Answer:
The ratio of stopping distances is 4 i.e by a factor 4 the stopping distances differ
Explanation:
Using 3rd equation of motion we have
For car 1
[tex]v_{1}^{^{2}}=u_{1}^{2}+2a_{1}s_{1}[/tex]
For car 2 [tex]v_{2}^{^{2}}=u_{2}^{2}+2a_{2}s_{2}[/tex]
Since the initial speed of both the cars are equal and when the cars stop the final velocities of both the cars become zero thus the above equations reduce to
[tex]u^{2}=-2a_{1}s_{1}\\\\s_{1}=\frac{-u^{2}}{2a_{1}}[/tex].............(i)
Similarly for car 2 we have
[tex]s_{2}=\frac{-u^{2}}{2a_{2}}[/tex]..................(ii)\
Taking ratio of i and ii we get
[tex]\frac{s_{1}}{s_{2}}=\frac{a_{2}}{a_{1}}[/tex]
Let
[tex]\frac{a_{2}}{a_{1}}=4[/tex]
Thus
[tex]\frac{s_{1}}{s_{2}}=4[/tex]
The ratio of stopping distances is 4
Final answer:
The stopping distances of two cars with different deceleration rates will differ by the inverse ratio of their decelerations. If one car's deceleration is a quarter of the other's, the car with the lower deceleration will require four times the stopping distance of the other car.
Explanation:
The question at hand is about how the stopping distances of two cars differ when their deceleration rates are different. If one car's deceleration is a quarter of the other's, then to find the factor by which the stopping distances differ, we can use the equation of motion [tex]v^2 = u^2 + 2as[/tex] v is the final velocity, u is the initial velocity, a is the deceleration, and s is the stopping distance.
Since the final velocity (v) for both cars will be zero (they come to a stop), and assuming the initial velocities (u) are the same for both cars, we can set the equation for both as follows: 0 = u^2 + 2a1s1 and 0 = u^2 + 2a2s2, with a1 being the higher deceleration of the first car and a2 being a quarter of that, s1 and s2 being the stopping distances respectively.
When we solve the equations for s1 and s2, we find that s1 is proportional to 1/a1 and s2 is proportional to 1/a2, so if a2 = 1/4 a1, then the ratio of the stopping distances s2/s1 is equal to the inverse ratio of the decelerations, which is 4. Hence, the car with the lower deceleration requires four times the distance to stop compared to the car with the higher deceleration.
You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8 × 107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 59.7 m/s2. What are the mass of (a) the planet and (b) the star?
Using the acceleration due to gravity on the planet, we calculate its mass to be 8.97 x 10²⁴ kg. For the mass of the star, using Kepler's third law, we find it to be approximately 1.99 x 10³⁰ kg, assuming the planet's orbit is about 1 AU.
To find the mass of the planet and the star, we can use Newton's form of Kepler's third law and Newton's universal law of gravitation. Let's denote the mass of the star as M and the mass of the planet as m.
Mass of the Planet
The acceleration due to gravity (g) on the surface of the planet is given by:
g = Gm / r²
Where:
G is the gravitational constant (approximately 6.674 x 10⁻¹¹ N m2 kg-²).
m is the mass of the planet.
r is the radius of the planet, which is half of its diameter.
Rearranging the formula to solve for m:
m = g r² / G
Substituting the given values (g = 59.7 m/s² and r = 9 x 106 m), we get:
m = 59.7 x (9 x 106)² / 6.674 x 10⁻¹¹
= 8.97 x 10²⁴ kg
Mass of the Star
Using Newton's version of Kepler's third law, we get:
M = 4π² a3 / G T²
Where: a is the semi-major axis of the planet's orbit, which is the average distance from the planet to the star.
T is the orbital period of the planet around the star.
Assuming that the planet's orbit is approximately 1 Astronomical Unit (AU) since the period is close to 1 Earth year, and converting 402 Earth days to seconds, we find M.
T = 402
days x 24 hours/day x 3600 seconds/hour = 3.47 x 10⁷ seconds
a = 1 AU = 1.496 x 10¹¹ m
Then the mass of the star M is:
M = 4π² (1.496 x 1011)3 / (6.674 x 10⁻¹¹ x (3.47 x 10⁷)²) = 1.99 x 10³⁰ kg
Find the heat that flows in 1.0 s through a lead brick 13 cm long if the temperature difference between the ends of the brick is 8.0°C. The cross-sectional area of the brick is 13 cm^2. Express your answer using two significant figures.
Final answer:
The heat flow through the lead brick can be calculated using Fourier's law of heat conduction. The thermal conductivity of lead is used along with the provided dimensions and temperature difference to find the heat flow, which is 2.2 W to two significant figures.
Explanation:
To calculate the heat flow through a lead brick, we can use Fourier's law of heat conduction, which states that the heat transfer rate (Q) through a material is proportional to the thermal conductivity of the material (k), the area (A) through which the heat is being transferred, the temperature difference (ΔT) across the material, and inversely proportional to the thickness (d) of the material.
