Answer:[tex]10.82 kg-m^2[/tex]
Explanation:
Given
Mass of solid uniform disk [tex]M=13 kg[/tex]
radius of disk [tex]r=1.25 m[/tex]
mass of lump [tex]m=1.7 kg[/tex]
distance of lump from axis [tex]r_0=0.63[/tex]
Moment of inertia is the distribution of mass from the axis of rotation
Initial moment of inertia of disk [tex]I_1=\frac{Mr^2}{2}[/tex]
[tex]I_1=\frac{13\times 1.25^2}{2}=10.15 kg-m^2[/tex]
Final moment of inertia [tex]I_f[/tex]=Moment of inertia of disk+moment of inertia of lump about axis
[tex]I_f=\frac{Mr^2}{2}+mr_0^2[/tex]
[tex]I_f=10.15+1.7\times 0.63^2[/tex]
[tex]I_f=10.15+0.674[/tex]
[tex]I_f=10.82 kg-m^2[/tex]
Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to a halt after traveling 1.50 m along the incline. (a) If the initial speed of the crate was 1.77 m/s and the angle of inclination is 30.0°, how much energy was dissipated by friction? (b) What is the coefficient of sliding friction?
Answer:(a)891.64 N
(b)0.7
Explanation:
Mass of crate [tex]m=100 kg[/tex]
Crate slows down in [tex]s=1.5 m[/tex]
initial speed [tex]u=1.77 m/s[/tex]
inclination [tex]\theta =30^{\circ}[/tex]
From Work-Energy Principle
Work done by all the Forces is equal to change in Kinetic Energy
[tex]W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2[/tex]
[tex]W_{gravity}=mg(0-h)=mgs\sin \theta [/tex]
[tex]W_{gravity}=-mgs\sin \theta[/tex]
[tex]W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N[/tex]
change in kinetic energy[tex]=\frac{1}{2}\times 100\times 1.77^2=156.64 J[/tex]
[tex]W_{friction}=156.64+735=891.645[/tex]
(b)Coefficient of sliding friction
[tex]f_r\cdot s=W_{friciton}[/tex]
[tex]891.645=f_r\times 1.5[/tex]
[tex]f_r=594.43 N[/tex]
and [tex]f_r=\mu mg\cos \theta [/tex]
[tex]\mu 100\times 9.8\times \cos 30=594.43[/tex]
[tex]\mu =0.7[/tex]
Final answer:
To answer part (a), calculate the work done by friction using the equation work = force × distance. Calculate the change in kinetic energy of the crate using the equation change in kinetic energy = 0.5 × mass × (final velocity)^2 - 0.5 × mass × (initial velocity)^2 and subtract it from the work done by friction. To answer part (b), use the equation frictional force = coefficient of friction × normal force and calculate the normal force using the equation weight = mass × acceleration due to gravity.
Explanation:
To answer part (a), we need to calculate the work done by friction. The work done by friction can be calculated using the equation work = force × distance. In this case, the force is the frictional force and the distance is the distance traveled by the crate. The work done by friction is equal to the change in kinetic energy of the crate, which can be calculated using the equation change in kinetic energy = 0.5 × mass × (final velocity)^2 - 0.5 × mass × (initial velocity)^2. Subtracting the change in kinetic energy from the work done by friction will give us the energy dissipated by friction.
To answer part (b), we can use the equation frictional force = coefficient of friction × normal force. The normal force is equal to the weight of the crate, which can be calculated using the equation weight = mass × acceleration due to gravity. By substituting the known values into these equations, we can find the coefficient of sliding friction.
A loop of wire enclosing an area of 0.19 m2 has a magnetic field passing through its plane at an angle to the plane. The component of the field perpendicular to the plane is 0.5 T and the component parallel to the plane is 0.8 T. What is the magnetic flux through this coil? Round the final answer to three decimal places.
To solve this exercise it is necessary to apply the concepts related to Magnetic Flow which is defined through the Gaus Law of Magnetism as the measure of the total magnetic field that passes through a given area.
Vectorially it can be defined as
[tex]\Phi = \vec{A}\vec{B}[/tex]
Where,
A = Area
B = Magnetic Field
A scalar mode can also be expressed as
[tex]\Phi = A*B Cos\theta[/tex]
Where,
[tex]\theta[/tex] is the angle between B and A, at this case the direction are perpendicular then
Our values are given as,
[tex]A = 0.19m^2[/tex]
[tex]B_1 = 0.5T[/tex]
[tex]B_2 = 0.8T[/tex]
The magnetic field component that interests us is the perpendicular therefore
[tex]\Phi = 0.19* 0.5cos(90)[/tex]
[tex]\Phi = 0.095Tm^2[/tex]
Therefore the magnetic flux through this coil is 0.095[tex]Tm^2[/tex]
Unpolarized light of intensity 800 W/m2 is incident on two ideal polarizing sheets that are placed with their transmission axes perpendicular to each other. An additional polarizing sheet is then placed between the two, with its transmission axis oriented at 30∘ to that of the first. What is the intensity of the light passing through the stack of polarizing sheets? (Express your answer to two significant figures.)
Answer:
75 W/m²
Explanation:
[tex]I_0[/tex] = Unpolarized light intensity = 800 W/m²
[tex]\theta[/tex] = Angle between filters = 30°
Intensity of light after passing through first polarizer
[tex]I=\frac{I_0}{2}\\\Rightarrow I=\frac{800}{2}\\\Rightarrow I=400\ W/m^2[/tex]
Intensity of light after passing through second polarizer
[tex]I_1=Icos^2\theta\\\Rightarrow I_1=400\times cos^2(30)\\\Rightarrow I_1=300\ W/m^2[/tex]
Intensity of light after passing through third polarizer
[tex]I_2=I_1cos^2\theta\\\Rightarrow I_2=300\times cos^2(90-30)\\\Rightarrow I_1=75\ W/m^2[/tex]
The intensity of the light passing through the stack of polarizing sheets is 75 W/m²
The intensity of the light passing through three polarizing sheets, with the first and the third being perpendicular and the second making an angle of 30 degrees with the first, is zero, due to the rules of light polarization.
Explanation:Unpolarized light is composed of many rays having random polarization directions. When it passes through a polarizing filter, it decreases its intensity by a factor of 2, thus the intensity after the first polarizer is 800 W/m^2 /2 = 400 W/m^2. According to Malus's Law, the intensity of polarized light after passing through a second polarizing filter is given as I = Io cos² θ, where Io is the incident intensity (from the first polarizer) and θ is the angle between the direction of polarization and the axis of the filter.
