To carry out this exercise, it is necessary to use the equations made to Centripetal Force and Gravitational Energy Conservation.
By definition we know that the Centripetal Force is estimated as
[tex]F_c = M\omega^2R[/tex]
Where,
M = mass
[tex]\omega =[/tex] Angular velocity
R = Radius
From the 'linear' point of view the centripetal force can also be defined as
[tex]F_c = \frac{GM^2}{R^2}[/tex]
PART A ) Equating both equations we have,
[tex]\frac{M}{\omega^2R}=\frac{GM^2}{R^2}[/tex]
Re-arrange to find \omega
[tex]\omega = \sqrt{\frac{Gm}{r^3}}[/tex]
Replacing with our values
[tex]\omega = \sqrt{\frac{(6.67*10^{-11})(4.6*10^{30})}{(1.9*10^{11})^3}}[/tex]
[tex]\omega = 2.115*!0^{-7}rad/s[/tex]
Therefore the angular speed is [tex]\omega = 2.115*!0^{-7}rad/s[/tex]
PART B) For energy conservation we have to
[tex]KE_{min} = PE_{cm}[/tex]
Where,
[tex]KE_{min} =[/tex] Minimus Kinetic Energy
[tex]PE_{cm} =[/tex] Gravitational potential energy at the center of mass
Then,
[tex]\frac{1}{2} mv^2_{min} = \frac{2GMm}{R}[/tex]
Re-arrange to find v,
[tex]v_min = \sqrt{\frac{4GM}{R}}[/tex]
[tex]v_min = \sqrt{\frac{4(6.67*10^{-11})(2.2*10^{30})}{(1.9*10^{11})}}[/tex]
[tex]v_min = 5.55*10^4m/s[/tex]
Therefore the minimum speed must it have at the center of mass if it is to escape to "infinity" from the two-star system is [tex]v_min = 5.55*10^4m/s[/tex]
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
Learn more about Conservation of Momentum here:https://brainly.com/question/33316833
#SPJ12
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
#SPJ3
In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A student heats 69.12 grams of tungsten to 98.93 °C and then drops it into a cup containing 85.45 grams of water at 23.82 °C. She measures the final temperature to be 25.63 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.56 J/°C. Assuming that no heat is lost to the surroundings calculate the specific heat of tungsten.
Answer:
[tex]c_{e1}[/tex] = 128.3 J / kg ° C
Explanation:
In this exercise we will use that the expression for heat is
Q = m [tex]c_{e}[/tex] ΔT
As they indicate that there are no losses with the medium, the heat transferred by the tungsten is equal to the heat absorbed by the water plus the calorimeter
Q assigned = QAbsorbed
Q hot = Q cold + Q calorimeter
The mass of tungsten (m₁ = 69.12 10⁻³ kg) with an initial temperature (T₁ = 98.93°C),
The mass of water (m₂ = 85.45 10⁻³ kg) at a temperature (T₂ = 23.82°C),
a calorimeter constant (C = 1.56 J/ °C)
m₁ [tex]c_{e1}[/tex] (T₁ - [tex]T_{f}[/tex]) = (m₂ [tex]c_{e2}[/tex] + C) ([tex]T_{f}[/tex] - T₂)
[tex]c_{e1}[/tex]= (m₂ ce2 + C) ([tex]T_{f}[/tex]-T₀) / (m₁ (T₁-[tex]T_{f}[/tex])
[tex]c_{e1}[/tex] = (85.45 10-3 4186 + 1.56) (25.63 - 23.82) / (69.12 10-3 (98.93 - 25.63))
[tex]c_{e1}[/tex] = (357.69 + 1.56) 1.81 / (69.12 10-3 73.3)
[tex]c_{e1}[/tex] = 650.24 / 5.0665
[tex]c_{e1}[/tex] = 128.3 J / kg ° C
A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10,000 kg, and two antennas projecting out from the center of mass of the main body that can be approximated with rods of length 3.0 m each and mass 10 kg. The antenna’s lie in the plane of rotation. What is the angular momentum of the satellite?