The formula to calculate heat flow is: Q = k × A × ΔT × t / d
First, we need to convert all measurements to SI units: the length from cm to m (13 cm = 0.13 m), the area from cm² to m² (13 cm² = 0.0013 m²), and the time from seconds to hours (1.0 s). The thermal conductivity of lead is approximately 35 W/m°C. Plugging in the values: Q = 35 W/m°C × 0.0013 m² × 8.0°C × 1.0 s / 0.13 m
By simplifying the equation, we find the heat flow through the lead brick:
Q = (35 × 0.0013 × 8.0 × 1.0) / 0.13 W = 2.215 W
To two significant figures, the heat flow is 2.2 W.
A hockey puck of mass m traveling along the x axis at 6.0 m/s hits another identical hockey puck at rest. If after the collision the second puck travels at a speed of 4.8 m/s at an angle of 30° above the x axis, what is the final velocity of the first puck
Answer:
Velocity is 3.02 m/s at an angle of 53.13° below X-axis.
Explanation:
Let unknown velocity be v.
Here momentum is conserved.
Initial momentum = Final momentum
Initial momentum = m x 6i + m x 0i = 6m i
Final momentum = m x (4.8cos 30 i + 4.8sin 30 j) + m x v = 4.16 m i + 2.4 m j + m v
Comparing
4.16 m i + 2.4 m j + m v = 6m i
v = 1.84 i - 2.4 j
Magnitude of velocity
[tex]v=\sqrt{1.84^2+(-2.4)^2}=3.02m/s[/tex]
Direction,
[tex]\theta =tan^{-1}\left ( \frac{-2.4}{1.8}\right )=-53.13^0[/tex]
Velocity is 3.02 m/s at an angle of 53.13° below X-axis.
To find the final velocity of the first puck, apply conservation of momentum and solve for v'. The final velocity of the first puck is 4.8 m/s.
Explanation:To find the final velocity of the first puck, we can apply the principle of conservation of linear momentum. The initial momentum of the system is given by m1v1 + m2v2, where m1 is the mass of the first puck, m2 is the mass of the second puck, v1 is the initial velocity of the first puck, and v2 is the initial velocity of the second puck.
Since the collision is elastic, the total momentum before and after the collision is conserved. So, m1v1 + m2v2 = (m1 + m2)v', where v' is the final velocity of both pucks after the collision. We can plug in the given values to find the final velocity of the first puck.
Let's solve for v': m1v1 + m2v2 = (m1 + m2)v' => (m1)(6.0 m/s) + (m2)(0 m/s) = (m1 + m2)(4.8 m/s).
From the given information, we know that the two pucks are identical, so m1 = m2. Substituting this into the equation, we get (m)(6.0 m/s) = 2(m)(4.8 m/s), where m is the mass of each puck. Simplifying this equation, we find the mass cancels out, leaving 6.0 m/s = 2(4.8 m/s). Solving for v', we find v' = 4.8 m/s.
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A dentist's drill starts from rest. After 3.50 s of constant angular acceleration, it turns at a rate of 2.49 104 rev/min. (a) Find the drill's angular acceleration. Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. rad/s2 (along the axis of rotation) (b) Determine the angle through which the drill rotates during this period. rad
Answer:
Part A)
[tex]\alpha = 745 rad/s^2[/tex]
Part B)
[tex]\theta = 4563.1 rad[/tex]
Explanation:
Drill starts from rest so its initial angular speed will be
[tex]\omega_i = 0[/tex]
now after 3.50 s the final angular speed is given as
[tex]f = 2.49 \times 10^4 rev/min[/tex]
[tex]f = {2.49 \times 10^4}{60} = 415 rev/s[/tex]
so final angular speed is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega_f = 2607.5 rad/s[/tex]
now we have angular acceleration given as
[tex]\alpha = \frac{\omega_f - \omega_i}{\Delta t}[/tex]
[tex]\alpha = \frac{2607.5 - 0}{3.50}[/tex]
[tex]\alpha = 745 rad/s^2[/tex]
Part b)
The angle through which it is rotated is given by the formula
[tex]\theta = \frac{(\omega_f + \omega_i)}{2}\delta t[/tex]
now we have
[tex]\theta = \frac{(2607.5 + 0)}{2}(3.50)[/tex]
[tex]\theta = 4563.1 rad[/tex]
American football is played on a 100 yd100 yd long field, excluding the end zones. What is the length ????L of the field in meters? Assume that 1 m equals 3.281 ft.3.281 ft.
Answer:
Answer to the question: 91.44m
Explanation:
L=100 yds = 300 ft = 91.44 m
To convert 100 yards to meters for an American football field, you multiply 100 yards by 3 to get 300 feet, then convert feet to meters using the given conversion factor, resulting in approximately 91.4 meters.
To convert the length of an American football field from yards to meters, we first need to understand the conversion factors between these units. We know that 1 yard equals 3 feet, and we are given that 1 meter equals 3.281 feet. Starting with the length of the field in yards, which is 100 yards, we convert yards to feet and then convert feet to meters.
Step-by-Step Conversion
Convert yards to feet: 100 yards imes 3 feet/yard = 300 feet.Convert feet to meters: 300 feet imes (1 meter/3.281 feet) akes approximately 91.4 meters.Therefore, the length of an American football field is roughly 91.4 meters, excluding the end zones.
