Answer:
47.17 degree C
Explanation:
mg = 3.5 kg, T1 = 94 degree C, sg = 129 J/kg C
mw = 0.2 kg, T2 = 22 degree C, sw = 4200 J/kg C
Let T be the temperature at equilibrium.
Heat given by the gold = Heat taken by water
mg x sg x (T1 - T) = mw x sw x (T - T2)
3.5 x 129 x (94 - T) = 0.2 x 4200 x (T - 22)
42441 - 451.5 T = 840 T - 18480
60921 = 1291.5 T
T = 47.17 degree C
A tree is 257 ft high. To the nearest tenth of a meter, how tall is it in meters? There are 3.28 ft in 1 m.
Answer:
Height of tree = 78.35 meters.
Explanation:
We have
1 meter = 3.28 feet
That is
[tex]1 ft = \frac{1}{3.28}=0.3048m[/tex]
Here height of tree = 257 ft
Height of tree = 257 x 0.3048 = 78.35 m
Height of tree = 78.35 meters.
A solenoid 81.0 cm long has a radius of 1.70 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.
Answer:
The magnitude of the magnetic field inside the solenoid is [tex]7.3\times10^{-3}\ T[/tex].
Explanation:
Given that,
Length = 81.0 cm
Radius = 1.70 cm
Number of turns = 1300
Current = 3.60 A
We need to calculate the magnetic field
Using formula of magnetic field inside the solenoid
[tex]B =\mu nI[/tex]
[tex]B =\mu\dfrac{N}{l}I[/tex]
Where, [tex]\dfrac{N}{l}[/tex]=Number of turns per unit length
I = current
B = magnetic field
Put the value into the formula
[tex]B =4\pi\times10^{-7}\times\dfrac{1300}{81.0\times10^{-2}}\times3.60[/tex]
[tex]B = 7.3\times10^{-3}\ T[/tex]
Hence, The magnitude of the magnetic field inside the solenoid is [tex]7.3\times10^{-3}\ T[/tex].
The slotted arm revolves in the horizontal plane about the fixed vertical axis through point O. The 2.2-lb slider C is drawn toward O at the constant rate of 3.6 in./sec by pulling the cord S. At the instant for which r = 7.5 in., the arm has a counterclockwise angular velocity ω = 6.3 rad/sec and is slowing down at the rate of 2.1 rad/sec 2. For this instant, determine the tension T in the cord and the force N exerted on the slider by the sides of the smooth radial slot. The force N is positive if side A contacts the slider, negative if side B contacts the slider.
Answer:
T = 2.5 lb
N= -0.33 lb
Explanation:
given
r = 9 in
[tex]\dot{r} =-3.6 in/s and\ \ddot{r} = 0[/tex]
[tex]\dot{\theta} = 6.3\ rad/s and\ \ddot{\theta} = 2.1\ rad/s^2[/tex]
[tex]-T = m a_r = m(\ddot{r} -r{\dot{\theta}^2)[/tex]
[tex]N= m a_{\theta} = m(r\ddot{\theta}+2\dot{r}\dot{\theta}})[/tex]
[tex]T= mr{\dot{\theta}^2 = \frac{3}{386.4}(9)(6)^2 =2.5lb[/tex]
[tex]N= m(r\ddot{\theta}+2\dot{r}\dot{\theta}})=\frac{3}{386.4}[9(-2)+2(-2)(6)]=-0.326 lb[/tex]
A gas sample has a volume of 0.225 L with an unknown temperature. The same gas has a volume of 0.180 L when the temperature is 35 ∘C, with no change in the pressure or amount of gas. Part A What was the initial temperature, in degrees Celsius, of the gas?
Answer:
The initial temperature of the gas was of T1= 112ºC .
Explanation:
T1= ?
T2= 35 ºC = 308.15 K
V1= 0.225 L
V2= 0.18 L
T2* V1 / V2 = T1
T1= 385.18 K = 112ºC
In the presence of a dielectric, the capacitance of a electric field inside the plates now is: a) Less b) More c) Same as the electric field in absence of the dielectric d) Zero
Answer:
Explanation:
As the dielectric is inserted between the plates of a capacitor, the capacitance becomes K times and the electric field between the plates becomes 1 / K times the original value. Where, K be the dielectric constant.
A record of travel along a straight path is as follows:
1. Start from rest with constant acceleration of 2.04 m/s2 for 11.0 s.
2. Maintain a constant velocity for the next 2.85 min.
3. Apply a constant negative acceleration of −9.73 m/s2 for 2.31 s.
(a) What was the total displacement for the trip?
(b) What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip?
