Explanation:
It is given that,
Let q₁ and q₂ are two small positively charged spheres such that,
[tex]q_1+q_2=5\times 10^{-5}\ C[/tex].............(1)
Force of repulsion between the spheres, F = 1 N
Distance between spheres, d = 2 m
We need to find the charge on the sphere with the smaller charge. The force is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]q_1q_2=\dfrac{F.d^2}{k}[/tex]
[tex]q_1q_2=\dfrac{1\ N\times (2\ m)^2}{9\times 10^9}[/tex]
[tex]q_1q_2=4.45\times 10^{-10}\ C[/tex]............(2)
On solving the system of equation (1) and (2) using graph we get,
[tex]q_1=0.0000384\ C=38.4\ \mu C[/tex]
[tex]q_2=0.0000116\ C=11.6\ \mu C[/tex]
So, the charge on the smaller sphere is 11.6 micro coulombs. Hence, this is the required solution.
Combined charge = 5.0 x 10-5 C
Distance between spheres = 2.0 m
Force = 1.0 N
Formula:
Coulomb's Law: F = k * (q1 * q2) / r2
Calculations:
q1 + q2 = 5.0 x 10-5 C (Given)
q1 = q2 - x (Assuming q2 is larger charge)
Substitute and solve for x
Answer:
The charge on the sphere with the smaller charge is 2.5 x 10-5 C or 25 micro-coulombs.
What is the frequency of oscillation for a mass on the end of spring with a period of motion of 2.6 seconds? Calculate answer to one decimal place.
Answer:
Frequency, f = 0.38 Hz
Explanation:
Time period of the spring, T = 2.6 seconds
We need to find the frequency of oscillation for a mass on the end of spring. The relation between the time period and the frequency is given by :
Let f is the frequency of oscillation. So,
[tex]f=\dfrac{1}{T}[/tex]
[tex]f=\dfrac{1}{2.6\ s}[/tex]
f = 0.38 Hz
or
f = 0.4 Hz
So, the frequency of oscillation for a mass on the end of a spring is 0.38 hertz. Hence, this is the required solution.
Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m/s.After the collision, ~vA = (−5.0ˆi + 20ˆj) m/s. What are (a) the final velocity of B and (b) the change in the total kinetic energy (including sign)?
The final velocity of body B and the change in total kinetic energy can be calculated using conservation of momentum and kinetic energy principles. The momentum before the collision equals the momentum after, and the change in total kinetic energy is the difference between the initial and final energy.
Explanation:Using physics principles, notably the law of conservation of momentum, we can calculate the final velocity of B and the change in the total kinetic energy. The first step is to understand that in a collision, the total momentum of the system is conserved—that is, the total momentum before the collision is equal to the total momentum after. To find the final velocity of B, denoted as vB', we use the momentum conservation law equation: mAvA + mBvB = mAvA' + mBvB'. Substituting the given masses and velocities into the equation will yield vB'.
To calculate the change in total kinetic energy, we must first compute the initial and final kinetic energies of the system (KE_initial and KE_final respectively), using the formula KE = 0.5 m v². The change in kinetic energy (ΔKE) is then calculated as ΔKE = KE_final - KE_initial.
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A rectangular coil with sides 0.10 m by 0.25 m has 500 turns of wire. It is rotated about its long axis in a magnetic field of 0.58 T directed perpendicular to the rotation axis. At what frequency must the coil be rotated for it to generate a maximum potential of 110 V?
Final answer:
The rectangular coil needs to be rotated at a specific frequency in order to generate a maximum potential of 110 V. The angular velocity of the coil can be calculated using the formula ω = V / (nBA), where V is the potential, n is the number of turns, B is the magnetic field strength, and A is the area of the coil.
Explanation:
To generate a maximum potential of 110 V, the rectangular coil needs to be rotated at a specific frequency. The potential generated by a rotating coil is given by the equation
V = nBAω
where V is the potential, n is the number of turns, B is the magnetic field strength, A is the area of the coil, and ω is the angular velocity. Rearranging the equation for ω, we have
ω = V / (nBA)
Substituting the given values, we get
ω = 110 V / (500 turns * 0.10 m * 0.25 m * 0.58 T)
Simplifying this expression will give you the required frequency at which the coil should be rotated to generate a maximum potential of 110 V.
A billiard ball strikes and rebounds from the cushion of a pool table perpendicularly. The mass of the ball is 0.38 kg The ball approaches the cushion with a velocity of +2.20 m/s and rebounds with a velocity of -1.70 m/s. The ball remains in contact with the cushion for a time of 3.40 x 10^-3 s. What is the average net force (magnitude and direction) exerted on the ball by the cushion?
Answer:
Force is 432.94 N along the rebound direction of ball.
Explanation:
Force is rate of change of momentum.
[tex]\texttt{Force}=\frac{\texttt{Final momentum-Initial momentum}}{\texttt{Time}}[/tex]
Final momentum = 0.38 x -1.70 = -0.646 kgm/s
Initial momentum = 0.38 x 2.20 = 0.836 kgm/s
Change in momentum = -0.646 - 0.836 = -1.472 kgm/s
Time = 3.40 x 10⁻³ s
[tex]\texttt{Force}=\frac{\texttt{Final momentum-Initial momentum}}{\texttt{Time}}=\frac{-1.472}{3.40\times 10^{-3}}\\\\\texttt{Force}=-432.94N[/tex]
Force is 432.94 N along the rebound direction of ball.