Two spherical objects with a mass of 6.22 kg each are placed at a distance of 1.02 m apart. How many electrons need to leave each object so that the net force between them becomes zero?
Answer:
The number of electrons need to leave each object is [tex]3.35\times10^{9}[/tex]
Explanation:
Given that,
Mass of object = 6.22 kg
Distance = 1.02 m
We need to calculate the number of electron
Using formula of electric force
[tex]F_{e}=\dfrac{k(q)^2}{r^2}[/tex]....(I)
We know that,
[tex]q = Ne[/tex]
Put the value of q in equation (I)
[tex]F_{e}=\dfrac{k(Ne)^2}{r^2}[/tex].....(II)
Using gravitational force
[tex]F_{G}=\dfrac{Gm^2}{r^2}[/tex].....(III)
Equating equation (II) and (III)
[tex]F_{e}=F_{G}[/tex]
[tex]\dfrac{k(Ne)^2}{r^2}=\dfrac{Gm^2}{r^2}[/tex]
[tex]N=\sqrt{\dfrac{G}{k}}\times\dfrac{m}{e}[/tex]....(IV)
Put the value in the equation(IV)
[tex]N=\sqrt{\dfrac{6.67\times10^{-11}}{9\times10^{9}}}\times\dfrac{6.22}{1.6\times10^{-19}}[/tex]
[tex]N=3.35\times10^{9}[/tex]
Hence, The number of electrons need to leave each object is [tex]3.35\times10^{9}[/tex]
A circular swimming pool has a diameter of 8 meters. The sides are 3 meters high and the depth of the water is 1.5 meters. How much work is required to pump all of the water over the side? Your answer must include the correct
The amount of work required to pump all of the water over the side of the pool is 471,238.9 Newton Meters or Joules.
Explanation:The work done to pump water out of a pool involves the concept of physics specifically related to potential energy, gravity, and volume. The work done to move a certain volume of water is given by the formula: Work = Weight x Height.
First, we need to find the volume of the water in the pool. The pool's shape resembles a cylinder, and the volume is given by the formula for a cylinder: Volume = pi × (diameter/2)²× height. Given a diameter of 8 meters and a height of 1.5 meters, the volume to be moved is pi * (8/2)² × 1.5 = 48pi cubic meters.
The weight of this water can be calculated by multiplying its volume by its density. The density of water is 1000 kg/m³. Therefore, the weight of the water is Volume x Density x g (acceleration due to gravity), which is 48pi × 1000 × 9.8 = 471,238.9 kg×m²/s² or Newton Meter (Nm) which is the unit of work.
So, the amount of work required to pump all the water over the side is 471,238.9 Nm or Joules (J).
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The work required to pump all the water over the side of a circular swimming pool of diameter 8 meters and water depth 1.5 meters is calculated using the formula Work = Force x Distance. The Force required is given by the weight of the water, which depends on its volume and density. The result is about 353,429.16 Joules.
Explanation:To calculate the work required to pump water out of a swimming pool, we can use the formula for work: Work = Force x Distance. The force required is equal to the weight of the water which depends on the volume of the water and its density.
First, let's calculate the volume of the water in the pool. Given that it's a circular pool with a diameter of 8 meters, the radius is 4 meters. The depth of the water is 1.5 meters. So, volume (V) = πr²h = π×(4m)²×(1.5m) = 24π cubic meters.
The density of water is 1000 kg/m³. Therefore, the weight of water = Volume x Density x Gravity = 24π m³ ×1000 kg/m³ × 9.8 m/s² = 235619.44 kg×m/s², or Newtons. This is the force we need to overcome to lift the water.
The distance that we want to lift this mass is the depth of the pool, assuming we are pumping the water over the side of the pool which is 1.5 meters high. So, Work = Force x Distance = 235619.44 N×1.5m = 353429.16 Joules.
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A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of heat output. Express your answer using two significant figures.
Answer:1200
Explanation:
Given data
Upper Temprature[tex]\left ( T_H\right )=240^{\circ}\approx 513[/tex]
Lower Temprature [tex]\left ( T_L\right )=20^{\circ}\approx 293[/tex]
Engine power ouput[tex]\left ( W\right )=910 W[/tex]
Efficiency of carnot cycle is given by
[tex]\eta =1-\frac{T_L}{T_H}[/tex]
[tex]\eta =\frac{W_s}{Q_s}[/tex]
[tex]1-\frac{293}{513}=\frac{910}{Q_s}[/tex]
[tex]Q_s=2121.954 W[/tex]
[tex]Q_r=1211.954 W[/tex]
rounding off to two significant figures
[tex]Q_r=1200 W[/tex]
Answer:
1200
Explanation:
The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.
Answer:
The density of the metal is 5200 kg/m³.
Explanation:
Given that,
Weight in air= 0.10400 N
Weight in water = 0.08400 N
We need to calculate the density of metal
Let [tex]\rho_{m}[/tex] be the density of metal and [tex]\rho_{w}[/tex] be the density of water is 1000kg/m³.
V is volume of solid.