(C)COMPLETE TRIP:
Answer:
a) Total displacement = 3986.54 m
b) Average speeds
Leg 1 -> 11.22 m/s
Leg 2 -> 22.44 m/s
Leg 3 -> 11.20 m/s
Complete trip -> 21.63 m/s
Explanation:
a) Leg 1:
Initial velocity, u = 0 m/s
Acceleration , a = 2.04 m/s²
Time, t = 11 s
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
s = 0 x 11 + 0.5 x 2.04 x 11²
s = 123.42 m
Leg 2:
We have equation of motion v = u + at
Initial velocity, u = 0 m/s
Acceleration , a = 2.04 m/s²
Time, t = 11 s
Substituting
v = 0 + 2.04 x 11 = 22.44 m/s
We have equation of motion s= ut + 0.5 at²
Initial velocity, u = 22.44 m/s
Acceleration , a = 0 m/s²
Time, t = 2.85 min = 171 s
Substituting
s= ut + 0.5 at²
s = 22.44 x 171 + 0.5 x 0 x 171²
s = 3837.24 m
a) Leg 3:
Initial velocity, u = 22.44 m/s
Acceleration , a = -9.73 m/s²
Time, t = 2.31 s
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
s = 22.44 x 2.31 + 0.5 x -9.73 x 2.31²
s = 25.88 m
Total displacement = 123.42 + 3837.24 + 25.88 = 3986.54 m
Average speed is the ratio of distance to time.
b) Leg 1:
[tex]v_{avg}=\frac{123.42}{11}=11.22m/s[/tex]
Leg 2:
[tex]v_{avg}=\frac{3837.24}{171}=22.44m/s[/tex]
Leg 3:
[tex]v_{avg}=\frac{25.88}{2.31}=11.20m/s[/tex]
Complete trip:
[tex]v_{avg}=\frac{3986.54}{11+171+2.31}=21.63m/s[/tex]
Ignoring the mass of the spring, a 5 kg mass hanging from a coiled spring having a constant k= 50 N/m will have a period of oscillation of about: (A) 10 sec., (B) 5 sec., (C) 2 sec., (D) 0.1 secC., (E) 1 min.
Answer:
Period of oscillation, T = 2 sec
Explanation:
It is given that,
Mass of the object, m = 5 kg
Spring constant of the spring, k = 50 N/m
This object is hanging from a coiled spring. We need to find the period of oscillation of the spring. The time period of oscillation of the spring is given by :
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
[tex]T=2\pi\sqrt{\dfrac{5\ kg}{50\ N/m}}[/tex]
T = 1.98 sec
or
T = 2 sec
So, the period of oscillation is about 2 seconds. Hence, this is the required solution.
A robot car drives off a cliff that is 11 meters above the water below. The car leaves the cliff horizontally with some initial speed and travels down to the water. The car hits the water a distance of 15 meters from the base of the cliff (it has no air resistance as it falls). What was the driving speed of the car? g =< 0,-9.8,0> N/kg.
Answer:
Driving speed of the car = 10 m/s
Explanation:
The car hits the water a distance of 15 meters from the base of the cliff.
Horizontal displacement = 11 m
A robot car drives off a cliff that is 11 meters above the water below.
Vertical displacement = 11 m
We have
s = ut + 0.5 at²
11 = 0 x t + 0.5 x 9.81 x t²
t = 1.50 s
So the car moves 15 meters in 1.50 seconds.
Velocity
[tex]v=\frac{15}{1.5}=10m/s[/tex]
Driving speed of the car = 10 m/s
What is the common trade name of the polymer polytetrafluoroethylene? For what is it commonly used?
Answer:
Common trade name of polytetrafluoroethylene is Teflon
Many uses are there some of them are given in explanation.
Explanation:
Common trade name of polytetrafluoroethylene is Teflon.
Main uses of polytetrafluoroethylene:
1) To coat non stick pans
2) Used as ski bindings
3) Used as fabric protector to repel stains on formal school-wear
4) Used to make to make a waterproof, breathable fabric in outdoor apparel.
An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.99 cm in a uniform magnetic field with B = 1.43 T. Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.
Answer:
a) [tex]4.1\times 10^{6} \frac{m}{s}[/tex]
b) [tex]9.2\times 10^{-8} s[/tex]
c) [tex]5.6\times 10^{-14} J[/tex]
d) 175000 volts
Explanation:
a)
[tex]q[/tex] = magnitude of charge on the alpha particle = 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C
[tex]m[/tex] = mass of alpha particle = 4 x 1.67 x 10⁻²⁷ kg = 6.68 x 10⁻²⁷ kg
[tex]r[/tex] = radius of circular path = 5.99 cm = 0.0599 m
[tex]B[/tex] = magnitude of magnetic field = 1.43 T
[tex]v[/tex] = speed of the particle
Radius of circular path is given as
[tex]r = \frac{mv}{qB}[/tex]
[tex]0.0599 = \frac{(6.68\times 10^{-27})v}{(3.2\times 10^{-19})(1.43)}[/tex]
[tex]v = 4.1\times 10^{6} \frac{m}{s}[/tex]
b)
Time period is given as
[tex]T = \frac{2\pi m}{qB}[/tex]
[tex]T = \frac{2(3.14)(6.68\times 10^{-27})}{(3.2\times 10^{-19})(1.43)}[/tex]
[tex]T = 9.2\times 10^{-8} s[/tex]
c)
Kinetic energy is given as
[tex]K = (0.5)mv^{2}[/tex]
[tex]K = (0.5)(6.68\times 10^{-27})(4.1\times 10^{6})^{2}[/tex]
[tex]K = 5.6\times 10^{-14} J[/tex]
d)
ΔV = potential difference
Using conservation of energy
q ΔV = K
(3.2 x 10⁻¹⁹) ΔV = 5.6 x 10⁻¹⁴
ΔV = 175000 volts
how large can the kinetic energy of an electron be that is localized within a distance (change in) x = .1 nmapproximately the diameter of a hydrogen atom (ev)
Answer:
The kinetic energy of an electron is [tex]1.54\times10^{-15}\ J[/tex]
Explanation:
Given that,
Distance = 0.1 nm
We need to calculate the momentum
Using uncertainty principle
[tex]\Delta x\Delta p\geq\dfrac{h}{4\pi}[/tex]
[tex]\Delta p\geq\dfrac{h}{\Delta x\times 4\pi}[/tex]
Where, [tex]\Delta p[/tex] = change in momentum
[tex]\Delta x[/tex] = change in position
Put the value into the formula
[tex]\Delta p=\dfrac{6.6\times10^{-34}}{4\pi\times10^{-10}}[/tex]
[tex]\Delta p=5.3\times10^{-23}[/tex]
We need to calculate the kinetic energy for an electron
[tex]K.E=\dfrac{p^2}{2m}[/tex]
Where, P = momentum
m = mass of electron
Put the value into the formula
[tex]K.E=\dfrac{(5.3\times10^{-23})^2}{2\times9.1\times10^{-31}}[/tex]
[tex]K.E=1.54\times10^{-15}\ J[/tex]
Hence, The kinetic energy of an electron is [tex]1.54\times10^{-15}\ J[/tex]
A vector has components Ax = 52.0 m and Ay = 41.0 m. Find: (a) The length of the vector A.