The second polarizer is making an angle of 30 degrees with the first polarizer. Therefore, the intensity after the second polarizer is I = 400 W/m^2 * cos²(30) ≈ 346 W/m^2. The third polarizer is perpendicular to the first polarizer (or, equivalently, it makes an angle of 90 degrees with the second polarizer), so the final intensity is I = 346 W/m^2 * cos²(90), which is 0 because the cosine of 90 degrees is zero. So, no light (intensity=0) passes through the stack of these polarizing sheets.
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You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. Unfortunately the archer stands on an elevated platform of unknown height. However, you find the arrow stuck in the ground 67.0 m away, making a 3.00 ∘ angle with the ground.How fast was the arrow shot?
To develop this problem it is necessary to apply the kinematic equations that describe displacement, velocity and clarification.
By definition we know that velocity is defined as the change of position due to time, therefore
[tex]V = \frac{d}{t}[/tex]
Where,
d = Distance
t = Time
Speed can also be expressed in vector form through its components [tex]V_x[/tex] and [tex]V_y[/tex]
In the case of the horizontal component X, we have to
[tex]V_x = \frac{d}{t}[/tex]
Here d means the horizontal displacement, then
[tex]t = \frac{d}{V_x}[/tex]
[tex]t = \frac{67}{V_x}[/tex]
At the same time we have that the vertical component of the velocity is
[tex]V_y = gt[/tex]
Here,
g = Gravity
Therefore using the relation previously found we have that
[tex]V_y = g \frac{67}{V_x}[/tex]
The relationship between the two velocities and the angle can be expressed through the Tangent, therefore
[tex]tan\theta = \frac{V_y}{V_x}[/tex]
[tex]tan \theta = \frac{g \frac{67}{V_x} }{V_x}[/tex]
[tex]tan 3 = \frac{9.8\frac{67}{V_x} }{V_x}[/tex]
[tex]tan 3 = \frac{9.8*67}{V_x^2}[/tex]
[tex]V_x^2 = \frac{9.8*67}{tan 3}[/tex]
[tex]V_x= \sqrt{ \frac{9.8*67}{tan 3}}[/tex]
[tex]V_x = 111.93m/s \hat{i}[/tex]
This is the horizontal component, we could also find the vertical speed and the value of the total speed with the information given,
Then [tex]V_y,[/tex]
[tex]V_y = g \frac{67}{V_x}[/tex]
[tex]V_y = 9.8*\frac{67}{111.93}[/tex]
[tex]V_y = 5.866m/s\hat{j}[/tex]
[tex]|\vec{V}| = \sqrt{111.93^2+5.866^2}[/tex]
[tex]|\vec{V}| = 112.084m/s[/tex]
Final answer:
To calculate the initial speed of the arrow in a projectile motion problem, one needs to find the time of flight using the horizontal range and the final impact angle and then solve for the initial horizontal and vertical velocity components.
Explanation:
To determine the initial speed of the arrow shot from the bow, we'll use the principle of projectile motion. First, we need to find the time of flight. The arrow makes a 3.00° angle with the ground upon impact, and since it was shot parallel to the ground, it maintains a constant horizontal velocity throughout its flight. So, we can calculate the initial horizontal speed using this final impact angle.
The horizontal range (R) is given by R = v0x * t, where v0x is the initial horizontal velocity and t is the time of flight. The arrow landed 67.0 m away, so R is 67.0 m.
To find t, we will use the vertical motion equation under constant acceleration due to gravity (g = 9.8 m/s2). We know the final vertical velocity component (vfy) can be related to the initial vertical velocity component (v0y = 0 since it was shot parallel to the ground) by vfy = v0y + g*t. We can also calculate vfy using the impact angle: vfy = tan(impact angle) * v0x. Combining these, we can solve for t and subsequently for the initial speed v0.
Calculate the angular momentum, in kg.m^2/s, of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.420 kg · m2. kg · m2/s (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia, in kg · m2, if his angular velocity drops to 1.95 rev/s. kg · m2 (c) Suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. What average torque, in N m, was exerted if this takes 23.0 seconds? (Indicate the direction with the sign of your answer. Assume that the skater's rotation is in the positive direction.) N.m
The angular momentum of the ice skater spinning at 6.00 rev/s is 15.82 kg · m²/s. When the skater reduces their rate of spin to 1.95 rev/s, the moment of inertia is 1.29 kg · m². If the skater keeps their arms in and allows friction with the ice to slow them to 3.00 rev/s in 23.0 seconds, the average torque exerted is -0.343 N · m (indicating the direction of the torque).
Explanation:(a) The angular momentum of the ice skater can be calculated by multiplying the angular velocity (in radians per second) by the moment of inertia (in kg · m²). First, we need to convert the angular velocity from rev/s to rad/s. Since 1 rev = 2π rad, the angular velocity in rad/s is 6.00 rev/s * 2π rad/rev = 37.70 rad/s. Now, we can calculate the angular momentum: Angular momentum = angular velocity * moment of inertia = 37.70 rad/s * 0.420 kg · m² = 15.82 kg · m²/s.
(b) To find the new moment of inertia when the angular velocity decreases to 1.95 rev/s, we can rearrange the formula for angular momentum to solve for moment of inertia. Angular momentum = angular velocity * moment of inertia. Rearranging the equation: Moment of inertia = angular momentum / angular velocity = 15.82 kg · m²/s / (1.95 rev/s * 2π rad/rev) = 1.29 kg · m².
(c) To calculate the average torque exerted when the skater slows to 3.00 rev/s, we can use the equation torque = change in angular momentum / time. First, we need to calculate the change in angular momentum by subtracting the initial angular momentum from the final angular momentum. The initial angular momentum is 6.00 rev/s * 2π rad/rev * 0.420 kg · m² = 15.82 kg · m²/s, and the final angular momentum is 3.00 rev/s * 2π rad/rev * 0.420 kg · m² = 7.91 kg · m²/s. The change in angular momentum is 7.91 kg · m²/s - 15.82 kg · m²/s = -7.91 kg · m²/s. Finally, we can calculate the average torque: Torque = change in angular momentum / time = -7.91 kg · m²/s / 23.0 s = -0.343 N · m (negative sign indicates the direction of the torque).
(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. The electrical conductivity of and density of gold are 4.3 X 107 (O-m)-1 and 19.32 g/cm3, respectively.
(b) Now compute the electron mobility for Au.