The weight of metal in air is
[tex]W =0.10400\ N[/tex]
[tex]mg=0.10400[/tex]
[tex]\rho V g=0.10400[/tex]
[tex]Vg=\dfrac{0.10400}{\rho_{m}}[/tex].....(I)
The weight of metal in water is
Using buoyancy force
[tex]F_{b}=0.10400-0.08400[/tex]
[tex]F_{b}=0.02\ N[/tex]
We know that,
[tex]F_{b}=\rho_{w} V g[/tex]....(I)
Put the value of [tex]F_{b}[/tex] in equation (I)
[tex]\rho_{w} Vg=0.02[/tex]
Put the value of Vg in equation (II)
[tex]\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02[/tex]
[tex]1000\times\dfrac{0.10400}{0.02}=\rho_{m}[/tex]
[tex]\rho_{m}=5200\ kg/m^3[/tex]
Hence, The density of the metal is 5200 kg/m³.
The density of the metal bracelet can be determined through the principles of Archimedes using the difference of its weight in air and water. After calculating the buoyant force, the volume of water displaced was determined, which was also the volume of the bracelet. Dividing its weight in the air by this volume gave us the density, which is approximately 5,434 kg/m³.
Explanation:In order to find the density of the metal bracelet, we use Archimedes' principle, which states that the buoyant force (force exerted on a submerged object) is equal to the weight of the fluid displaced by the object.
The first step is to calculate the difference in weight in air and water, which gives the buoyant force, i.e., the weight of the water displaced. So, the difference is 0.10400 N - 0.08400 N = 0.02000 N.
Then, we can find the volume of the water displaced by using the formula for weight, w = mg. Here, w is the buoyant force, and g is the acceleration due to gravity (9.8 m/s² on Earth). So, m = w / g = 0.02000 N / 9.8 m/s² = 0.00204 kg. This is the mass of the water displaced, which is also the volume since 1 kg of water is 1 liter (or 1000 cm³).
Finally, the density of the metal bracelet can be calculated by dividing its weight in air by the volume of water displaced, and by gravity. That is, density = 0.10400 N / (0.00204 kg x 9.8 m/s²) = 5,434.4 kg/m³. So the density of the metal bracelet is approximately 5,434 kg/m³.
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An alpha particle, which has charge 3.20 ✕ 10−19 C, is moved from point A, where the electric potential is 2.40 ✕ 103 J/C, to point B, where the electric potential is 4.95 ✕ 103 J/C. Calculate the work in electron volts done by the electric field on the alpha particle.
Answer:
5100 eV
Explanation:
q = 3.2 x 10^-19 C
Va = 2.4 x 10^3 J/C
Vb = 4.95 x 10^3 J/C
Work done = q (Vb - Va)
W = 3.2 x 10^-19 x (4.95 - 2.4 ) x 10^3 = 8.16 x 10^-16 J
A we know that 1 eV = 1.6 x 10^-19 J
So, W = (8.16 x 10^-16) / (1.6 x 10^-19) = 5100 eV
A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.2 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.) rad/s
Answer:0.2 rad/s
Explanation:
Given data
Velocity of the bottom point of the ladder=1.2Ft/s
Length of ladder=10ft
distance of the bottom most point of ladder from origin=8ft
From the data the angle θ with ladder makes with horizontal surface is
Cosθ=[tex]\frac{8}{10}[/tex]
θ=36.86≈37°
We have to find rate of change of θ
From figure we can say that
[tex]x^{2}[/tex]+[tex]y^{2}[/tex]=[tex]AB^{2}[/tex]
Differentiating above equation we get
[tex]\frac{dx}{dt}[/tex]=-[tex]\frac{dy}{dt}[/tex]
i.e [tex]{V_A}=-{V_B}=1.2ft/s[/tex]
[tex]{at\theta}={37}[/tex]
[tex]Y=6ft[/tex]
[tex]and\ about\ Instantaneous\ centre\ of\ rotation[/tex]
[tex]{\omega r_A}={V_A}[/tex]
[tex]{\omega=\frac{1.2}{6}[/tex]
ω=0.2rad/s
i.e.Rate of change of angle=0.2 rad/s
A 900 kg car initially going 15 m/s only in the x-direction runs into a stationary 1500 kg truck. After the collision the car is going 5.0 m/s at an angle of 40 degrees above the x- axis. What is the magnitude and direction of the velocity of the truck right after the collision?
Answer:
6.97 m/s, 344 degree
Explanation:
mass of car, m = 900 kg, uc = 15 m/s, vc = 5 m/s, θ = 40 degree
mass of truck, M = 1500 kg, uT = 0, vT = ?, Φ = ?
Here, vT be the velocity of truck after collision and Φ its direction above x axis.
Use conservation of momentum in X axis
900 x 15 + 1500 x 0 = 900 x 5 Cos 40 + 1500 x vT Cos Φ
13500 - 3447.2 = 1500 vT CosΦ
vT CosΦ = 6.7 .....(1)
Use conservation of momentum in y axis
0 + 0 = 900 x 5 Sin 40 + 1500 vT SinΦ
vT SinΦ = - 1.928 .....(2)
Squarring both the equations and then add
vT^2 = 6.7^2 + (-1.928)^2
vT = 6.97 m/s
Dividing equation 2 by 1
tan Φ = - 1.928 / 6.7
Φ = - 16 degree
Angle from + X axis = 360 - 16 = 344 degree
First, let us consider an object launched vertically upward with an initial speed v. Neglect air resistance. Part B) At the top point of the flight, what can be said about the projectile's kinetic and potential energy?