(b) The angle it makes with the x-axis (in degrees).
Answer:
Part a)
A = 66.2 m
Part b)
Angle = 38.35 degree
Explanation:
Part a)
Length of the vector is the magnitude of the vector
here we know that
[tex]A_x = 52.0 m[/tex]
[tex]A_y = 41.0 m[/tex]
now we have
[tex]A = \sqrt{A_x^2 + A_y^2}[/tex]
[tex]A = \sqrt{52^2 + 41^2}[/tex]
[tex]A = 66.2 m[/tex]
Part b)
Angle made by the vector is given as
[tex]tan\theta = \frac{A_y}{A_x}[/tex]
[tex]tan\theta = \frac{41}{52}[/tex]
[tex]\theta = 38.25 degree[/tex]
A 1.2-kg ball drops vertically onto the floor, hitting with a speed of 25 m/s. Consider the impulse during this collision. Would the magnitude of the impulse be greater: (i) if the ball rebounded with a speed of 10 m/s (the ball was made of rubber), or (ii) if the ball stuck to the floor (the ball was made of clay)? Support your answer with a calculation.
Answer:
3kg
Explanation:
impulse = MV
then
m1v1=m2v2
when the values are subtitude
then
m2=1.2*25/10
m2=30kg//
What is the hydrostatic pressure at 20,000 leagues under the sea? (a league is the distance a person can walk in one hour) ?) 40 kPa b) 100 ps? c) 1300 Pad) 2000 psi e) none of these answers
Answer:
alternative E- none of these answers
Explanation:
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at a given point within the fluid, due to the force of gravity. Hydrostatic pressure increases in proportion to depth measured from the surface because of the increasing weight of fluid exerting downward force from above.
The formula is :
P= d x g x h
p: hydrostatic pressure (N/m²)
d: density (kg/m³) density of seawater is 1,030 kg/m³
g: gravity (m/s²) ≅ 9.8m/s²
h: height (m)
The hydrostatic pressure at 20,000 leagues under the sea is approximately 1,002,500,000,000 Pa.
Explanation:The hydrostatic pressure at 20,000 leagues under the sea can be calculated using the equation for pressure in a fluid, which is given by P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth below the surface.
Since a league is the distance a person can walk in one hour, we need to convert it to meters. Assuming an average walking speed of 5 km/h, a league is equal to 5 km. Therefore, 20,000 leagues is equal to 100,000 km.
The pressure at this depth can be calculated using the known values: density of seawater is about 1025 kg/m³ and acceleration due to gravity is 9.8 m/s². Plugging in these values, we get P = (1025 kg/m³)(9.8 m/s²)(100,000,000 m) = 1,002,500,000,000 Pa.
Therefore, the correct answer is none of the provided options. The hydrostatic pressure at 20,000 leagues under the sea is approximately 1,002,500,000,000 Pa.
With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons of fuel in its tank? The slope is 5.7
Answer: 2561.7 pounds
Explanation:
If we assume the total weight of an airplane (in pounds units) as a linear function of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:
[tex]m=\frac{Y-Y_{1}}{X-X_{1}}[/tex] (1)
Where:
[tex]m=5.7[/tex] is the slope of the line
[tex]Y_{1}=2390.7pounds[/tex] is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)
[tex]X_{1}=51gallons[/tex] is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)
This means we already have one point of the graph, which coordinate is:
[tex](X_{1},Y_{1})=(51,2390.7)[/tex]
Rewritting (1):
[tex]Y=m(X-X_{1})+Y_{1}[/tex] (2)
As Y is a function of X:
[tex]Y=f_{(X)}=m(X-X_{1})+Y_{1}[/tex] (3)
Substituting the known values:
[tex]f_{(X)}=5.7(X-51)+2390.7[/tex] (4)
[tex]f_{(X)}=5.7X-290.7+2390.7[/tex] (5)
[tex]f_{(X)}=5.7X+2100[/tex] (6)
Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):
[tex]f_{(81)}=5.7(81)+2100[/tex] (7)
[tex]f_{(81)}=2561.7[/tex] (8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.