Answer:
605447.7066 kgm²/s
Explanation:
[tex]m_1[/tex] = Mass of sphere = 10000 kg
[tex]m_2[/tex] = Mass of rod = 10 kg
r = Radius of sphere = 2 m
l = Length of antenna = 3 m
Angular speed
[tex]\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s[/tex]
Angular momentum is given by
[tex]L=I\omega[/tex]
Moment of inertia of the satellite is
[tex]I_s=\frac{2}{5}m_1r^2[/tex]
Moment of antenna of the satellite is
[tex]I_a=\frac{1}{3}m_2l^2[/tex]
The angular momentum of the system is
[tex]L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s[/tex]
The angular momentum of the satellite is 605447.7066 kgm²/s
The angular momentum of a rotating satellite can be calculated by considering the moment of inertia of its components, including the main body and antennas. The moment of inertia is calculated based on the mass and distribution of mass in each component. The total angular momentum is the sum of the angular momentum of the main body and the angular momentum of the antennas.
Explanation:The angular momentum of a rotating object can be calculated by multiplying its moment of inertia with its angular velocity. In this case, the satellite consists of a main body in the shape of a sphere and two antennas projecting out from the center of mass. The moment of inertia can be calculated by considering the mass and distribution of mass in each component of the satellite. The angular momentum is the sum of the angular momentum of the main body and the angular momentum of the antennas.
For the main body of the satellite, the moment of inertia can be calculated as: I = (2/5)×m × r², where m is the mass and r is the radius of the sphere. For the antennas, the moment of inertia can be calculated as: I = (1/12) × m × L², where m is the mass and L is the length of each antenna.
Finally, the total angular momentum can be calculated by summing the angular momentum of the main body and the angular momentum of the antennas: L_total = L_main body + 2× L_antennas, since there are two antennas.
Learn more about angular momentum here:https://brainly.com/question/37906446
#SPJ3
A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg. The child grabs and clings to a bar that is 1.55 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 35.0 rpm to 19.0 rpm. What is the moment of inertia of the merry‑go‑round with respect to its central axis?
Answer:
72.75 kg m^2
Explanation:
initial angular velocity, ω = 35 rpm
final angular velocity, ω' = 19 rpm
mass of child, m = 15.5 kg
distance from the centre, d = 1.55 m
Let the moment of inertia of the merry go round is I.
Use the concept of conservation of angular momentum
I ω = I' ω'
where I' be the moment of inertia of merry go round and child
I x 35 = ( I + md^2) ω'
I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19
35 I = 19 I + 1164
16 I = 1164
I = 72.75 kg m^2
Thus, the moment of inertia of the merry go round is 72.75 kg m^2.
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]
Learn more about waves and wavelength:
brainly.com/question/5354733
brainly.com/question/9077368
#LearnwithBrainly
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)
A DC current of 60 mA can cause paralysis of the body's respiratory muscles and hence interfere with breathing, but only 15 mA (rms) of AC current will do the same thing. Suppose a person is working with electrical power lines on a warm humid day and therefore has a low body resistance of 1000 OHMS. What DC and AC (amplitude and rms) potentials would it take to cause respiratory paralysis?
Answer:
DC: 60V
AC: 15Vrms (21.2V amplitude)
Explanation:
Since we know the current values for each case, we can calculate voltage in both scenarios.
For DC voltage:
Vdc = Idc * R = 60mA * 1000 Ohms = 60Vdc
For AC voltage:
Vac = Iac * R = 15mA * 1000 Ohms = 15 Vac (rms) the amplitude of this signal would be Vmax = 15Vac * 1.414 = 21.2V
Assume that the total cholesterol levels for adults are normally distributed with mean cholesterol level of 51.6 mg/dL and standard deviation 14.3 mg/dL. Find the probability that an individual will have a cholesterol level greater than 60 mg/dL.
Answer:
therefore, the probability that an individual will have a cholesterol level greater than 60 mg/dL.= 0.27
Explanation:
given data
Normal distribution
mean cholesterol level μ= 51.6 mg/dL
Standard deviation σ= 14.3 mg/dL
x= 60 mg/dL
We have to find out P(x>60)
We Know that [tex]P(x>a) =P(Z>(a-\mu)/\sigma)[/tex]
therefore, [tex]P(x>60) =P(Z>(60-51.3)/14.3)[/tex]
= P(Z>0.61)
= 1 - P(Z<0.61)
= 1 - 0.7291
= 0.27
therefore, the probability that an individual will have a cholesterol level greater than 60 mg/dL.= 0.27
To find the probability that an individual will have a cholesterol level greater than 60 mg/dL, we calculate the z-score and find the corresponding probability using a Z-table or calculator.