Answer:
Part A:
[tex]n=8.85*10^{28}m^{-3}[/tex]
Part B:
[tex]Electron Mobility=3.03*10^{-3} m^2/V[/tex]
Explanation:
Part A:
To calculate the number of free electrons n we use the following formula::
n=1.5N-Au
Where N-Au is number of gold atoms per cubic meter
[tex]N-Au=\frac{Density*Avogadro Number}{atomic weight}[/tex]
[tex]Density = 19.32g/cm^3[/tex]
[tex]Avogadro Number=6.02*10^{23} atoms/mol[/tex]
[tex]Atomic weight=196.97g/mol[/tex]
So:
[tex]n=1.5*\frac{Density*Avogadro Number}{atomic weight}[/tex]
[tex]n=1.5*\frac{19.32*6.02*10^{23}}{196.97}[/tex]
[tex]n=8.85*10^{28}m^{-3}[/tex]
Part B:
[tex]Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}[/tex]
n is calculated above which is 8.85*10^{28}m^{-3}
Charge on electron=1.602*10^{-19}
Elec- Conductivity= 4.3*10^{7}
[tex]Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}[/tex]
[tex]Electron Mobility=3.03*10^{-3} m^2/V[/tex]
A garden hose is attached to a water faucet on one end and a spray nozzle on the other end. The water faucet is turned on, but the nozzle is turned off so that no water flows through the hose. The hose lies horizontally on the ground, and a stream of water sprays vertically out of a small leak to a height of 0.78 m.
What is the pressure inside the hose?
To solve this problem it is necessary to use the concepts related to pressure depending on the depth (or height) in which the object is on that fluid. By definition this expression is given as
[tex]P = P_{atm} +\rho gh[/tex]
Where,
[tex]P_{atm} =[/tex] Atmospheric Pressure
[tex]\rho =[/tex]Density, water at this case
g = Gravity
h = Height
The equation basically tells us that under a reference pressure, which is terrestrial, as one of the three variables (gravity, density or height) increases the pressure exerted on the body. In this case density and gravity are constant variables. The only variable that changes in the frame of reference is the height.
Our values are given as
[tex]P_{atm} = 1.013*10^5Pa[/tex]
[tex]\rho = 1000Kg/m^3[/tex]
[tex]g = 9.8m/s^2[/tex]
[tex]h = 0.78m[/tex]
Replacing at the equation we have,
[tex]P = P_{atm} +\rho gh[/tex]
[tex]P = 1.013*10^5 +(1000)(9.8)(0.78)[/tex]
[tex]P = 108944Pa[/tex]
[tex]P = 0.1089Mpa[/tex]
Therefore the pressure inside the hose is 0.1089Mpa
A graduated cylinder is half full of mercury and half full of water. Assume the height of the cylinder is 0.25 m and don't forget about atmospheric pressure. 1) Calculate the pressure at the bottom of a graduated cylinder. (Express your answer to two significant figures.)
To develop this problem it is necessary to apply the concepts related to the definition of absolute, manometric and atmospheric pressure.
The pressure would be defined as
[tex]P = P_0 + P_w +P_m[/tex]
Where,
[tex]P_0 =[/tex] Standard atmosphere pressure
[tex]P_W =[/tex]Pressure of Water
[tex]P_m =[/tex]Pressure of mercury
Therefore we have that
[tex]P = P_0 \rho_w g \frac{h}{2} + \rho_m g\frac{h}{2}[/tex]
Here g is the gravity, \rho the density at normal conditions and h the height of the cylinder.
Replacing with our values we have that,
[tex]P = 1.013*10^5+10^3*9.81*\frac{0.25}{2}+13.6*10^3*9.81*\frac{0.25}{2}[/tex]
[tex]P = 1.21*10^5Pa \approx 0.121Mpa[/tex]
Therefore the pressure at the Bottom of a graduated cylinder is 0.121Mpa
The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.
A graduated cylinder is half full of mercury and half full of water. If the height of the cylinder is 0.25 m, the height of each column of liquid is:
[tex]\frac{0.25m}{2} = 0.13 m[/tex]
We can calculate the pressure at the bottom of the cylinder (P) using the following expression.
[tex]P = P_{atm} + P_{Hg} + P_w[/tex]
where,
[tex]P_{atm}[/tex]: atmospheric pressure (101,325 Pa)[tex]P_{Hg}[/tex]: pressure exerted by Hg[tex]P_w[/tex]: pressure exerted by the waterWe can calculate the pressure exerted by each liquid (P) as:
[tex]P = \rho \times g \times h[/tex]
where,
ρ: density of the liquidg: gravity (9.81 m/s²)h: height of the liquid column (0.13 m)The pressure exerted by the column of Hg is:
[tex]P_{Hg} = \frac{13.6 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 17 \times 10^{3} Pa[/tex]
The pressure exerted by the column of water is:
[tex]P_{w} = \frac{1.00 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 1.3 \times 10^{3} Pa[/tex]
The pressure at the bottom of the cylinder is:
[tex]P = P_{atm} + P_{Hg} + P_w = 101325 Pa + 17 \times 10^{3} Pa + 1.3 \times 10^{3} Pa = 1.2 \times 10^{5} Pa[/tex]
The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.
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For the following reaction, 24.8 grams of diphosphorus pentoxide are allowed to react with 13.2 grams of water . diphosphorus pentoxide(s) + water(l) phosphoric acid(aq) What is the maximum mass of phosphoric acid that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams
The maximum mass of phosphoric acid that can be formed is 34.2 g.
The formula for the limiting reagent is [tex]\bold{P_4O_1_0}[/tex]
The excess reagent that remains after the reaction is complete is 3.76 g.
What is phosphoric acid?Phosphoric acid (H3PO4), also known as orthophosphoric acid, is the most important oxygen acid of phosphorus.
It is a crystalline solid and is used to generate fertilizer phosphate salts.
Given,
The balanced chemical equation is
[tex]P_4O_1_0 + 6 H_2O =4 H_3PO_4[/tex]
The molar mass of the following
[tex]\bold{P_4O_1_0}[/tex] = 284 g
[tex]\bold{H_2O}[/tex] = 16 g
[tex]\bold{H_3PO_4}[/tex] = 98 g
Calculating the number of moles
[tex]\dfrac{\bold{P_4O_1_0}}{284 g} = 0.0873\; mol/P_4O_1_0[/tex]
[tex]\dfrac{\bold{13.2\;g\;H_2O}}{18 g} = 0.733\; mol/P_4O_1_0[/tex]
From equation, 1 mol of [tex]\bold{P_4O_1_0}[/tex] reacts with 6 mol of water, then for 0.0873 mol of [tex]\bold{P_4O_1_0}[/tex] will react
[tex]0.0873\; mol\; P_4O_1_0 \times 6 mol H_2O/ 1 molP_4O_1_0 = 0.524\; mol\; H_2O[/tex]
The number of moles of water remaining are:
0.733 mol - 0.524 mol = 0.209 mol
Convert the amount of water into mass unit
0.209 mol · 18 g / mol = 3.76 g
The number of moles of water remaining is 3.76 g
Assuming that the reaction has a yield of 100%
Since 1 mol of [tex]\bold{P_4O_1_0}[/tex] produces 4 mol of phosphoric acid, 0.0873 mol [tex]\bold{P_4O_1_0}[/tex] will produce:
[tex]0.0873\; mol\; P_4O_1_0 \times 4 mol H_3PO_4/ molP_4O_1_0 = 0.349\; mol\; of\;phosphoric\;acid[/tex]
Then, the maximum mass of phosphoric acid that can be formed is
[tex]0.349\; mol\; H_3PO_4 \times 98\;g/ mol = 34.2\; g[/tex]
Thus, The maximum mass of phosphoric acid is 34.2 g, The formula for the limiting reagent is [tex]\bold{P_4O_1_0}[/tex], and the excess reagent that remains after the reaction is complete is 3.76 g.