Final answer:
To find the weight of the airplane with 81 gallons of fuel, calculate the additional fuel weight (30 gallons
* 5.7 pounds/gallon = 171 pounds) and add it to the initial weight (2390.7 pounds + 171 pounds = 2561.7 pounds).
Explanation:
The question asks to calculate the weight of an airplane with a different amount of fuel in its tank, given the weight with a specific amount and the slope of weight increase per gallon of fuel added. To find the new weight, we first calculate the weight increase due to the additional fuel, then add this increase to the original weight of the airplane.
Initial weight with 51 gallons: 2390.7 pounds
Fuel increase: 81 gallons - 51 gallons = 30 gallons
Slope (rate of weight increase): 5.7 pounds per gallon
Additional weight from extra fuel: 30 gallons
* 5.7 pounds/gallon = 171 pounds
New weight with 81 gallons: 2390.7 pounds + 171 pounds = 2561.7 pounds
A pair of eyeglass frames are made of an epoxy plastic (coefficient of linear expansion = 1.30 ✕ 10−4°C−1). At room temperature (20.0°C), the frames have circular lens holes 2.34 cm in radius. To what temperature must the frames be heated if lenses 2.35 cm in radius are to be inserted into them? °C
Answer:
Final temperature = 52.44 °C
Explanation:
We have equation for thermal expansion
ΔL = LαΔT
We have change in length = Circumference of 2.35 cm radius - Circumference of 2.34 cm radius = 2π x 2.35 - 2π x 2.34 = 0.062 cm
Length of eyeglass frame = 2π x 2.34 = 14.70 cm
Coefficient of linear expansion, α = 1.30 x 10⁻⁴ °C⁻¹
Substituting
0.062 = 14.70 x 1.30 x 10⁻⁴ x ΔT
ΔT = 32.44°C
Final temperature = 32.44 + 20 = 52.44 °C
To fit lenses of 2.35 cm radius into eyeglass frames with lens holes of 2.34 cm radius at room temperature, the frames made of an epoxy plastic should be heated to about 52°C. This fact is obtained using the physics concept of thermal expansion.
Explanation:The question relates to the concept of thermal expansion typically studied in physics. The change in radius due to thermal expansion in a one-dimensional system like the eyeglass frames can be given by the formula Δr = αr(ΔT), where Δr is the change in radius, α is the coefficient of expansion, r is the initial radius, and ΔT is the change in temperature. Upon heating, the frames will expand and their lens holes will become larger. Here, we are trying to determine the temperature needed to increase the hole radius from 2.34 cm to 2.35 cm. Using the above formula:
0.01 cm = 1.30 × 10−4°C−1 * 2.34 cm * ΔT
Solving for ΔT (the change in temperature), we get ΔT = about 32°C. Thus, the frames need to be heated to about 32°C above room temperature, i.e., 20°C + 32°C = 52°C.
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What are the basic primitive solids?
Answer:
A primitive solid is a 'building block' that you can use to work with in 3D. Rather than extruding or revolving an object, AutoCAD has some basic 3D shape commands at your disposal.
Explanation:
The motion of a particle is defined by the relation x = t3 – 12t2 +36t +32, where x and t are expressed in feet and seconds, respectively. Determine the time, position, and acceleration of the particle when v = 0 ft/s.
1) Time: 2 s, 6 s
The position of the particle is given by:
[tex]x=t^3 -12t^2 +36t+32[/tex]
where t is the time in seconds and x is the position in feet.
The velocity of the particle can be found by differentiating the position:
[tex]v(t)=x'(t)=3t^2 -24t+36[/tex]
and it is expressed in ft/s.
In order to find the time at which the velocity is v=0 ft/s, we substitute v=0 into the previous equation:
[tex]0=3t^2-24t+36\\0=t^2 -8t+12\\0=(t-2)(t-6)[/tex]
So the two solutions are
t = 2 s
t = 6 s
2) Position: x = 64 ft and x = 32 ft
The position at which the velocity of the particle is v = 0 can be found by susbtituting t = 2 and t = 6 into the equation for the position.
For t = 2 s, we have
[tex]x=(2)^3-12(2)^2 +36(2)+32=64[/tex]
For t = 6 s, we have
[tex]x=(6)^3-12(6)^2 +36(6)+32=32[/tex]
So the two positions are
x = 64 ft
x = 32 ft
3) Acceleration: [tex]-12 ft/s^2[/tex] and [tex]+12 ft/s^2[/tex]
The acceleration of the particle can be found by differentiating the velocity. We find:
[tex]a(t)=v'(t)=6t-24[/tex]
And substituting t = 2 and t = 6, we find the acceleration when the velocity of the particle is zero:
[tex]a(2)=6(2)-24=-12[/tex]
[tex]a(6)=6(6)-24=12[/tex]
So the two accelerations are
[tex]a=-12 ft/s^2[/tex]
[tex]a=12 ft/s^2[/tex]
The time, position, and acceleration of the particle when v = 0 ft/s are t = 2 s and t = 6 s, x = 24 ft and x = 184 ft, and a = -12 ft/s² and a = 12 ft/s².