Explanation:To find the probability that an individual will have a cholesterol level greater than 60 mg/dL, we need to use the z-score formula. The z-score is calculated by subtracting the mean from the value of interest and dividing it by the standard deviation. In this case, the z-score is (60 - 51.6) / 14.3 = 0.589. Using a Z-table or a calculator, we can find that the probability of a z-score greater than 0.589 is approximately 0.279.
Learn more about Normal distribution here:https://brainly.com/question/34741155
#SPJ12
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.
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.
Learn more about phosphoric acid, here
https://brainly.com/question/8156365
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.
1 - A submarine has a circular porthole with a radius of 0.10 m. If a force of more than 2x106 N is applied, the porthole will fail. What is the maximum pressure that this window can support? To what maximum depth can the submarine dive, assuming that the interior is maintained at atmospheric pressure?
In order to solve this problem, it is necessary to apply the concepts related to the Pressure according to the Force and the Area as well as to the pressure depending on the density, gravity and height.
In the first instance we know that the pressure can be defined as
[tex]P = \frac{F}{A}[/tex]
Where
F= Force
A = Mass
In the second instance the pressure can also be defined as
[tex]P = \rho gh[/tex]
Where,
[tex]\rho=[/tex]Density of Fluid at this case Water
g = Gravitational Acceleration
h = Height
If we develop the problem to find the pressure then,
[tex]P = \frac{F}{A}[/tex]
[tex]P = \frac{F}{\pi r^2}[/tex]
[tex]P = \frac{2*10^6}{\pi (0.1)^2}[/tex]
[tex]P = 6.37*10^{7} Pa[/tex]
Through the second equation we can find the depth to which it can be submerged,
[tex]P = \rho gh[/tex]
Re-arrange to find h
[tex]h = \frac{P}{\rho g}[/tex]
[tex]h = \frac{6.37*10^{7}}{1000*9.8}[/tex]
[tex]h = 6499.4m[/tex]
A transformer connected to a 120-v(rms) at line is to supply 13,000V(rms) for a neon sign.
To reduce shock hazard, a fuse is to beinserted in the primary circuit;the fuse is to blow when the rmscurrent in the secondary circuit exceeds 8.50 mA.
(a) What is theratio of secondary to primary turns of the transformer?
(b) Whatpower must be supplied to the transformer when the rms secondarycurrent is 8.50 mA?
(c) What current rating should the fuse in theprimary circuit have?
Answer:
(a) 108
(b) 110.500 kW
(c) 920.84 A
Solution:
As per the question:
Voltage at primary, [tex]V_{p} = 120\ V[/tex] (rms voltage)
Voltage at secondary, [tex]V_{s} = 13000\ V[/tex] (rms voltage)
Current in the secondary, [tex]I_{s} = 8.50\ mA[/tex]
Now,
(a) The ratio of secondary to primary turns is given by the relation:
[tex]\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}[/tex]
where
[tex]N_{p}[/tex] = No. of turns in primary
[tex]N_{s}[/tex] = No. of turns in secondary
[tex]\frac{N_{s}}{N_{p}} = \frac{13000}{120}[/tex] ≈ 108
(b) The power supplied to the line is given by:
Power, P = [tex]V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW[/tex]
(c) The current rating that the fuse should have is given by:
[tex]\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}[/tex]
[tex]\frac{13000}{120} = \frac{I_{p}}{8.50}[/tex]
[tex]I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A[/tex]
uppose one gallon of gasoline produces 1.17×108 J of energy, and this energy is sufficient to operate a car for 18.3 miles. An aspirin tablet has a mass of 319 mg. If the aspirin could be converted completely into thermal energy, how many miles could the car go on a single tablet?
Answer:
4.5 x 10^6 miles
Calculations can be viewed on the snapshot attached to this reply.
Thanks
A bullet, m = 0.500 kg, traveling with some initial velocity strikes and embeds itself in the bob of a ballistic pendulum, M = 9.50 kg. a) The combined masses rise to a height h = 1.28 m. The speed Vf of the combined masses immediately following impact is ____
Answer:
5 m/s
Explanation:
mass of bullet, m = 0.5 kg
initial velocity of bullet is u.
mass of pendulum, M = 9.50 kg
height raised, h = 1.28 m
The kinetic energy of the bullet and pendulum system is equal to the potential energy of the system
Let vf be the final velocity after the impact
KE = PE
1/2 (M + m) Vf^2 = (M + m) x g x h
vf^2 = 2 x 9.8 x 1.28 = 25.088
vf = 5 m/s
Thus, the final speed of the combined mass system after the impact s 5 m/s.