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Final answer:
The maximum mass of phosphoric acid that can be formed is 35.6 grams, with diphosphorus pentoxide being the limiting reagent and 8.77 grams of water remaining as excess.
Explanation:
The maximum mass of phosphoric acid that can be formed is 35.6 grams. The formula for the limiting reagent is diphosphorus pentoxide (P4O6). After the reaction is complete, 8.77 grams of water will remain as the excess reagent.
Usually, to find the maximum mass of the product formed (phosphoric acid), and the mass of the excess reagent that remains after the reaction, it involves determining the limiting reagent by calculating the moles of reactants and using stoichiometry. The limiting reagent is the reactant that will be completely used up first during the chemical reaction, limiting the amount of product formed.
A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 9.1 m/s. After the collision, the 0.20-kg puck has a speed of 5.5 m/s at an angle of θ = 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision, (b) Find the fraction of kinetic energy lost in the collision.
The problem presents a collision scenario where principles of momentum conservation and kinetic energy are applied to determine the final velocity of a striking puck and the fraction of kinetic energy lost during the collision.
Explanation:The question describes a collision scenario in physics involving two pucks, where conservation of momentum and kinetic energy principles are applied. The question can be resolved by dividing it into two sections: (a) Determination of the velocity of the 0.30-kg puck after the collision, and (b) Determination of the fraction of kinetic energy lost in the collision.
For part (a), we use the law of conservation of momentum, which states that the total linear momentum of an isolated system remains constant, regardless of whether the objects within the system are at rest or in motion. To find the final velocity of the 0.30-kg puck, we subtract the momentum vector of the 0.20-kg puck after the collision from the momentum vector before the collision. For part (b), we first find the initial and final total kinetic energy of the system. The fraction of kinetic energy lost is given by the difference between the initial and final kinetic energies, divided by the initial kinetic energy.
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The principle of conservation of momentum and kinetic energy principles are employed to figure out the final velocity of the 0.30-kg puck and the fraction of kinetic energy lost in the collision. Initial and final velocities of the pucks, as well as their masses, are used in the process.
Explanation:To solve this problem, we need to apply the principle of conservation of momentum, which states that the total momentum of a system before a collision is equal to the total momentum after the collision. To find out the velocity of the 0.30-kg puck after the collision (a), the momentum equation (m₁v₁_initial + m₂v₂_initial = m₁v₁_final + m₂v₂_final) is applied. It is known that initial velocities of 0.30-kg and 0.20-kg pucks are 0 and 9.1 m/s respectively; the final velocity of the 0.20-kg puck is 5.5 m/s at 53° to the x-axis.
To calculate the fractional kinetic energy lost in the collision (b), firstly, the kinetic energy before and after the collision is calculated using the equation KE = 1/2mv₂. The fraction of kinetic energy lost is then (KE_initial - KE_final)/KE_initial.
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A piece of curved glass has a radius of curvature of r = 10.8 m and is used to form Newton's rings, as in the drawing. Not counting the dark spot at the center of the pattern, there are one hundred dark fringes, the last one being at the outer edge of the curved piece of glass. The light being used has a wavelength of 652 nm in vacuum. What is the radius R of the outermost dark ring in the pattern? (Hint: Note that r is much greater than R, and you may assume that tan(θ) = θ for small angles, where θ must be expressed in radians.)
Answer:
Radius of the outer most dark fringe is 2.65 cm
Solution:
As per the question:
Radius of curvature of the glass, r = 10.8 m
No. of dark fringes, n = 100
Wavelength of light, [tex]\lambda = 652\ nm = 652\times 10^{- 9}\ m[/tex]
Now,
To calculate the radius R of the outermost ring:
Radius of the dark fringe of nth order is given by:
[tex]R^{2} = nr\lambda = 100\times 10.8\times 652\times 10^{- 9} = 7.042\times 10^{- 4}[/tex]
[tex]R = \sqrt{7.042\times 10^{- 4}} = 0.0265\ m = 2.65\ cm[/tex]
how does the gravitational force between two objects change if the distance between the objects is cut in half?
A. the gravitational force doubles
B. the gravitational force decreases by half
C. the gravitational force decreases by a factor of 4
D. the gravitational force increases by a factor of 4
D. the gravitational force increases by a factor of 4
Explanation:
The magnitude of the gravitational force between two objects is given by:
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
Here let's call F the initial gravitational force between the two objects. Later, the distance between the objects is halved, so the new distance is:
[tex]r'=\frac{r}{2}[/tex]
Substituting into the equation, we find the new force:
[tex]F'=G\frac{m_1 m_2}{(r/2)^2}=4(G\frac{m_1 m_2}{r^2})=4F[/tex]
Therefore, the gravitational force increases by a factor of 4.
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Final answer:
The gravitational force between two objects doubles if the distance between them is cut in half.
Explanation:
The gravitational force between two objects is inversely proportional to the square of the distance between them. This means that if the distance between the objects is cut in half, the force will increase by a factor of 4.
For example, let's say the original distance is 10 units. If we cut it in half, the new distance would be 5 units. The force would increase by a factor of (10/5)² = 4, meaning the gravitational force would quadruple.
So the correct answer is, A. the gravitational force doubles.
For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is p1. If the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?
Final answer:
The new pressure drop p2 will be equal to the original pressure drop p1 at the original mass flow rate.
Explanation:
The relation between the new pressure drop p2 and the original pressure drop p1 at the original mass flow rate can be determined using Poiseuille's equation. Poiseuille's equation states that the pressure drop p2 - p1 is equal to the product of the resistance R and the flow rate Q.
When the length of the pipe is doubled, the resistance R remains constant because the friction factor is constant at high Reynolds numbers. Therefore, the new pressure drop p2 will be equal to the original pressure drop p1.
A 10.0 g bullet is fired into a stationary 5.00 kg block of wood. The bullet lodges inside the block. The speed of the block plus bullet system immediately after the collision is measured at 0.600 m/s.