Explanation:To find the time, position, and acceleration of the particle when the velocity is 0 ft/s, we need to determine the values of t, x, and a when v = 0.
Given the relation x = t³ - 12t² + 36t + 32, we need to solve for t when v = 0. We can use the equation v = dx/dt to find the velocity function and set it equal to 0.
By differentiating x with respect to t, we get v = 3t² - 24t + 36. Setting v = 0, we can solve the quadratic equation 3t² - 24t + 36 = 0 to find the values of t. The solutions are t = 2 and t = 6.
Therefore, when v = 0, the time is t = 2 s and t = 6 s. We can substitute these values into the position function to find the corresponding positions. When t = 2 s, x = (2)³ - 12(2)² + 36(2) + 32 = 24 ft. When t = 6 s, x = (6)³ - 12(6)² + 36(6) + 32 = 184 ft.
To find the acceleration, we can differentiate the velocity function with respect to t. By differentiating v = 3t² - 24t + 36, we get a = 6t - 24. Substituting t = 2 s and t = 6 s into this equation, we get a = 6(2) - 24 = -12 ft/s² and a = 6(6) - 24 = 12 ft/s².
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A tennis ball bounces on the floor three times. If each time it loses 11% of its energy due to heating, how high does it rise after the third bounce, provided we released it 4.4 m from the floor?
Answer:
h = 3.10 m
Explanation:
As we know that after each bounce it will lose its 11% of energy
So remaining energy after each bounce is 89%
so let say its initial energy is E
so after first bounce the energy is
[tex]E_1 = 0.89 E[/tex]
after 2nd bounce the energy is
[tex]E_2 = 0.89(0.89 E)[/tex]
After third bounce the energy is
[tex]E_3 = (0.89)(0.89)(0.89)E[/tex]
here initial energy is given as
[tex]E = mgH_o[/tex]
now let say final height is "h" so after third bounce the energy is given as
[tex]E_3 = mgh[/tex]
now from above equation we have
[tex]mgh = (0.89)(0.89)(0.89)(mgH)[/tex]
[tex]h = 0.705H[/tex]
[tex]h = 0.705(4.4 m)[/tex]
[tex]h = 3.10 m[/tex]
What is the angular momentum of a 3-kg uni- form cylindrical grinding wheel of radius 0.2 m when rotating at 1500 rpm (Rotational Inertia of a cylinder is mR^2/2).
Answer:
[tex]L = 9.42 kg m^2/s[/tex]
Explanation:
Angular speed of the cylinder is given as
[tex]f = 1500 rpm[/tex]
[tex]f = 1500 round/60 s[/tex]
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(25) = 50 \pi[/tex]
now moment of inertia of the cylinder is given as
[tex]I = \frac{1}{2}mR^2[/tex]
[tex]I = \frac{1}{2}(3)(0.2)^2[/tex]
[tex]I = 0.06 kg m^2[/tex]
now we have
[tex]L = I\omega[/tex]
[tex]L = (0.06)(50\pi)[/tex]
[tex]L = 9.42 kg m^2/s[/tex]
A cylinder is being flattened so that its volume does not change. Find the rate of change of radius when r = 2 inches and h = 5 inches, if the height is decreasing at 0.7 in/sec. Hint: what is the rate of change of volume?
Answer:
[tex]\frac{dr}{dt} = 0.14 in/s[/tex]
Explanation:
As the volume of the cylinder is constant here so we can say that its rate of change in volume must be zero
so here we can say
[tex]\frac{dV}{dt} = 0[/tex]
now we have
[tex]V = \pi r^2 h[/tex]
now find its rate of change in volume with respect to time
[tex]\frac{dV}{dt} = 2\pi rh\frac{dr}{dt} + \pi r^2\frac{dh}{dt}[/tex]
now we know that
[tex]\frac{dV}{dt} = 0 = \pi r(2h \frac{dr}{dt} + r\frac{dh}{dt})[/tex]
given that
h = 5 inch
r = 2 inch
[tex]\frac{dh}{dt} = - 0.7 in/s[/tex]
now we have
[tex]0 = 2(5) \frac{dr}{dt} + 2(-0.7)[/tex]
[tex]\frac{dr}{dt} = 0.14 in/s[/tex]
Final answer:
The change in radius δr/δt of a cylinder with a constant volume is found to be 0.28 inches per second when the height is decreasing at 0.7 inches per second and r = 2 inches, h = 5 inches.
Explanation:
The question involves using calculus to find the rate of change of the radius of a cylinder given a constant volume and a known rate of change in height. The rate of change of volume for a cylinder, which is 0 because the volume doesn't change, can be described as δV/δt = (πr²) (δh/δt) + (2πrh) (δr/δt) = 0. Given the height is decreasing at 0.7 in/sec, we can find the rate of change of the radius δr/δt. Using the known values of r = 2 inches and h = 5 inches, we can solve for δr/δt.