A). Both kinetic and potential energy are at their maximum values.
B).Both kinetic and potential energy are at their minimum values.
C).Kinetic energy is at a maximum; potential energy is at a minimum.
D).Kinetic energy is at a minimum; potential energy is at a maximum.
Answer:
Option d is the correct option Kinetic energy is minimum while as potential energy is maximum
Explanation:
At the top most point of the flight since it cannot reach any further up in the vertical direction thus the potential energy at this position shall be maximum. Now since the total energy of the projectile is conserved so the remaining kinetic energy shall be minimum at that point so as the sum of the kinetic and potential energies remain constant.
A box contains 15 apples. How many different selections of 3 apples can be made sco as to include a particular apple?
Answer:
Total number of different selections = 455
Explanation:
Number of selections from n objects if we select r things [tex]=^nC_r[/tex]
Here we need to find number off selections of 3 apples from 15 apples.
Number off selections of 3 apples from 15 apples
[tex]=^{15}C_3=\frac{15\times 14\times 13}{1\times 2\times 3}=455[/tex]
Total number of different selections = 455
If the area of an iron rod is 10 cm by 0.5 cm and length is 35 cm. Find the value of resistance, if 11x10^-8 ohm.m be the resistivity of iron.
Answer:
Resistance of the iron rod, R = 0.000077 ohms
Explanation:
It is given that,
Area of iron rod, [tex]A=10\ cm\times 0.5\ cm=5\ cm^2 = 0.0005\ m^2[/tex]
Length of the rod, L = 35 cm = 0.35 m
Resistivity of Iron, [tex]\rho=11\times 10^{-8}\ \Omega-m[/tex]
We need to find the resistance of the iron rod. It is given by :
[tex]R=\rho\dfrac{L}{A}[/tex]
[tex]R=11\times 10^{-8}\times \dfrac{0.35\ m}{0.0005\ m^2}[/tex]
[tex]R=0.000077 \Omega[/tex]
So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a friction force of 12 N. How far does it slide on the snow before coming to rest?
Answer:The sled slides 16.875m before rest.
Explanation:
[tex]a=\frac{F}{m} =\frac{12N}{20kg}[/tex]
a=0.6 m/s²
[tex]Vf=0=Vi-a.t[/tex]
[tex]t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg[/tex]
[tex]d= Vi.t - \frac{a.t^{2}}{2}[/tex]
[tex]d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m[/tex]
A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal into it. (a) What is the final velocity of the loaded freight car? (b) How much kinetic energy is lost?
Explanation:
The mass of freight car, m₁ = 30,000 kg
Velocity of freight car, u₁ = 0.85 m/s
Mass of hopper, m₂ = 110,000 kg
(a) Let v is the final velocity of the loaded freight car. Initial momentum of the car before the dump, [tex]p_i=30000\ kg\times 0.850\ m/s=25500\ kg-m/s[/tex]
Final momentum, [tex]p_f=(30000\ kg+110000\ kg)v=140000\ v[/tex]
According to the conservation of momentum,
initial momentum = final momentum
25500 kg-m/s = 140000 v
v = 0.182 m/s
So, the final velocity of the loaded freight car is 0.182 m/s.
(b) Initial kinetic energy, [tex]k_i=\dfrac{1}{2}\times 30000\ kg\times (0.850\ m/s)^2=10837.5\ J[/tex]
Final kinetic energy, [tex]k_f=\dfrac{1}{2}(m_1+m_2)v^2[/tex]
[tex]k_f=\dfrac{1}{2}\times (30000\ kg+110000\ kg)\times (0.182\ m/s)^2=2318.68\ J[/tex]
So, loss in kinetic energy, [tex]\Delta k=k_f-k_i[/tex]
[tex]\Delta k=2318.68\ J-10837.5\ J=-8518.82\ J[/tex]
So, 8518.82 J of kinetic energy is lost after collision. Hence, this is the required solution.
Using the conservation of momentum, we find the final velocity of the loaded freight car to be 0.182 m/s. By comparing the initial and final kinetic energies, we find that 8,459.89 Joules of kinetic energy is lost during the process.
Explanation:This problem is a classic example of conservation of momentum, and to a lesser extent, conservation of kinetic energy. Firstly, we calculate the initial momentum of the system. Momentum is mass multiplied by velocity, so the initial momentum of the freight car is 30,000 kg * 0.850 m/s = 25,500 kg*m/s. After the scrap metal is dumped into the freight car, the total mass of the system is now 30,000 kg + 110,000 kg = 140,000 kg. Because momentum must be conserved, we calculate the final velocity of the freight car as the total momentum divided by the total mass, 25,500 kg*m/s / 140,000 kg = 0.182 m/s.