What was the original speed of the bullet?
Answer:
v = 300.6 m/s
Explanation:
given,
mass of bullet(m) = 10 g = 0.01 kg
mass of block of wood (M)= 5 Kg
speed of the block plus bullet after collision(V) = 0.6 m/s
speed of wood(u) = 0 m/s
original speed of the bullet(v) = ?
using conservation of momentum
m v + Mu = (M + m)V
0.01 x v + 5 x 0 = (5 + 0.01) x 0.6
0.01 x v = (5.01) x 0.6
[tex]v = \dfrac{3.006}{0.01}[/tex]
v = 300.6 m/s
the original speed of the bullet before collision is equal to v = 300.6 m/s
The original speed of the bullet can be determined using the law of conservation of momentum. Applying the formula and solving for the original bullet speed, the bullet was initially traveling at 3 m/s.
Explanation:This problem is an application of the law of conservation of momentum. The principle states that the total momentum before a collision is equal to the total momentum after the collision. In this case, before the collision, only the bullet has momentum as the block is stationary. After the collision, the momentum is shared between the block and the lodged bullet.
To find the original speed of the bullet, we can set up this equation representing the law of conservation of momentum: m1v1 (before collision) = (m1 + m2) v2 (after collision). Here, m1 = mass of the bullet, v1 = original speed of the bullet, m2 = mass of the block, v2 = final speed of the block and bullet together.
Substituting the given values: (10 g)x = (10 g + 5.00 kg)(0.600 m/s). Solving for x (original bullet speed) yields a value of x = 3 m/s.
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A transformer has 500 primary turns and 9.7 secondary turns.
(a) If Vp is 120 V (rms), what is Vs with an open circuit?
(b) If the secondary now has a resistive load of 13 Ω, what is the current in the primary and secondary?
Answer: a) 2.33V b) Ip = 0.0035A, Is = 0.179A
Explanation:
The relationship formula between the Voltage, Number of turns and current in a transformer is given as
Vp/Vs = Np/Ns = Is/Ip
Vp is voltage in the primary coil
Vs is voltage in the secondary coil
Ns is number of turns in the secondary coil
Np is number of turns in the primary coil
Is is current in the secondary coil
Ip is current in the primary coil
a) substituting the values in the formula
Vp/Vs = Np/Ns
120/Vs = 500/9.7
Vs = 120×9.7/500
Vs = 2.33V
b) since secondary now has a resistive load of 13 Ω, Using ohm's law to get current in the secondary
V=IR
Is = Vs/R
Is =2.33/13
Is = 0.179A
To get Ip, we will use the relationship Np/Ns = Is/Ip
500/9.7= 0.179/Ip
Ip = 9.7×0.179/500
Ip = 0.0035A
A bungee jumper of mass m = 65 kg jumps from a bridge.
A bungee cord with unstretched length L is attached to her leg.
After falling a distance L, the jumper is moving at speed v = 26 m/s and the bungee cord begins stretching and acts like an ideal spring with spring constant k.
Neglect air resistance and the mass of the bungee cord.
Which choice best represents the unstretched length of the bungee cord, L?A. 5.9 m
B. 26 m
C. 34 m
D. 65 m
E. None of these
To solve the problem it is necessary to apply energy conservation.
By definition we know that kinetic energy is equal to potential energy, therefore
PE = KE
[tex]mgh = \frac{1}{2}mv^2[/tex]
Where,
m = mass
g = gravitaty constat
v = velocity
h = height
Re-arrange to find h,
[tex]h=\frac{v^2}{2g}[/tex]
Replacing with our values
[tex]h=\frac{26^2}{2*9.8}[/tex]
[tex]h = 34.498\approx 34.5m[/tex]
Therefore the correct answer is C.
When light of wavelength 350 nm is incident on a metal surface, the stopping potential of the photoelectrons is measured to be 0.500 V. What is the work function of this metal? (c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s, 1 eV = 1.60 × 10-19 J) 3.05 eV 4.12 eV 3.54 eV 0.500 eV
The work function of the metal can be calculated using Einstein's photoelectric equation and the given information about the wavelength of light, stopping potential, Planck's constant, and speed of light. The initial step is to calculate the energy of the photon; thereafter, calculate the kinetic energy of the photoelectron with the stopping potential, and finally find the work function by subtracting the kinetic energy from the photon energy.
Explanation:To calculate the work function of the metal, we need to first figure out the energy of the photon hitting the metal surface. This can be found using Einstein's photoelectric equation: E = hc/λ. Here, h is Planck's constant (6.626 x 10-34 J · s), c is the speed of light (3.00 x 108 m/s), and λ is the wavelength of the light (350 nm or 350 x 10-9 m).
After calculating the energy of the photon, we can determine the energy required to eject the photoelectrons, known as the stopping potential, which in this case is 0.500 V. Note this is the extra energy required to stop the photoelectron and is equal to the kinetic energy of the photoelectron. Hence, 0.500V = 0.500 eV because 1V = 1eV for a single electron with charge e. To convert this to joules, we use the conversion 1eV = 1.60 x 10-19J.
Finally, in order to find the work function, we use Einstein's photoelectric equation again which states that the energy of the incident photon is equal to the work function (φ) plus the kinetic energy of the photoelectron. That is, E = φ + KE. From this, we can rearrange the equation to solve for φ, the work function of the metal where: φ = E - KE.
By substituting the appropriate values into this equation, the work function of the metal can be found out. But remember, work function is a property of the metal only and is independent of the incident radiation. Also, it determines the threshold frequency or cutoff wavelength of the metal beyond which photoelectric effect will not occur.
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A spring has a 12 cm length. When a 200-g mass is hung from the spring, it extends to 27 cm. The hanging mass were pulled downward a further 5 cm. What is the time-dependent function of the position, in centimeters, of the mass, assuming that the phase angle LaTeX: \phi=0ϕ=0?
LaTeX: x(t)=15\cos{(8.08\textrm{ }t)}x(t)=15cos(8.08 t), where the coefficient inside the trigonometric function has units of rad/s.
LaTeX: x(t)=5\cos{(8.08\textrm{ }t)}x(t)=5cos(8.08 t), where the coefficient inside the trigonometric function has units of rad/s.
LaTeX: x(t)=5\cos{(1.29\textrm{ }t)}x(t)=5cos(1.29 t)x(t)=5cos(1.29 t), where the coefficient inside the trigonometric function has units of rad/s.
LaTeX: x(t)=15\cos{(1.29\textrm{ }t)}x(t)=15cos(1.29 t)x(t)=15cos(1.29 t), where the coefficient inside the trigonometric function has units of rad/s.