Starting with the equation for the volume of a cylinder V = πr²h, since the volume is constant, we take the derivative with respect to time to obtain 0 = π(2×r×δr/δt×h + r²×δh/δt). Substituting the known values gives 0 = π(2×2×δr/δt×5 + 2²×(-0.7)), which simplifies to 0 = 20πδr/δt - 5.6π. From this we can solve for δr/δt = 5.6π / 20π = 0.28 in/sec.
The rate of change of the radius is 0.28 inches per second when the height is decreasing at 0.7 inches per second, and the radius is 2 inches while the height is 5 inches.
The position vector of a particle of mass 1.70 kg as a function of time is given by r with arrow = (6.00 î + 5.70 t ĵ), where r with arrow is in meters and t is in seconds. Determine the angular momentum of the particle about the origin as a function of time.
Final answer:
The angular momentum of a particle about the origin is given by the cross product of its position vector and linear momentum. In this case, the position vector of the particle is (6.00 î + 5.70 t ĵ) and the mass of the particle is 1.70 kg. To find the angular momentum, calculate the linear momentum and take the cross product with the position vector.
Explanation:
The angular momentum of a particle about the origin is given by the cross product of its position vector and linear momentum. In this case, the position vector of the particle is represented by r = (6.00 î + 5.70 t ĵ) and the mass of the particle is 1.70 kg. To find the angular momentum, we need to calculate the linear momentum first.
The linear momentum of a particle is given by the product of its mass and velocity. The velocity vector is given by the derivative of the position vector with respect to time, which in this case is v = (0 î + 5.70 ĵ) m/s. Substituting the values, we can find the linear momentum, which is p = (1.70 kg)(0 î + 5.70 ĵ) m/s.
To find the angular momentum, we take the cross product of the position vector and linear momentum:
ŕ × p = (6.00 î + 5.70 t ĵ) × (1.70 kg)(0 î + 5.70 ĵ) m/s = (0 î + 34.65 î t + 9.69 ĵ) kg·m²/s
Therefore, the angular momentum of the particle about the origin as a function of time is (0 î + 34.65 î t + 9.69 ĵ) kg·m²/s.
The angular momentum of the particle about the origin as a function of time is L(t) = 58.14 t k kg·m²/s. This result was obtained by calculating the cross product of the position and linear momentum vectors.
To determine the angular momentum of a particle about the origin as a function of time, we start by using the given position vector r(t) = (6.00 î + 5.70t ĵ) in meters.
The linear momentum p of the particle is given by p = m v, where m is the mass and v is the velocity. Since mass m = 1.70 kg, we first need the velocity. The velocity v(t) is the derivative of the position vector:
v(t) = d(r(t))/dt = 0 î + 5.70 ĵ = 5.70 ĵ m/s
Now, the linear momentum p(t) is:
p(t) = m v(t) = 1.70 kg * 5.70 ĵ m/s = 9.69 ĵ kg·m/s
The angular momentum L about the origin is given by L = r × p. Performing the cross product calculation:
r = 6.00 î + 5.70t ĵ
p = 9.69 ĵ
r × p = (6.00 î + 5.70t ĵ) × (9.69 ĵ)
Calculating the cross product, we get:
i-component: 0j-component: 0k-component: 6.00 * 9.69 - 0 = 58.14 t kThus, the angular momentum as a function of time is:
L(t) = 58.14 t k kg·m²/s
A top-fuel dragster starts from rest and has a constant acceleration of 42.0 m/s2. What are (a) the final velocity of the dragster at the end of 1.8 s, (b) the final velocity of the dragster at the end of of twice this time, or 3.6 s, (c) the displacement of the dragster at the end of 1.8 s, and (d) the displacement of the dragster at the end of twice this time, or 3.6 s
Answer:
a) Final velocity of the dragster at the end of 1.8 s = 75.6 m/s
b) Final velocity of the dragster at the end of 3.6 s = 151.2 m/s
c) The displacement of the dragster at the end of 1.8 s = 68.04 m
d) The displacement of the dragster at the end of 3.6 s = 272.16 m
Explanation:
a) We have equation of motion v = u + at
Initial velocity, u = 0 m/s
Acceleration , a = 42 m/s²
Time = 1.8 s
Substituting
v = u + at
v = 0 + 42 x 1.8 = 75.6 m/s
Final velocity of the dragster at the end of 1.8 s = 75.6 m/s
b) We have equation of motion v = u + at
Initial velocity, u = 0 m/s
Acceleration , a = 42 m/s²
Time = 3.6 s
Substituting
v = u + at
v = 0 + 42 x 3.6 = 75.6 m/s
Final velocity of the dragster at the end of 3.6 s = 151.2 m/s
c) We have equation of motion s= ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration , a = 42 m/s²
Time = 1.8 s
Substituting
s= ut + 0.5 at²
s = 0 x 1.8 + 0.5 x 42 x 1.8²
s = 68.04 m
The displacement of the dragster at the end of 1.8 s = 68.04 m
d) We have equation of motion s= ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration , a = 42 m/s²
Time = 3.6 s
Substituting
s= ut + 0.5 at²
s = 0 x 3.6 + 0.5 x 42 x 3.6²
s = 272.16 m
The displacement of the dragster at the end of 3.6 s = 272.16 m
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 10-5 m. (c = 3.00 × 108 m/s) (a) At what angle away from the central bright spot does the third bright fringe past the central bright spot occur? (b) At what angle does the second dark fringe occur?