Final answer:
The final speed of the combined masses after the collision can be found using the conservation of momentum and mechanical energy principles, resulting in a final speed of approximately 1.20 m/s.
Explanation:
The final speed Vf of the combined masses after the collision can be calculated using the principle of conservation of momentum and conservation of mechanical energy.
By equating the initial momentum of the system before the collision to the final momentum of the system after the collision, you can determine the final speed Vf.
In this case, the final speed Vf is approximately 1.20 m/s.
Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is moving upward along the y axis at 2.0 m/s and ball B is moving to the right along the x axis with speed 3.7 m/s. After the collision ball B is moving along the positive y axis. a. What is the speed of ball A and B after the collision? b. What is the direction of motion of ball A after the collision? b. What is the total momentum and kinetic energy of the two balls after the collision?
Answer:
Speed of ball A after collision is 3.7 m/s
Speed of ball B after collision is 2 m/s
Direction of ball A after collision is towards positive x axis
Total momentum after collision is m×4·21 kgm/s
Total kinetic energy after collision is m×8·85 J
Explanation:
If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the systemLet the mass of each ball be m kg
v[tex]_{1}[/tex] be the velocity of ball A along positive x axis
v[tex]_{2}[/tex] be the velocity of ball A along positive y axis
u be the velocity of ball B along positive y axis
Conservation of momentum along x axis
m×3·7 = m× v[tex]_{1}[/tex]
∴ v[tex]_{1}[/tex] = 3.7 m/s along positive x axis
Conservation of momentum along y axis
m×2 = m×u + m× v[tex]_{2}[/tex]
2 = u + v[tex]_{2}[/tex] → equation 1
Assuming that there is no permanent deformation between the balls we can say that it is an elastic collisionAnd for an elastic collision, coefficient of restitution = 1∴ relative velocity of approach = relative velocity of separation
-2 = v[tex]_{2}[/tex] - u → equation 2
By adding both equations 1 and 2 we get
v[tex]_{2}[/tex] = 0
∴ u = 2 m/s along positive y axis
Kinetic energy before collision and after collision remains constant because it is an elastic collision
Kinetic energy = (m×2² + m×3·7²)÷2
= 8·85×m J
Total momentum = m×√(2² + 3·7²)
= m× 4·21 kgm/s
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 large aquarium of height 6 m is filled with fresh water to a depth of D = 1.50 m. One wall of the aquarium consists of thick plastic with horizontal length w = 8.40 m. By how much does the total force on that wall increase if the aquarium is next filled to a depth of D = 4.30 m? (Note: use g = 9.81 m/s2 and rho = 998 kg/m3.)
Answer
given,
height of aquarium = 6 m
Depth of fresh water = D = 1.50 m
horizontal length of the aquarium(w) = 8.40 m
total force increased when liquid is filled to depth = 4.30 m
g = 9.81 m/s²
ρ = 998 Kg/m³
force in the aquarium.
dF = PdA
[tex]F = \int PdA[/tex]
[tex]F = \int \rho\ g\ y\ (wdy)[/tex]
[tex]F = \rho\ g\ w \int ydy[/tex]
[tex]F = \rho\ g\ w\dfrac{y^2}{2}[/tex]
[tex]F = \dfrac{\rho\ g\ w\ y^2}{2}[/tex]
At D = 1.5 m
[tex]F = \dfrac{980\times 9.8\times 8.4\times 1.5^2}{2}[/tex]
F = 9.08 x 10⁴ N
At D = 4.30 m
[tex]F = \dfrac{980\times 9.8\times 8.4\times 4.3^2}{2}[/tex]
F = 7.46 x 10⁵ N
Total force on the wall increased by
ΔF = 74.6 x 10⁴ - 9.08 x 10⁴
ΔF = 65.52 x 10⁴ N
The universe is filled with photons left over from the Big Bang that today have an average energy of about 4.9 ✕10-4 (corresponding to a temperature of 2.7 K).
What is the number of available energy states per unit volume for these photons in an interval of 4 ✕10-5eV?