Secondly, before the scrap metal is added, the initial kinetic energy of the freight car is 0.5 * 30,000 kg * (0.850 m/s)² = 10,781.25 Joules. After the metal is added, the final kinetic energy is 0.5 * 140,000 kg * (0.182 m/s)² = 2,321.36 Joules. The change in kinetic energy can then be obtained by subtracting the final kinetic energy from the initial one, which gives us 8,459.89 Joules as the amount of kinetic energy that is lost.
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A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a) What is the acceleration of the ball while it is in flight? (b) What is the velocity of the ball when it reaches its maximum height? Find (c) the initial velocity of the ball and (d) the maximum height it reaches
Answer:
Part a)
a = -9.81 m/s/s
Part b)
v = 0
Part c)
v = 9.81 m/s
Part d)
[tex]H = 4.905 m[/tex]
Explanation:
Part a)
During the motion of ball it will have only gravitational force on the ball
so here the acceleration of the ball is only due to gravity
so it is given as
[tex]a = g = 9.81 m/s^2[/tex]
Part b)
As we know that ball is moving against the gravity
so here the velocity of ball will keep on decreasing as the ball moves upwards
so at the highest point of the motion of the ball the speed of ball reduce to zero
[tex]v_f = 0[/tex]
Part c)
We know that the total time taken by the ball to come back to the initial position is T = 2 s
so in this time displacement of the ball will be zero
[tex]\Delta y = 0 = v_y t + \frac{1}{2} at^2[/tex]
[tex]0 = v_y (2) - \frac{1}{2}(9.81)(2^2)[/tex]
[tex]v_y = 9.81 m/s[/tex]
Part d)
at the maximum height position we know that the final speed will be zero
so we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
here we have
[tex]0 - (9.81^2) = 2(-9.81)H[/tex]
[tex]H = 4.905 m[/tex]
The acceleration of the ball is -9.8 m/s², the velocity of the ball at max height is 0 m/s, the initial velocity is 9.8 m/s and the max height is 4.9 m.
Explanation:The questions refer to the motion of a tennis ball under gravity. (a) The acceleration of the ball while it is in flight is -9.8 m/s², due to Earth's gravity. This will be the case throughout the ball's flight, regardless of whether it's moving up or down. (b) The velocity of the ball when it reaches its maximum height is 0 m/s, as the ball stops moving vertically for an instant before it starts coming down. (c) We can calculate the initial velocity using the equation v = u + at. Here, v is final velocity, u is initial velocity, a is acceleration and t is time. As v = 0 m/s at max height, a = -9.8 m/s² and t = 1 s (time taken to reach max height, half of total time), we find u to be 9.8 m/s. (d) The maximum height can be calculated using the equation h = ut + 0.5at². Here, h is height, u is initial velocity, t is time and a is acceleration. Subsituting u = 9.8 m/s, a = -9.8 m/s², t = 1 s, we get h to be 4.9 m.
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If a pendulum has a period of 4 seconds at the north pole with gravity force 9.83 ms^-2 and a 3.97m pendulum length, how could the same pendulum be adjusted to have the same period as the equator with 9.78ms^-2 gravity force?
Answer:
The same pendulum could be adjusted to have the same period, in the equator must have a length of 3.949m.
Explanation:
Tnp= 4 sec
gnp= 9.83 m/sec²
Lnp= 3.97m
Tequ= 4 sec
gequ= 9.78 m/sec²
Lequ=?
Lequ= (Lnp* gequ) / gnp
Lequ= 3.949 m
A 63.9-kg wrecking ball hangs from a uniform, heavy-duty chain having a mass of 20.5 kg . (Use 9.80 m/s2 for the gravitational acceleration at the earth's surface.)
Find the maximum tension in the chain.
Find the minimum tension in the chain.
What is the tension at a point three-fourths of the way up from the bottom of the chain?
Answer:
a) Tmax=827.12N
b) Tmin = 626.22N
c) 776.895 N
Explanation:
Given:
Mass of wrecking ball M1=63.9 Kg
Mass of the chain M2=20.5 Kg
acceleration due to gravity, g=9.8m/s²
Now,
(a)The Maximum Tension generated in the chain,
Tmax=(M1+M2)×g)
Tmax=(M1+M2)×(9.8 m/s²)
Tmax=(63.9+ 20.5)×(9.8 m/s²)
Tmax=827.12N
(b) The Minimum Tension Tmin will be due to the weight of the wrecking ball only
Mathematically,
Tmin=weight of the wrecking ball
Tmin = 63.9kg×9.8m/s²
Tmin = 626.22N
(c)Now. the tension at 3/4 from the bottom of the chain
In this part we have to use only 75% of the chain i.e the weight acting below the point of consideration
thus, the tension will be produced by the weight of the 3/4 part of the chain and the wrecking ball
Therefore, the weight of the 3/4 part of the chain = [tex]\frac{3}{4}\times 20.5\times 9.8 N[/tex]
= 150.675 N
Hence, the tension at a point 3/4 of the way up from the bottom of the chain will be = 150.675 N + (63.9×9.8) N = 776.895 N
A child is twirling a 0.0154-kg ball on a string in a horizontal circle whose radius is 0.149 m. The ball travels once around the circle in 0.639 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, by what factor does the centripetal force increase?