Answer:
The equation of the time-dependent function of the position is [tex]x(t)=5\cos(8.08t)[/tex]
(b) is correct option.
Explanation:
Given that,
Length = 12 cm
Mass = 200 g
Extend distance = 27 cm
Distance = 5 cm
Phase angle =0°
We need to calculate the spring constant
Using formula of restoring force
[tex]F=kx[/tex]
[tex]mg=kx[/tex]
[tex]k=\dfrac{mg}{x}[/tex]
[tex]k=\dfrac{200\times10^{-3}\times9.8}{(27-12)\times10^{2}}[/tex]
[tex]k=13.06\ N/m[/tex]
We need to calculate the time period
Using formula of time period
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Put the value into the formula
[tex]T=2\pi\sqrt{\dfrac{0.2}{13.6}}[/tex]
[tex]T=0.777\ sec[/tex]
At t = 0, the maximum displacement was 5 cm
So, The equation of the time-dependent function of the position
[tex]x(t)=A\cos(\omega t)[/tex]
Put the value into the formula
[tex]x(t)=5\cos(2\pi\times f\times t)[/tex]
[tex]x(t)=5\cos(2\pi\times\dfrac{1}{T}\times t)[/tex]
[tex]x(t)=5\cos(2\pi\times\dfrac{1}{0.777}\times t)[/tex]
[tex]x(t)=5\cos(8.08t)[/tex]
Hence, The equation of the time-dependent function of the position is [tex]x(t)=5\cos(8.08t)[/tex]
When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 45.5 N and it is acting with an effective lever arm of 2.45 cm, what is the torque that the muscle produces on the wrist?
Answer:
Torque on the rocket will be 1.11475 N -m
Explanation:
We have given that muscles generate a force of 45.5 N
So force F = 45.5 N
This force acts on the is acting on the effective lever arm of 2.45 cm
So length of the lever arm d = 2.45 cm = 0.0245 m
We have to find torque
We know that torque is given by [tex]\tau =F\times d=45.5\times 0.0245=1.11475N-m[/tex]
So torque on the rocket will be 1.11475 N -m
Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e., protons with one orbiting electron) are fed into a chamber where there is a strong magnetic field. Electrons in this chamber are trapped in tight orbits, which greatly increases the chance that they will collide with a hydrogen atom and ionizes it. One such source uses a magnetic field of 70 mT, and the electrons' kinetic energy is 1.2 eV. If the electrons travel in a plane perpendicular to the field, what is the radius of their orbits?
Answer:
[tex]r=5.278\times 10^{-4}\ m[/tex]
Explanation:
Given that:
magnetic field intensity, [tex]B=0.07\ T[/tex]kinetic energy of electron, [tex]KE=1.2\ eV= 1.2\times 1.6\times 10^{-19}\ J= 1.92\times 10^{-19}\ J[/tex]we have mass of electron, [tex]m=9.1\times 10^{-31}\ kg[/tex]Now, form the mathematical expression of Kinetic Energy:
[tex]KE= \frac{1}{2} m.v^2[/tex]
[tex]1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2[/tex]
[tex]v^2=4.2198\times 10^{11}[/tex]
[tex]v=6.496\times 10^6\ m.s^{-1}[/tex]
from the relation of magnetic and centripetal forces we have the radius as:
[tex]r=\frac{m.v}{q.B}[/tex]
[tex]r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}[/tex]
[tex]r=5.278\times 10^{-4}\ m[/tex]
The radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.
Force of the electron
The magnetic force and centripetal force of the electron is calculated as follows;
qvB = mv²/r
qB = mv/r
r = mv/qB
where;
m is mass of electronv is speed of the electronq is the magnitude of the chargeB is magnetic field strengthKinetic energy of the electronThe kinetic energy of the electron is given as;
K.E = ¹/₂mv²
mv² = 2K.E
v² = 2K.E/m
v² = (2 x 1.2 x 1.6 x 10⁻¹⁹)/(9.11 x 10⁻³¹)
v² = 4.215 x 10¹¹
v = √4.215 x 10¹¹
v = 6.5 x 10⁵ m/s
Radius of their orbitThe radius of their orbit is calculated as follows;
r = mv/qB
r = (9.11 x 10⁻³¹ x 6.5 x 10⁵) / (1.6 x 10⁻¹⁹ x 70 x 10⁻³)
r = 5.287 x 10⁻⁵ m
Thus, the radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.
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A block is attached to a spring, with spring constant k, which is attached to a wall. It is initially moved to the left a distance d (at point A) and then released from rest, where the block undergoes harmonic motion. The floor is frictionless.The points labelled A and C are the turning points for the block, and point B is the equilibrium point.1)Which of these quantities are conserved for the spring and block system?mechanical energyx-direction linear momentumneither energy nor momentum
Answer:
Explanation:
When a spring attached with a block is stretched and released , the block starts moving under SHM. The elastic energy stored in it initially at the time of initial stretch is repeatedly converted into kinetic energy and vice - versa while the body keeps moving under SHM. So we can tell that the mechanical energy of the system remains constant.
In the whole process the velocity of the body keeps changing in terms of both magnitude and direction . It happens because a variable force acts on the body constantly towards the equilibrium point so its momentum also keeps changing all the time
A block of mass 11.8 kg slides from rest down a frictionless 35.6° incline and is stopped by a strong spring with k = 3.11 x 10^4 N/m. The block slides 2.17 m from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?
Answer:
x = 6.58 m
Explanation:
For this exercise let's use the concept of conservation of mechanical energy. Let's look for energy at two points the highest and when the spring is compressed
Initial
Em₀ = U = m g and
Final
[tex]E_{mf}[/tex] = ke = ½ k x2
As there is no friction the mechanical energy is conserved
Em₀ = [tex]E_{mf}[/tex]
m g y = ½ k x²
Let's use trigonometry to face height
sin θ = y / L
y = L sin θ
x = √ 2mg (L synth) / k
x = √ (2 11.8 9.8 2.17 sin35.6 / 3.11 104)
x = 6.58 m
Sound waves are a longitudinal wave that have a speed of about 340 m/s when traveling through room temperature air. What is the wavelength of a sound wave that has a frequency of 6,191 Hz?
The wavelength of the wave is 0.055 m
Explanation:
The relationship between speed, frequency and wavelength of a wave is given by the wave equation:
[tex]v=f\lambda[/tex]
where
v is the speed
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the sound wave in this problem we have
v = 340 m/s is the speed
f = 6,191 Hz is the frequency
Solving for [tex]\lambda[/tex], we find the wavelength:
[tex]\lambda=\frac{v}{f}=\frac{340}{6191}=0.055 m[/tex]
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An antitank weapon fires a 3.00-kg rocket which acquires a speed of 50.0 m/s after traveling 90.0 cm down a launching tube. Assuming the rocket was accelerated uniformly, what is the average force that acted on it?