Final answer:
Using Newton's second law and the change in the ball's momentum, the average net force exerted on the billiard ball by the cushion is 436.47 N, directed away from the cushion.
Explanation:
The question relates to the concept of Newton's second law of motion, which states that the force on an object is equal to the mass of the object multiplied by the acceleration (F = ma). In this situation, a billiard ball strikes the cushion perpendicularly, changing its velocity and hence experiencing acceleration. To calculate the average net force exerted on the ball by the cushion, we can use the change in velocity and the time of impact in the following steps:
Calculate the change in momentum of the ball (Δp = m−v_f - m−v_i), where m is the mass, v_f is the final velocity, and v_i is the initial velocity.
Divide the change in momentum by the time of impact (Δt) to get the average force (F_avg). Use the formula F_avg = Δp / Δt.
Now let's apply these steps to the given values:
The change in momentum is Δp = 0.38 kg * (-1.70 m/s) - 0.38 kg * (+2.20 m/s) = -0.38 kg * (-1.70 - 2.20) m/s = -0.38 kg * (-3.90) m/s = 1.482 kg−m/s.
The average force is F_avg = 1.482 kg−m/s / 3.40 x 10^-3 s = 436.47 N.
The average net force exerted on the ball by the cushion is 436.47 N, directed away from the cushion since the ball rebounded after the collision.
The percent by which the fundamental frequency changed if the tension is increased by 30 percent is ? a)-20.04% b)-40.12% c)-30% d)-14.02%
Answer:
Percentage increase in the fundamental frequency is
d)-14.02%
Explanation:
As we know that fundamental frequency of the wave in string is given as
[tex]f_o = \frac{1}{2L}\sqrt{\frac{T}{\mu}}[/tex]
now it is given that tension is increased by 30%
so here we will have
[tex]T' = T(1 + 0.30)[/tex]
[tex]T' = 1.30T[/tex]
now new value of fundamental frequency is given as
[tex]f_o' = \frac{1}{2L}\sqrt{\frac{1.30T}{\mu}}[/tex]
now we have
[tex]f_o' = \sqrt{1.3}f_o[/tex]
so here percentage change in the fundamental frequency is given as
[tex]change = \frac{f_o' - f_o}{f_o} \times 100[/tex]
% change = 14.02%
When the frequency of the driving force equals the natural frequency of the system, the system is said to be in. (a) Resonance (b) Equilibrium (c) Elastic limit (d) None
Answer:
Option A is the correct answer.
Explanation:
When the frequency of the driving force equals the natural frequency of the system, the system is said to be in resonance. At resonance the system vibrates in maximum amplitude.
Marching soldiers are cautioned to break stride on a bridge is because of resonance, if the frequency of soldiers stride is equal to frequency of bridge, the bridge will vibrate with maximum amplitude. This will in turn collapse bridge.
Option A is the correct answer.
On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.87 m/s2. (a) How long does such a car, initially traveling at 23.1 m/s, take to stop? (b) How far does it travel in this time?
Answer:
(a): the car takes to stop 4.74 seconds.
(b): the car travels 54.78 meters in this time.
Explanation:
a= 4.87 m/s²
Vi= 23.1 m/s
Vf= Vi - a*t
t= Vi/a
t= 4.74 sec
d= Vi*t - (a*t²)/2
d= 54.78m
a rod of some material 0.20 m long elongates 0.20 mm on heating from 21 to 120°c. determine the value of the linear coefficient of thermal expansion [in (degrees c)^-1] for this material.
Answer:
The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹
Explanation:
Li = 0.2m
ΔL = 0.2 mm = 0.0002m
T1 = 21ºC
T2 = 120ºC
ΔT =99ºC
α =ΔL/(Li*ΔT)
α =0.0002m /(0.2m * 99ºC)
α = 1.01 *10⁻⁵ (ºC)⁻¹
The correct value of the linear coefficient of thermal expansion for the material is [tex]1.01 \times 10^{-5} \ (^\circ{C})^{-1}}\)[/tex].
To determine the linear coefficient of thermal expansion [tex](\(\alpha\))[/tex] for the material, we can use the formula:
[tex]\[\alpha = \frac{\Delta L}{L_0 \Delta T}\][/tex]
where:
[tex]\(\Delta L\)[/tex] is the change in length,
[tex]\(L_0\)[/tex] is the original length, and
[tex]\(\Delta T\)[/tex] is the change in temperature.
Given:
The original length [tex]\(L_0\)[/tex] is 0.20 m,
The elongation [tex]\(\Delta L\)[/tex] is 0.20 mm, which we need to convert to meters to match the units of [tex]\(L_0\)[/tex].