Answer:
The answers are: a)Fcp=0,23N b)As Fcp=0,93N, it increases 4 times when the speedis doubled
Explanation:
Let´s explain what´s the centripetal force about: It´s the force applied over an object moving on a curvilinear path. This force is directed to the rotation center.
This definition is described this way:
[tex]Fcp*r=m*V^2[/tex] where:
Fcp is the Centripetal Force
r is the horizontal circle radius
m is the ball mass
V is the tangencial speed, same as the rotation speed
w is the angular speed
Here we need to note that the information we have talks about 1cycle/0,639s (One cycle per 0,639s). We need to express this in terms of radians/seconds. To do it we define that 1cycle is equal to 2pi, so we can find the angular speed this way:
[tex]w=(1cycle/0,639s)*(2pi/1cycle)[/tex]
So the angular speed is [tex]w=9,83rad/s[/tex]
Now that we have this information, we can find the tangencial speed, which will be the relation between the angular speed and the circle radius, this way:
[tex]V=w*r[/tex] so the tangencial speed is:
[tex]V=(9,83rad/s)*(0,149m)[/tex]
[tex]V=1,5m/s[/tex]
Now we have all the information to find the Centripetal Force:
[tex]Fcp=(m*V^2)/(r)[/tex]
[tex]Fcp=((0,0154kg)*(1,5m/s)^2)/(0,149m)[/tex]
a) So the Centripetal Force is: [tex]Fcp=0,23N[/tex]
b) If the tangencial speed is doubled, its new value will be 3m/s. replacing this information we will get the new Centripetal Force is:
[tex]Fcp=((0,0154kg)*(3m/s)^2)/(0,149m)[/tex]
The Centripetal Force is: [tex]Fcp=0,93N[/tex]
Here we can see that if the speed is doubled, the Centripetal Force will increase four times.
Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticable, but upon reaching shore, they can rise up to the astonishing height of a six-story building. One tsunami, generated off the Aleutian islands in Alaska, had a wavelength of 646 km and traveled a distance of 3410 km in 4.84 h. (a) What was the speed (in m/s) of the wave? For reference, the speed of a 747 jetliner is about 250 m/s. Find the wave's (b) frequency and (c) period.
Answer:
a) V = 195.70 m/s
b) f=3.02 × 10⁻⁴ Hz
c) T = 3311.25 seconds
Explanation:
Given:
Wavelength, λ = 646 Km = 646000 m
Distance traveled = 3410 Km = 3410000 m
Time = 4.84 h = 4.84 × 3600 s = 17424 seconds
a) The speed (V) of the wave is given as
V = distance / time
V = 3410000 m/ 17424 seconds
or
V = 195.70 m/s
b) The frequency (f) of the wave is given as:
f = V / λ
f= 195.70 / 646000
f=3.02 × 10⁻⁴ Hz
c) The time period (T) is given as:
T = 1/ f
T = 1/ (3.02 × 10⁻⁴) Hz
T = 3311.25 seconds
If given both the speed of light in a material and an incident angle. How can you find the refracted angle?
Answer:
[tex]r = Sin^{-1}\left ( \frac{v Sini}{c} \right )[/tex]
Explanation:
Let the speed of light in vacuum is c and the speed of light in medium is v. Let the angle of incidence is i.
By using the definition of refractive index
refractive index of the medium is given by
n = speed of light in vacuum / speed of light in medium
n = c / v ..... (1)
Use Snell's law
n = Sin i / Sin r
Where, r be the angle of refraction
From equation (1)
c / v = Sin i / Sin r
Sin r = v Sin i / c
[tex]r = Sin^{-1}\left ( \frac{v Sini}{c} \right )[/tex]
In a two-slit experiment, monochromatic coherent light of wavelength 500 nm passes through a pair of slits separated by 1.30 x 10^-5 m. At what angle away from the centerline does the first bright fringe occur? O 1.56° O 2.20° O 3.85° O 2.73° O 4.40°
Answer:
2.20°
Explanation:
λ = wavelength of the coherent light = 500 x 10⁻⁹ m
d = slits separation = 1.30 x 10⁻⁵ m
n = order of the fringe = 1
θ = angle made by the first bright fringe with the center line = ?
For first bright fringe, Using the equation
d Sinθ = n λ
(1.30 x 10⁻⁵) Sinθ = (1) (500 x 10⁻⁹)
[tex]Sin\theta =\frac{500\times 10^{-9}}{1.30\times 10^{-5}}[/tex]
Sinθ = 0.0385
θ = 2.20°
Final answer:
To find the angle of the first bright fringe in a double-slit experiment, the formula for constructive interference is used. By substituting the provided measurements into the formula and calculating the inverse sine, the angle is found to be approximately 2.20°.
Explanation:
The question refers to a Young's double-slit experiment, where monochromatic light of a known wavelength is used to produce an interference pattern on a screen. To find the angle at which the first bright fringe occurs, we can use the formula for constructive interference in a double-slit setup, which is given by:
d sin(θ) = mλ
Where d is the separation between the slits, θ is the angle of the fringe from the centerline, m is the order number of the fringe (m=1 for the first bright fringe), and λ is the wavelength of the light.