Answer:
Force in the rocket will be 4166.64 N
Explanation:
We have given mass of the rocket m = 3 kg
Rocket acquires a speed of 50 m sec so final speed v = 50 m/sec
Initial speed u = 0 m/sec
Distance traveled s = 90 cm = 0.9 m
From third equation of motion we know that [tex]v^2=u^2+2as[/tex]
[tex]50^2=0^2+2\times a\times 0.9[/tex]
[tex]a=1388.88m/sec^2[/tex]
From newton's law we F = ma
So force will be [tex]F=1388.88\times 3=4166.64N[/tex]
So force in the rocket will be 4166.64 N
The average force that acted on the rocket can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 3.00 kg and the acceleration can be determined by dividing the change in velocity by the distance traveled. The average force that acted on the rocket is 166.8 Newtons.
Explanation:The average force that acted on the rocket can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 3.00 kg and the acceleration can be determined by dividing the change in velocity by the distance traveled. The change in velocity is 50.0 m/s and the distance traveled is 90.0 cm, which is equivalent to 0.90 m.
So, the acceleration of the rocket is 50.0 m/s divided by 0.90 m, which gives us 55.6 m/s². Now, we can use Newton's second law to calculate the force: F = m * a, where F is the force, m is the mass, and a is the acceleration.
Plugging in the values, we get F = 3.00 kg * 55.6 m/s² = 166.8 N. Therefore, the average force that acted on the rocket is 166.8 Newtons.
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Police lieutenants, examining the scene of an accident involving two cars, measure the skid marks of one of the cars, which nearly came to a stop before colliding, to be 78 m long. The coefficient of kinetic friction between rubber and the pavement is about 0.90. Estimate the initial speed of that car assuming a level road.
Answer:
37.11231 m/s
Explanation:
u = Initial velocity
v = Final velocity = 0
s = Displacement = 78 m
g = Acceleration due to gravity = 9.81 m/s²
[tex]\mu[/tex] = Coefficient of kinetic friction = 0.9
Acceleration is given by
[tex]a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g[/tex]
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=-2\mu gs\\\Rightarrow -u^2=2(-\mu g)s-v^2\\\Rightarrow u=\sqrt{v^2-2(-\mu g)s}\\\Rightarrow u=\sqrt{0^2-2\times (-0.9\times 9.81)\times 78}\\\Rightarrow u=37.11231\ m/s[/tex]
The initial speed of that car is 37.11231 m/s
On which of the following does the moment of inertia of an object depend?Check all that apply.linear speedlinear accelerationangular speedangular accelerationtotal massshape and density of the objectlocation of the axis of rotation
Explanation:
Moment of Inertia in physics is the measure of the rotational inertia of an object. It is nothing but the opposition that the body offers to having its speed of rotation about an axis changed by application some external torque.
I= MR^2
M= mass of object R= radius of rotation.
Therefore, MOI depends upon
a)total mass
b)shape and density of the object
c)location of the axis of rotation
The moment of inertia of an object depends on its total mass, the distribution of this mass in relation to the rotation axis (shape and density), and the location of the rotation axis. The higher the moment of inertia, the greater the object's resistance to changes in rotational rate. The parallel axis theorem provides a tool to calculate the moment of inertia for different rotation axes.
Explanation:The moment of inertia of an object depends on certain factors. Specifically, its total mass, the shape and density of the object, and the location of the axis of rotation. The moment of inertia is defined as the measure of the object's resistance to changes in its rotation rate. The moment of inertia varies depending on how mass is distributed in relation to the axis of rotation.
For instance, an object with more mass distributed farther from its axis of rotation will have a higher moment of inertia. This is seen in a hollow cylinder versus a solid cylinder of the same mass and size. The hollow cylinder has a greater moment of inertia because more of its mass is distributed at a greater distance from its axis of rotation.
A principle known as the parallel axis theorem can be used to find the moment of inertia about a new axis of rotation once it is known as a parallel axis. This theorem involves the center of mass and the distance from the initial axis to the parallel axis.
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Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass=0.36 kg) and the other of metal (mass=0.82 kg) They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum
Answer:
a) T = 0.579 s , b) T = 0.579 s
Explanation:
The great advantage of wave mechanics is that the general equations for different systems are the same, what changes are the physical parameters involved, the equation of motion is
x = A cos (wt +φ)
Where w is the angular velocity that is this case for being a solid body is
w = √ (mg d / I)
Where I is the moment of inertia and d the distance to the pivot point
The moment of inertia for a ruler hold one end is
I = 1/12 M L²
The lost is related to the frequency and is with the angular velocity
T = 1 / f
w = 2π f
w = 2π / T
T = 2π / w
T = 2π √ I / mgd
For our case
d = L
T = 2π √(1/12 M L²) / M g L)
T = 2π √(L/(12 g))
a) wood suppose it is one meter long (L = 1m)
T = 2π √ (1 / (12 9.8))
T = 0.579 s
b) metal length (L = 1m)
T = 2pi RA (1 / (12 9.8))
T = 0.579 s
The period does not depend on mass but on length
A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-glass prism perpendicular to one face and then refracts through the hypotenuse back into the air. The angle between these two faces is 37 ∘.
Answer:
The angle between the red and blue light is 1.7°.
Explanation:
Given that,
Wavelength of red = 656 nm
Wavelength of blue = 486 nm
Angle = 37°
Suppose we need to find the angle between the red and blue light as it leaves the prism
[tex]n_{r}=1.572[/tex]
[tex]n_{b}=1.587[/tex]
We need to calculate the angle for red wavelength
Using Snell's law,
[tex]n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}[/tex]
Put the value into the formula
[tex]1.572\sin37=1\times\sin\theta_{r}[/tex]
[tex]\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})[/tex]
[tex]\theta_{r}=71.0^{\circ}[/tex]
We need to calculate the angle for blue wavelength
Using Snell's law,
[tex]n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}[/tex]
Put the value into the formula
[tex]1.587\sin37=1\times\sin\theta_{b}[/tex]
[tex]\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})[/tex]
[tex]\theta_{b}=72.7^{\circ}[/tex]
We need to calculate the angle between the red and blue light
Using formula of angle
[tex]\Delta \theta=\theta_{b}-\theta_{r}[/tex]
Put the value into the formula
[tex]\Delta \theta=72.7-71.0[/tex]
[tex]\Delta \theta=1.7^{\circ}[/tex]
Hence, The angle between the red and blue light is 1.7°.
