intensity at the distance R from the source can be formulated
[tex]\rm I=\dfrac{Power}{Area}=\dfrac{P}{\pi .R^2}[/tex]
From the equation above shows the intensity of the wave is inversely proportional to the square of the distance from the source.
[tex]\rm I\approx \dfrac{1}{R^2}[/tex]
The farther the wave spreads, the smaller the intensity
Cancerous tissue area:
[tex]\rm A=\dfrac{1}{4}\pi d^2\\\\A=\dfrac{1}{4}\pi(2.10^{-3})^2\Rightarrow d=2~mm=2.10^{-3}\:m\\\\A=\frac{1}{4}\pi 4.10^{-6}\\\\A=\pi .10^{-6}\\\\A=3.14.10^{-6}[/tex]
So that the intensity
[tex]\rm 0.9I(only~90\%~absorbed)=\dfrac{W}{A.t}\\\\0.9I=\dfrac{500}{3.14.10^{-6}.4}\\\\I=4.42.10^7~\dfrac{W}{m^2}[/tex]
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An electric device delivers a current of 5 a to a device.
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A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output power of the charger in unit of watt?
Answer:
Output power of the charger is 9.85 watts.
Explanation:
It is given that,
Output current of the USB charger, I = 1.97 A
The standard USB output voltage, V = 5 V
We need to find the output power of the charger. It can be determined using the following formula as :
P = V × I
[tex]P=5\ V\times 1.97\ A[/tex]
P = 9.85 watts
The output power of the charger is 9.85 watts. Hence, this is the required solution.
The USB charger's output power, in watts, is calculated using the formula P = IV. Substituting the given values of current (1.97 A) and voltage (5 V) gives a power output of 9.85 W.
Explanation:The power output of any device is calculated using the formula P = IV where I is the current in amperes and V is the voltage in volts. Given that the quick USB charger has an output current I of 1.97 amps and a standard USB output voltage V of 5 volts, we can substitute these values into our formula.
Thus, the power is P = IV = (1.97 A)(5 V) = 9.85 W. This means the output power of the quick USB charger is 9.85 watts. It's crucial to remember that power is measured in watts, where 1 watt is equivalent to 1 Ampere Volt (1 A.V = 1W).
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Suppose a force of 60 N is required to stretch and hold a spring 0.1 m from its equilibrium position. a. Assuming the spring obeys Hooke's law, find the spring constant k. b. How much work is required to compress the spring 0.5 m from its equilibrium position? c. How much work is required to stretch the spring 0.6 m from its equilibrium position? d. How much additional work is required to stretch the spring 0.1 m if it has already been stretched 0.1 m from its equilibrium? a. kequals 600
a. 600 N/m
Hooke's law states that:
F = kx
where
F is the force applied
k is the spring constant
x is the stretching/compression of the spring relative to the equilibrium position
In this problem we have
F = 60 N
x = 0.1 m
So the spring constant is
[tex]x=\frac{F}{x}=\frac{60 N}{0.1 m}=600 N/m[/tex]
b. 75 J
The work required to stretch a spring is equal to the elastic potential energy stored in the spring:
[tex]W=\frac{1}{2}kx^2[/tex]
where
k is the spring constant
x is the stretching/compression of the spring
Here we have
k = 600 N/m
x = 0.5 m
So the work done is
[tex]W=\frac{1}{2}(600 N/m)(0.5 m)^2=75 J[/tex]
c. 108 J
We can use the same formula used in the previous part:
[tex]W=\frac{1}{2}kx^2[/tex]
where here we have
k = 600 N/m
x = 0.6 m
So the work done is
[tex]W=\frac{1}{2}(600 N/m)(0.6 m)^2=108 J[/tex]
d. 9 J
In this case, the additional work required is the difference between the elastic potential energy in the two situations
[tex]W=\frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2[/tex]
where
k = 600 N/m
[tex]x_i = 0.1 m[/tex] is the initial stretching
[tex]x_f = 0.1 m + 0.1 m = 0.2 m[/tex] is the final stretching
Solving the equation,
[tex]W=\frac{1}{2}(600 N/m)(0.2 m)^2 - \frac{1}{2}(600 N/m)(0.1 m)^2=9 J[/tex]
A “cooling bag” is commonly used for storing water in hot, arid outdoor environments. The bag made of a thin porous fabric that allows water vapor, but not liquid water, to pass through it. A small amount of water (species A) diffuses through the fabric and evaporates from the surface of the bag. The rate of evaporation is controlled by convective mass transfer from the outer surface of the fabric to the surrounding dry air. The energy for evaporation is supplied by the hot air surrounding the outer surface of the bag. The evaporation of the water cools the remaining liquid water within the bag and a steady-state temperature driving force is established. If the surface temperature of the bag is 283 K, determine the temperature of the surrounding hot air. It may be assumed that the surrounding arid air contains no moisture
At what distance does a 100 Watt lightbulb deliver the same power per unit surface area as a 75 Watt lightbulb produces 10 m away from the bulb? (Assume both have the same efficiency for converting electrical energy in the circuit into emitted electromagnetic energy.). Recall that Watts = Joules/second = power = energy per unit time. Assume that the power of the electromagnetic waves spreads uniformly in all directions (i.e. spreads out over the area of a sphere) and use the formula for the surface area of a sphere.
Answer:
At 11.5 m
Explanation:
The power per unit area corresponds to the intensity, which is given by
[tex]I=\frac{P}{4\pi r^2}[/tex]
where
P is the power
[tex]4\pi r^2[/tex] is the area irradiated at a distance r from the source (it corresponds to the surface area of a sphere of radius r)
Here we want the intensity of the two light bulbs to be the same, so
[tex]I_1 = I_2\\\frac{P_1}{4 \pi r_1^2}=\frac{P_2}{4\pi r_2^2}[/tex]
where we have
P1 = 100 W is the power of the first light bulb
P2 = 75 W is the power of the second light bulb
r2 = 10 m is the distance from the second light bulb
Solving for r1, we find
[tex]r_1 = r_2 \sqrt{\frac{P_1}{P_2}}= (10 m) \sqrt{\frac{100 W}{75 W}} = 11.5 m[/tex]
The distance at which a 100 Watt lightbulb delivers the same power per unit surface area as a 75 Watt lightbulb at 10m is approximately 14.6 meters.
Explanation:The question is asking for the distance at which a 100 Watt lightbulb delivers the same power per unit surface area as a 75 Watt lightbulb at 10 m away. The intensity (power per unit area) of the light from a bulb diminishes with the square of the distance from the source. This is because the power (in Watts) is distributed over the surface area of a sphere, and the surface area of a sphere increases with the square of the radius (distance), following the formula 4πr².
First, we calculate the intensity I75 of the 75 Watt bulb at 10 m distance. That gives us: I75 = P75/4πr² = 75W/4π(10m)² ≈ 0.60 W/m².
Now, we want to find the distance r100 at which the 100 Watt bulb would deliver this same intensity. Solving the similar formula for r100 gives us: r100 = sqrt(P100/4πI75) = sqrt(100W/4π(0.60 W/m²)) ≈ 14.6 m.
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