Plugging in the values provided:
d = 1.30 x 10-5 m
λ = 500 x 10-9 m
m = 1
Now we solve for θ:
sin(θ) = mλ / d
sin(θ) = (1)(500 x 10-9 m) / (1.30 x 10-5 m)
sin(θ) ≈ 0.038462
Using a calculator to find the inverse sine, we get:
θ ≈ 2.20°
Certain planes of a crystal of halite have a spacing of 0.399 nm. The crystal is irradiated by a beam of x-rays. First order constructive interference occurs when the beam makes an angle of 20° with the planes. What is the wavelength of the x-rays?
Answer:
The wavelength of x-ray is 0.272 nm.
Explanation:
Inter planer spacing, d = 0.399 nm = 3.99 × 10⁻¹⁰ m
First order constructive interference occurs when the beam makes an angle of 20° with the planes. We need to find the wavelength of the x-rays. The condition for constructive interference is given by :
[tex]n\lambda=2d\ sin\theta[/tex]
Here, n = 1
[tex]\lambda=2\times 3.99\times 10^{-10} m\ sin(20)[/tex]
[tex]\lambda=2.72\times 10^{-10}\ m[/tex]
[tex]\lambda=0.272\ nm[/tex]
So, the wavelength of the x-rays is 0.272 nm. Hence, this is the required solution.
If the specific gravity of copper is 8.91, it weighs:
A. 55.6 lb/ft^3
B. 238.8 lb/ft^3
C. 23.88 lb/ft^3
D. 556 lb/ft^3
Answer:
Option D is the correct answer.
Explanation:
The specific gravity of copper is 8.91.
We have
[tex]\frac{\texttt{Density of copper}}{\texttt{Density of water}}=8.91\\\\\texttt{Density of copper}=8.91\times 1000kg/m^3=8910kg/m^3[/tex]
We also have
1 kg = 2.205 lb
1 m = 3.28 ft
Substituting
[tex]\texttt{Density of copper}=8910kg/m^3=\frac{8910\times 2.205}{3.28^3}=556.76lb/ft^3[/tex]
Option D is the correct answer.
Answer:
Option D is the correct answer.
A 3.8 L volume of neon gas (Ne) is at a pressure of 6.8 atm and a temperature of 470 K. The atomic mass of neon is 20.2 g/mol, and the ideal gas constant is R=8.314 J/mol*K. The mass of neon is closest to what?
Answer:
The mass of neon is 13.534 g.
Explanation:
Given that,
Volume = 3.8 L
Pressure = 6.8 atm
Temperature = 470 K
Atomic mass of neon =20.2 g/mol
Gas constant R = 8.314 = 0.082057 L atm/mol K
We need to calculate the mass of neon
Using equation of gas
[tex]PV=nRT[/tex]
[tex]6.8\times3.8=n\times0.082057\times470[/tex]
[tex]n=\dfrac{6.8\times3.8}{0.082057\times470}[/tex]
[tex]n= 0.670[/tex]
We know that,
[tex]n = \dfrac{m}{molar\ mass}[/tex]
[tex]m = n\times molar\ mass[/tex]
[tex]m=0.670\times20.2[/tex]
[tex]m=13.534\ g[/tex]
Hence, The mass of neon is 13.534 g.
Constants Part A Two small charged spheres are 9.20 cm apart. They are moved, and the force on each of them is found to have been tripled How far apart are they now? Express your answer to three significant figures and include the appropriate units.
Answer:
5.3 cm
Explanation:
Let the charges on the spheres be q and Q.
r = 9.20 cm
F = k Q q / (9.20)^2 ..... (1)
Let the new distance be r' and force F'
F' = 3 F
F' = k Q q / r'^2
3 F = k Q q / r'^2 ...... (2)
Divide equation (1) by (2)
F / 3 F = r'^2 / 84.64
1 / 3 = r'^2 / 84.64
r' = 5.3 cm
A ladder that is 4.6 m long is leaning against a wall at an angle of 66.0º with respect to the ground. If the base of the ladder is moved 0.31 m away from the wall, how far will the top of the ladder go down?
Answer:
The top of the ladder go down by 0.15 m.
Explanation:
Here we have right angled triangle.
Hypotenuse = 4.6 m
Bottom angle = 66º
Length from ladder bottom to wall bottom = 4.6 cos66 = 1.87 m
Length from ladder top to wall bottom = 4.6 sin66 = 4.20 m
New length from ladder bottom to wall bottom = 1.87 + 0.31 = 2.18 m
By Pythagoras theorem
New length from ladder top to wall bottom is given by
[tex]\sqrt{4.6^2-2.18^2}=4.05m[/tex]
Distance the top of the ladder go down = 4.20 - 4.05 = 0.15 m
The _____ is used to express absolute temperatures in the English system of measurement. A. Fahrenheit scale B. Kelvin scale C. Celsius scale D. Rankine scale
Answer:
B). Kelvin Scale
Explanation:
Kelvin scale is the absolute scale which is used to express temperature in English System
We have different temperature scales
1) Fahrenheit Scale
Generally use in British system of units
2) Celcius Scale
It is used to given temperature of different scales and its relation with kelvin
[tex]^0 C = K - 273[/tex]
3) Rankine Scale
It is used in thermodynamic scales with large temperature range