Explanation:
It is given that,
Frequency of monochromatic light, [tex]f=5\times 10^{14}\ Hz[/tex]
Separation between slits, [tex]d=2.2\times 10^{-5}\ m[/tex]
(a) The condition for maxima is given by :
[tex]d\ sin\theta=n\lambda[/tex]
For third maxima,
[tex]\theta=sin^{-1}(\dfrac{n\lambda}{d})[/tex]
[tex]\theta=sin^{-1}(\dfrac{n\lambda}{d})[/tex]
[tex]\theta=sin^{-1}(\dfrac{nc}{fd})[/tex]
[tex]\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})[/tex]
[tex]\theta=4.69^{\circ}[/tex]
(b) For second dark fringe, n = 2
[tex]d\ sin\theta=(n+1/2)\lambda[/tex]
[tex]\theta=sin^{-1}(\dfrac{5\lambda}{2d})[/tex]
[tex]\theta=sin^{-1}(\dfrac{5c}{2df})[/tex]
[tex]\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})[/tex]
[tex]\theta=3.90^{\circ}[/tex]
Hence, this is the required solution.
(a) The angle of the third bright fringe (θ₃) is approximately 4.69 degrees (b) The angle of the second dark fringe (θ₂) is approximately 3.90 degrees.
To solve this problem, we can use the formula for the angles of the maxima and minima in a Double-Slit Experiment diffraction pattern. For the third bright fringe, the condition for constructive interference is given by:
[tex]\[ d \sin(\theta_m) = m \lambda \][/tex]
where:
- d is the slit separation,
- [tex]\( \theta_m \)[/tex] is the angle of the mth bright fringe,
- m is the order of the fringe
- λ is the wavelength of the light.
For the second dark fringe, the condition for destructive interference is given by:
[tex]\[ d \sin(\theta_n) = \left( n + \frac{1}{2} \right) \lambda \][/tex]
where:
- [tex]\( \theta_n \)[/tex] is the angle of the nth dark fringe,
- n is the order of the dark fringe
Given that the frequency f is related to the wavelength λ by the speed of light (c) as c = fλ, we can express λ in terms of f.
Part (a): Third Bright Fringe
[tex]\[ d \sin(\theta_m) = m \lambda \][/tex]
[tex]\[ \sin(\theta_3) = \frac{3 \lambda}{d} \][/tex]
[tex]\[ \sin(\theta_3) = \frac{3 c}{d f} \][/tex]
[tex]\[ \theta_3 = \sin^{-1}\left(\frac{3 c}{d f}\right) \][/tex]
Part (b): Second Dark Fringe
[tex]\[ d \sin(\theta_n) = \left( n + \frac{1}{2} \right) \lambda \][/tex]
[tex]\[ \sin(\theta_2) = \frac{(2 + 0.5) \lambda}{d} \][/tex]
[tex]\[ \sin(\theta_2) = \frac{2.5 c}{d f} \][/tex]
[tex]\[ \theta_2 = \sin^{-1}\left(\frac{2.5 c}{d f}\right) \][/tex]
Now, plug in the values:
Given:
- [tex]\( d = 2.20 \times 10^{-5} \)[/tex] m,
- [tex]\( f = 5.00 \times 10^{14} \)[/tex] Hz,
- [tex]\( c = 3.00 \times 10^8 \)[/tex] m/s,
- m = 3,
- n = 2.
[tex]\[ \theta_3 = \sin^{-1}\left(\frac{3 \times 3.00 \times 10^8}{2.20 \times 10^{-5} \times 5.00 \times 10^{14}}\right) \][/tex]
[tex]\[ \theta_3 \approx 4.69^\circ \][/tex] (rounded to two decimal places, as given)
[tex]\[ \theta_2 = \sin^{-1}\left(\frac{2.5 \times 3.00 \times 10^8}{2.20 \times 10^{-5} \times 5.00 \times 10^{14}}\right) \][/tex]
[tex]\[ \theta_2 \approx 3.90^\circ \][/tex] (rounded to two decimal places, as given)
So, the correct answers are indeed [tex]\( \theta_3 \approx 4.69^\circ \)[/tex] for part (a) and [tex]\( \theta_2 \approx 3.90^\circ \)[/tex] for part (b).
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A 0.5 kg air-hockey puck is initially at rest. What will its kinetic energy be after a net force of 0.4 N acts on it for a distance of 0.7 m?
Answer:
0.28 J
Explanation:
Since the air-hockey puck was initially at rest
KE₀ = initial kinetic energy of the air-hockey puck = 0 J
KE = final kinetic energy of the air-hockey puck
m = mass of air-hockey puck 0.5 kg
F = net force = 0.4 N
d = distance moved = 0.7 m
Using work-change in kinetic energy
F d = (KE - KE₀)
(0.4) (0.7) = KE - 0
KE = 0.28 J
A burnt paper on the road has a picture, which shows a speed boat runs fast on the lake and produces V-like water waves. This remind you of Moessbauer Effect and Cherenkov Radiation. What are these?