Answer:
The number of available energy is [tex]4.820\times10^{45}[/tex]
Explanation:
Given that,
Energy [tex]E=4.9\times10^{-4}\ J[/tex]
Temperature = 2.7 K
Energy states per unit volume [tex]dE=4\times10^{-5}\ eV[/tex]
We need to calculate the number of available energy
Using formula of energy
[tex]N=g(E)dE[/tex]
[tex]N=\dfrac{8\pi\times E^2 dE}{(hc)^3}[/tex]
Where, h = Planck constant
c = speed of light
E = energy
Put the value into the formula
[tex]N=\dfrac{8\pi\times(4.9\times10^{-4})^2\times4\times10^{-5}\times1.6\times10^{-19}}{(6.67\times10^{-34}\times3\times10^{8})^3}[/tex]
[tex]N=4.820\times10^{45}[/tex]
Hence, The number of available energy is [tex]4.820\times10^{45}[/tex]
A water tunnel has a circular cross-section with a radius of 1.8 m in its main section. In a test section of the tunnel the radius constricts to a value of 0.60 m. If the speed of water flow is 3.0 m/s in the main section, determine the speed of water flow in the test section.
To solve this problem it is necessary to apply the concepts related to the concepts to continuity.
From the continuity equation we know that
[tex]A_1V_1 = A_2 V_2[/tex]
Where,
A = Cross-sectional Area
V = Velocity
In a circle the area is given by
[tex]A_1 = \pi r_1^2\\A_2 = \pi r_2^2[/tex]
Therefore replacing with our values we have that
[tex]A_1 = \pi r_1^2\\A_1 = \pi 1.8^2\\A_1 = 10.278m^2[/tex]
While the area two is defined as
[tex]A_2 = \pi r_2^2\\A_2 = \pi 0.6^2\\A_2 = 1.13m^2[/tex]
Applying the continuity equation we have to
[tex]A_1V_1 = A_2 V_2[/tex]
[tex](10.278)(3)= (1.13) V_2[/tex]
[tex]V_2 = 27.28m/s[/tex]
Therefore the speed of water flow in the test section is 27.28m/s
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.
Learn more about average force on a rocket here:https://brainly.com/question/19860505
#SPJ11
When you go out to your car one cold winter morning you discover a 0.70-cm thick layer of ice on the windshield, which has an area of 1.0 m2 . If the temperature of the ice is -4.0 ∘C, and its density is 917 kg/m3, find the heat required to melt all the ice.
To find the heat required to melt the ice on the windshield, calculate the mass of the ice using its thickness, area, and density, then use the formula for heat transfer to find the heat required using the latent heat of fusion for ice.
Explanation:In this Physics problem, we're trying to calculate the heat needed to melt the -4.0 °C ice from a windshield. The heat required to melt ice is given by the formula Q = m * Lf, where Q is the heat, m is the mass of the substance (ice in this case), and Lf is the latent heat of fusion, which is the amount of heat necessary to change the state of the substance from solid to liquid at its melting point.
To calculate the mass of the ice, we use the density (ρ) given and the volume of the ice, which can be found by multiplying the thickness of the ice by the area of the windshield and converting cm to m. Volume of the ice would be Area * thickness = 1.0 m^2 * 0.007 m (from 0.7 cm) = 0.007 m^3. Plug this into ρ = m/V to get the mass m = ρ * V = 917 kg/m^3 * 0.007 m^3 = 6.42 kg.
Finally, we use Q = m * Lf which is Q = 6.42 kg * 334 kJ/kg (given that the heat of fusion for water is 334 kJ/kg), we get Q = 2143 kJ required to melt all the ice.
Learn more about Heat Transfer here:https://brainly.com/question/13433948
#SPJ12
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 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.
Learn more about Dispersion of Light here:https://brainly.com/question/20259452
#SPJ3
At an accident scene on a level road, investigators measure a car's skid mark to be 98 m long. It was a rainy day and the coefficient of friction was estimated to be 0.40. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.
Answer:
The speed of the car was 28 m/s.
Explanation:
Hi there!
The initial kinetic energy of the car, KE, is equal to the negative work, W, done by friction to bring the car to stop. Let´s write the work-energy theorem:
W = ΔKE = final kinetic energy - initial kinetic energy
In this case, the final kinetic energy is zero, then:
W = - initial kinetic energy
Since the work done by friction is negative (the work is done in opposite direction to the movement of the car), then:
Wfr = initial kinetic energy
The work done by friction is calculated as follows:
Wfr = Fr · d
Where:
Fr = friction force.
d = distance.