The Hydrogen discharge lamp emits light of specific wavelengths which pass through a flint-glass prism, refract (or bend), and display a color spectrum due to dispersion. The red and blue lights have different refraction angles because of their different wavelengths. The distinct color bands indicate discrete energy transitions in the hydrogen gas.
Explanation:Light emission and its properties are central to this question. The light emitted by a hydrogen discharge lamp, which contains hydrogen gas at low pressure, indicates that there are discrete energies emitted, thus producing light of specific wavelengths. These distinct wavelengths are a result of H2 molecules being broken apart into separate H atoms when an electric discharge passes through the tube.
When this light enters a prism, it refracts or bends, leading to a phenomenon called dispersion. This happens because not all colors or wavelengths of light are bent the same amount: the degree of bending depends on the wavelength of the light, as well as the properties of the material (in this case, flint-glass prism). For instance, red and blue light have different refraction angles due to their different wavelengths of 656 nm and 486 nm respectively.
So, in the context of a hydrogen discharge lamp, the two prominent wavelengths, 656 nm (red) and 486 nm (blue), will refract differently when passing through the flint-glass prism. Further, the separation (or dispersion) of light into these distinct colored bands as it exits the prism gives us a line spectrum - visual evidence of the discrete energy transitions happening at the atomic level in the hydrogen gas of the lamp.
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A 93 kg man holding a 0.653 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 11.3 m/s (relative to the ground) and then catches the ball after it rebounds from the wall. How fast is he moving after he catches the ball? Ignore the projectile motion of the ball, and assume that it loses no energy in its collision with the wall. Answer in units of m/s.
The 93 kg man will be moving at a speed of 0.161 m/s after catching the ball that he threw at a wall and rebounded back to him, based on the principle of conservation of momentum.
Explanation:This question involves the principle of conservation of momentum. According to the law of conservation of momentum, the total momentum of the system (man and the ball) is conserved before and after the collision with the wall. The total initial momentum, before the ball is thrown, is zero as everything is at rest.
When the man throws the ball, the ball gains momentum in one direction, and to keep the total momentum zero, the man must gain an equal amount of momentum in the opposite direction. On returning, the ball has changed its direction, so it has lost the initial momentum and gained the same amount in the opposite direction.
The man catches the ball, meaning he gains the momentum that the ball brought back, causing him to move. To determine how fast he is moving, we set the final momentum of the system (which should still be zero) equal to the initial momentum.
To calculate the speed (v) of the man we'll use the equation: v = 2 × ball_mass × ball_speed / man_mass = 2 × 0.653 kg × 11.3 m/s / 93 kg = 0.161 m/s
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The man will be moving at approximately 0.078 m/s after he catches the ball. This is found using the conservation of momentum, considering the ball's momentum before and after it collides with the wall and bounces back to the man.
Explanation:The student's question involves the conservation of momentum, a fundamental concept in physics. When the man throws the ball and catches it after it bounces off the wall, both the ball and the man are involved in an isolated system where external forces are negligible. As momentum is conserved, the velocity of the man after catching the ball can be calculated using the conservation of momentum principle.
To solve this problem, we first consider the momentum before and after the collision with the wall. The man throws the ball at 11.3 m/s. The direction of the ball's momentum changes after it bounces back, but the speed remains the same due to the 'elastic' nature of the collision as mentioned in the problem statement (no energy loss).
The initial momentum of the system (man and ball) is zero because they're both at rest initially. When the man throws the ball, the ball gains momentum in the forward direction, and the man gains momentum in the opposite direction. This momentum for the man (in the opposite direction) is what will be calculated.
The initial momentum of the ball and man is:
Initial Ball: 0 kg·m/s (at rest)Initial Man: 0 kg·m/s (at rest)The final momentum after throwing the ball is:
Final Ball: -0.653 kg * 11.3 m/s (the negative sign indicates the direction change after bouncing)Final Man: 93 kg * v (where v is the velocity we want to calculate)By conservation of momentum:
Initial momentum = Final momentum
0 = -0.653 kg * 11.3 m/s + 93 kg * v
Solving for v gives:
v = (0.653 kg * 11.3 m/s) / 93 kg
v ≈ 0.078 m/s
This is the speed of the man after catching the ball, moving opposite to the direction in which he threw the ball.
Learn more about Conservation of Momentum here:https://brainly.com/question/33316833
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You have a glass ball with a radius of 2.00 mm and a density of 2500 kg/m3. You hold the ball so it is fully submerged, just below the surface, in a tall cylinder full of glycerin, and then release the ball from rest. Take the viscosity of glycerin to be 1.5 Pa s and the density of glycerin to be 1250 kg/m3. Use g = 10 N/kg = 10 m/s2. Also, note that the drag force on a ball moving through a fluid is: Fdrag = 6πηrv . (a) Note that initially the ball is at rest. Sketch (to scale) the free-body diagram of the ball just after it is released, while its velocity is negligible. (b) Calculate the magnitude of the ball’s initial acceleration. (c) Eventually, the ball reaches a terminal (constant) velocity. Sketch (to scale) the free-body diagram of the ball when it is moving at its terminal velocity. (d) Calculate the magnitude of the terminal velocity. (e) What is the magnitude of the ball’s acceleration, when the ball reaches terminal velocity? (f) Let’s say that the force of gravity acting on the ball is 4F, directed down. We can then express all the forces in terms of F.
Answer:
(a) check attachment
(b)5 m/s²
Explanation:
Given: radius = 2.00mm: density = 2500kg/m³: viscosity of glycerin = 1.5pa: decity of glycerin = 1250kg/m³: g = 10N/kg = 10m/s²: Fdrag = 6πnrv
(a) for answer check attachment.
(b) For the magnitude of the balls initial acceleration:
Initial net force(f) = mg - upthrust
= [tex]mg - (\frac{m}{p} )pg.g[/tex]
acceleration (a) = [tex]Acceleration(a)=\frac{f}{m}\\=g - (\frac{pg}{p})g\\=g(1-\frac{pg}{p} )\\=10(1-\frac{1250}{2500} )\\a=10(1-0.5)\\a=5 m/s²[/tex]
c.) fromthe force diagram in the attachment; when the ball attains terminal velocity the net force will be zero(0)
[tex]mg=6πnrv + upthrust[/tex]
d.) For the magnitude of terminal velocity:
[tex]mg=6πnrv + (\frac{m}{p})pg.g\\\\(\frac{4}{3}πr^{3} p)g=6πnrv +\frac{4}{3}πr^3pg.g\\\\V = \frac{2}{9}.\frac{(2*10^{-3})^{2}*(2500-1250)*10}{1.5}\\\\=0.79cm/s[/tex]
e.) when the ball reaches terminal velocity, the acceleration is zero (0)