Answer:
Moessbauer Effect = eggy eggs
Explanation:
A proton moves through a region of space where there is a magnetic field B⃗ =(0.64i+0.40j)T and an electric field E⃗ =(3.3i−4.5j)×103V/m. At a given instant, the proton's velocity is v⃗ =(6.6i+2.8j−4.8k)×103m/s.
Determine the components of the total force on the proton.
Express your answers using two significant figures. Enter your answers numerically separated by commas.
Answer:
[tex]F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k[/tex]
Explanation:
When a charge is moving in constant magnetic field and electric field both then the net force on moving charge is vector sum of force due to magnetic field and electric field both
so first the force on the moving charge due to electric field is given by
[tex]\vec F_e = q\vec E[/tex]
[tex]\vec F_e = (1.6 \times 10^{-19})(3.3 \hat i - 4.5 \hat j) \times 10^3[/tex]
[tex]\vec F_e = (5.28 \times 10^{-16}) \hat i - (7.2 \times 10^{-16}) \hat j[/tex]
Now force on moving charge due to magnetic field is given as
[tex]\vec F_b = q(\vec v \times \vec B)[/tex]
[tex]\vec F_b = (1.6 \times 10^{-19})((6.6 \hat i+2.8 \hat j−4.8 \hat k) \times 10^3 \times (0.64 \hat i + 0.40 \hat j) )[/tex]
[tex]\vec F_b = (4.22 \times 10^{-16})\hat k - (2.87 \times 10^{-16})\hat k - (4.92 \times 10^{-16})\hat j + (3.07 \times 10^{-16}) \hat i[/tex]
[tex]\vec F_b = (3.07\times 10^{-16})\hat i - (4.92 \times 10^{-16})\hat j + (1.35 \times 10^{-16})\hat k[/tex]
Now net force due to both
[tex]F = F_e + F_b[/tex]
[tex]F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k[/tex]
An electric and magnetic field exerts force on a proton moving with velocity in that field. The total force can be calculated from the Lorentz Force equation, which requires knowledge of the charge of the proton, its velocity, and the electric and magnetic fields it is experiencing.
Explanation:In Physics, the total force acting on a charged particle moving through an electric field E and a magnetic field B is given by the Lorentz Force equation: F = q(E + v × B), where q is the charge of the particle, and v is its velocity.
By given that the proton's charge is q = 1.6×10^-19 C and proton's velocity v = (6.6i+2.8j-4.8k)x10^3 m/s, electric field E = (3.3i-4.5j)x10^3 V/m, and magnetic field B = (0.64i+0.40j)T, we can plug these values into the equation.
To find the cross product of v and B, we use the determinant of a 3x3 matrix. The value for the F vector can be calculated as follows: F = q [E + v × B] and the cross product 'v × B' is calculated as the determinant of a 3x3 matrix. This yields force values that are expected to be in i, j, and k components. These calculations need to be done carefully to ensure accuracy, but are straightforward with the use of any standard physics formula sheet.
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Two ships leave a harbor at the same time, traveling on courses that have an angle of 110∘ between them. If the first ship travels at 22 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 1.2 hours?
Answer:
49.07 miles
Explanation:
Angle between two ships = 110° = θ
First ship speed = 22 mph
Second ship speed = 34 mph
Distance covered by first ship after 1.2 hours = 22×1.2 = 26.4 miles = b
Distance covered by second ship after 1.2 hours = 34×1.2 = 40.8 miles = c
Here the angle between the two sides of a triangle is 110° so from the law of cosines we get
a² = b²+c²-2bc cosθ
⇒a² = 26.4²+40.8²-2×26.4×40.8 cos110
⇒a² = 2408.4
⇒a = 49.07 miles
Choose the statement(s) that is/are true about the ratio \frac{C_p}{C_v} C p C v for a gas? (Ii) This ratio is the same for all gases. (ii) This ratio has a value 1.67 for a monatomic gas (iii) This ratio has an approximate value of 1.4 for diatomic gases. (iii) This ratio has a value 8.314\:J.mol^{-1}.K^{-1}
Answer:
(i) false
(ii) true
(iii) true
(iv) false
Explanation:
(i) The ratio of Cp and Cv is not constant for all the gases. It is because the value of cp and Cv is different for monoatomic, diatomic and polyatomic gases.
So, this is false.
(ii) For monoatomic gas
Cp = 5R/2, Cv = 3R/2
So, thier ratio
Cp / Cv = 5 / 3 = 1.67
This statement is true.
(iii) for diatomic gases
Cp = 7R/2, Cv = 5R/2
Cp / Cv = 7 / 5 = 1.4
This statement is true.
(iv) It is false.
A 12-V battery maintains an electric potential difference between two parallel metal plates separated by 10 cm. What is the electric field between the plates? a. 1.2 V/m b. 12 V/m c. 120 V/m d. zero
Answer:
The electric field between the plates is 120 V/m.
(c) is correct option.
Explanation:
Given that,
Potential difference = 12 volt
Distance = 10 cm = 0.1 m
We need to calculate the electric field between the plates
Using formula of electric field
[tex]E = \dfrac{V}{d}[/tex]
Where, V = potential difference
d = distance between the plates
Put the formula
[tex]E =\dfrac{12}{0.1}[/tex]
[tex]E=120\ V/m[/tex]
Hence, The electric field between the plates is 120 V/m.