The friction force is calculated as follows:
Fr = N · μ
Where:
N = normal force.
μ = coefficient of friction
Since the only vertical forces acting on the car are the weight of the car and the normal force, and the car is not being accelerated in the vertical direction, the normal force has to be equal to the weight of the car (with opposite sign).
Then the friction force can be written as follows:
Fr = m · g · μ
Where:
m = mass of the car.
g = acceleration due to gravity (9.8 m/s²)
The work done by friction will be:
W = m · g · μ · d
The equation of kinetic energy is the following:
KE = 1/2 · m · v²
Where
m = mass of the car.
v = speed.
Then:
W = KE
m · g · μ · d = 1/2 · m · v²
2 · g · μ · d = v²
2 · 9.8 m/s² · 0.40 · 98 m = v²
v = 28 m/s
The speed of the car was 28 m/s.
A point charge q1 = 4.10 nC is placed at the origin, and a second point charge q2 = -2.90 nC is placed on the x-axis at x = + 20.0 cm. A third point charge q3 = 2.00 nC is to be placed on the x-axis between q1 and q2. (Take as zero the potential energy of the three charges when they are infinitely far apart.)
1. What is the potential energy of the system of the three charges if q3 is placed at x = + 11.0 cm?2. Where should q3 be placed between q1 and q2 to make the potential energy of the system equal to zero?
q3 should be placed at x = + 6.9 cm to make the potential energy of the system equal to zero.
To calculate the potential energy of the system of three charges, we use the formula:
PE = k * q1 * q2 / r12 + k * q2 * q3 / r23 + k * q3 * q1 / r31
where:
k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)
q1, q2, and q3 are the charges of the three point charges (4.10 nC, -2.90 nC, and 2.00 nC, respectively)
r12, r23, and r31 are the distances between the three point charges (20.0 cm, 9.0 cm, and 11.0 cm, respectively)
Plugging in the values, we get:
PE = (8.98755 x 10^9 N·m^2/C^2) * (4.10 x 10^-9 C) * (-2.90 x 10^-9 C) / 0.20 m + (8.98755 x 10^9 N·m^2/C^2) * (-2.90 x 10^-9 C) * (2.00 x 10^-9 C) / 0.09 m + (8.98755 x 10^9 N·m^2/C^2) * (2.00 x 10^-9 C) * (4.10 x 10^-9 C) / 0.11 m
PE = 1.19 x 10^-6 J
Therefore, the potential energy of the system of three charges if q3 is placed at x = + 11.0 cm is 1.19 x 10^-6 J.
To make the potential energy of the system equal to zero, we need to place q3 at a point where the forces from q1 and q2 cancel each other out. This means that the vectors representing the forces must be equal and in opposite directions.
The force from q1 on q3 is given by:
F13 = k * q1 * q3 / r31^2
where:
k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)
q1 is the charge of q1 (4.10 nC)
q3 is the charge of q3 (2.00 nC)
r31 is the distance between q1 and q3
The force from q2 on q3 is given by:
F23 = k * q2 * q3 / r23^2
where:
k is Coulomb's constant (8.98755 x 10^9 N·m^2/C^2)
q2 is the charge of q2 (-2.90 nC)
q3 is the charge of q3 (2.00 nC)
r23 is the distance between q2 and q3
For the forces to cancel each other out, we need:
F13 = -F23
Substituting in the equations for the forces, we get:
k * q1 * q3 / r31^2 = -k * q2 * q3 / r23^2
Solving for r31, we get:
r31 = r23 * sqrt(q2/q1)
Substituting in the values, we get:
r31 = 0.09 m * sqrt(-2.90 nC / 4.10 nC)
r31 = 0.069 m
Therefore, q3 should be placed at x = + 6.9 cm to make the potential energy of the system equal to zero.
for such more question on potential energy
https://brainly.com/question/1455245
#SPJ3
During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a velocity of 4.1 m/s, grabs and holds onto him so that they move off together with a velocity of 2.2 m/s. If the mass of the tackler is 100 kg, determine the mass of the receiver in kilograms. Assume momentum is conserved.
Answer:m=86.36 kg
Explanation:
Given
mass of tackler [tex]m_T=100 kg[/tex]
initial velocity of Tackler [tex]u_t=4.1 m/s[/tex]
Final velocity of combined system [tex]v=2.2 m/s[/tex]
Let m be the mass of receiver
conserving momentum
[tex]m\times 0+m_t\times u_t=(m+m_t)v[/tex]
[tex]0+100\times 4.1=(100+m)\cdot 2.2[/tex]
[tex]410=(100+m)\cdot 2.2[/tex]
[tex]100+m=186.36[/tex]
[tex]m=86.36 kg[/tex]
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]
A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/s. A second piece has a mass of 300 g and flies off along the y-axis with a speed of 45.0 m/s. What are the magnitude and direction of the total momentum of these two pieces?A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along the -axis with a speed of 82.0 m/s. A second piece has a mass of 300 g and flies off along the -axis with a speed of 45.0 m/s. What are the magnitude and direction of the total momentum of these two pieces?361 kg·m/s at 0.983° from the x-axis361 kg·m/s at 56.3° from the x-axis93.5 kg·m/s at 28.8° from the x-axis21.2 kg·m/s at 39.5° from the x-axis21.2 kg·m/s at 56.3° from the x-axis
Answer:
The magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.
Explanation:
Hi there!
The total momentum is calculated as the sum of the momenta of the pieces.
The momentum of each piece is calculated as follows:
p = m · v
Where:
p = momentum.
m = mass.
v = velocity.
The momentum is a vector. The 200 g-piece flies along the x-axis then, its momentum will be:
p = (m · v, 0)
p = (0.200 kg · 82.0 m/s, 0)
p = (16.4 kg m/s, 0)
The 300 g-piece flies along the y-axis. Its momentum vector will be:
p =(0, m · v)
p = (0, 0.300 kg · 45.0 m/s)
p = (0, 13.5 kg m/s)
The total momentum is the sum of each momentum:
Total momentum = (16.4 kg m/s, 0) + (0, 13.5 kg m/s)
Total momentum = (16.4 kg m/s + 0, 0 + 13.5 kg m/s)
Total momentum = (16.4 kg m/s, 13.5 kg m/s)
The magnitude of the total momentum is calculated as follows:
[tex]|p| = \sqrt{(16.4 kgm/s)^2+(13.5 kg m/s)^2}= 21.2 kg m/s[/tex]
The direction of the momentum vector is calculated using trigonometry:
cos θ = px/p
Where px is the horizontal component of the total momentum and p is the magnitude of the total momentum.
cos θ = 16.4 kg m/s / 21.2 kg m/s
θ = 39.3 (39.5° if we do not round the magnitude of the total momentum)
Then, the magnitude of the total momentum is 21.2 kg m/s and its direction is 39.5° from the x-axis.
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.
A block is released from rest and slides down an incline. The coefficient of sliding friction is 0.38 and the angle of inclination is 60.0°. Use energy considerations to find how fast the block is sliding after it has traveled a distance of 38.7 cm along the incline.
Answer:
2.27m/s
Explanation:
The coefficient of sliding friction = 0.38, the incline angle = 60o
Coefficient of sliding friction = frictional force/ force of normal (fn)
The normal force = mgcos60 where m is the mass of the body and g is acceleration due to gravity
Frictional force = coefficient of friction*fn = 0.38*9.81*m*cos 60 = 1.8639m
Force tending to push the body down the incline plane = mgsin60 = 8.496m
Net force on the body = force pushing downward - Frictional force
Net force = 8.496m - 1.8639m = 6.632m
Kinetic energy of the body = 1/2 mv^2 = work done by the net force pushing the body down = net force * distance travelled
1/2mv^2 = 6.632m * 0.387in meters
Cancel mass on both side leaves 1/2v^2 = 6.632*0.387
V^2 = 2.566584*2 = 5.133
V = √5.133 = 2.27m/s
The speed of the block after it has traveled a distance of 38.7 cm is 1.2 m/s.
How to calculate the speed of the block?The speed of the block is calculated by applying the following formula.
Using work energy theorem;
The work done by the force of friction is equal to the change in kinetic energy of the box.
W = K.E
μmgd cosθ = ¹/₂mv²
μgdcosθ = ¹/₂v²
v² = (2gd cosθ)μ
v = √ (2gd cosθμ)
where;
d is the distance g is gravityθ is angle of inclinationμ is coefficient of frictionv = √ (2gd cosθμ)
v = √ (2 x 9.8 x 0.387 x cos(60) x 0.38)
v = 1.2 m/s
Learn more about speed of sliding blocks here: https://brainly.com/question/30440096
#SPJ3