Since 1 m = 1000 mm, 0.20 mm = [tex]\(0.20 \times 10^{-3}\)[/tex] m,
The change in temperature [tex]\(\Delta T\)[/tex] is from 21°C to 120°C, so [tex]\(\Delta T = 120^\circ{C} - 21^\circ{C} = 99^\circ{C}\)[/tex].
Now we can plug these values into the formula:
[tex]\[\alpha = \frac{0.20 \times 10^{-3} \ m}{0.20 \ m \times 99^\circ{C}}\][/tex]
[tex]\[\alpha = \frac{0.20}{0.20 \times 99} \times 10^{-3} \ (^\circ{C})^{-1}\][/tex]
[tex]\[\alpha = \frac{1}{99} \times 10^{-3} \ (^\circ{C})^{-1}\][/tex]
[tex]\[\alpha \approx 1.0101 \times 10^{-5} \ (^\circ{C})^{-1}\][/tex]
Rounding to two significant figures, we get:
[tex]\[\alpha \approx 1.01 \times 10^{-5} \ (^\circ{C})^{-1}\}[/tex]
A cylinder 8 inches in diameter and 3 ft long is concentric with a pipe of 8.25 inches i.d. Between cylinder and pipe there is an oil film. What force is required to move the cylinder along the pipe at a constant velocity of 3 fps? The kinematic viscosity of the oil is 0.006 ft2 /s; the specific gravity is 0.92.
Answer:
Required force equals 623.498 lb
Explanation:
We shall use newton's law of viscosity to calculate the shear force that acts on the cylinder
By Newton's law of viscosity we have
[tex]\tau =\mu \frac{dv}{dy}\\\\where \tau[/tex] is shear stress that acts on the internal surface
[tex]\mu[/tex] is dynamic viscosity of the fluid
[tex]\frac{dv}{dy}[/tex] is the velocity gradient that exists across the flow
The dynamic viscosity is calculated as follows
[tex]\mu =\rho \nu[/tex]
[tex]\mu =\rho \nu \\\rho[/tex] is density of the fluid
[tex]\nu =[/tex] kinematic viscosity of the fluid
By no slip boundary condition the fluid in contact with the stationary cylinder shall not have any velocity while as the fluid in contact with the moving cylinder shall have velocity equal to that of the cylinder itself. This implies a velocity gradient shall exist across the gap in between the cylinders.
Applying values of the quantities we can calculate shear stress as follows
The density of fluid is [tex]\rho=G\times \rho_{w}[/tex]
G = specific gravity of fluid
[tex]\rho_{w}[/tex]is density of water
[tex]\tau =\rho \nu \frac{dv}{dy}\\\\\tau=62.42\times 0.92\times 0.006\times \frac{3}{\frac{0.125inches}{12inch}}\\\\\tau=99.23lb/ft^{2}[/tex]
This pressure shall oppose the motion of the internal cylinder hence the force of opposition = [tex]F=\tau\times Area[/tex]
Using the area of internal cylinder we get total force
F=[tex]2\pi rl\times \tau\\\\F=2\pi\times \frac{4}{12}ft\times 3ft\times \tau\\\\ F=623.498lb[/tex]
The force required to move the cylinder along the pipe can be calculated using the drag force formula. F = 6π(0.006)(1/3)(3) ≈ 0.113 ft-lbf
Explanation:To calculate the force required to move the cylinder along the pipe at a constant velocity, we need to consider the drag force acting on the cylinder.
The drag force can be calculated using the formula:
F = 6πηrv
Where F is the force, η is the viscosity of the oil, r is the radius of the cylinder, and v is the velocity of the cylinder.
In this case, the radius of the cylinder is half of the diameter, so the radius is 4 inches or 1/3 ft. The velocity is given as 3 fps.
Plugging in the values, we get:
F = 6π(0.006)(1/3)(3) ≈ 0.113 ft-lbf
You can harness visionary language by using:
Answer:
Metaphor
Explanation:
We can harness visionary language by using metaphor.
In The framing theory when we compare two unlike things in figure of speech. The comparison influences us on unconscious level. The metaphor causes us to make an association. If we change the metaphor we change how the other thinks of the subject.
Complex metaphor forms the basis of narratives or stories.
Benny wants to estimate the mean lifetime of Energizer batteries, with a confidence level of 97%, and with a margin of error not exceeding ±10 hours. If the standard deviation of the lifetime is known to be 55 hours, how many batteries does Benny need to sample?
Answer:
143 batteries does Benny need to sample
Explanation:
Given data
confidence level = 97%
error = ±10 hours
standard deviation SD = 55 hours
to find out
how many batteries does Benny need to sample
solution
confidence level is 97%
so a will be 1 - 0.97 = 0.03
the value of Z will be for a 0.03 is 2.17 from standard table
so now we calculate no of sample i.e
no of sample = (Z× SD/ error)²
no of sample = (2.16 × 55 / 10)²
no of sample = 142.44
so 143 batteries does Benny need to sample
An AC generator consists of eight turns of wire, each of area 0.0775 m2 , and total resistance of 8.53 Ω. The loop rotates in the external magnetic field of strength 0.222 T at a constant frequency of 51 Hz. What is the maximum induced emf? Answer in units of V.
Answer:
44.08 Volt
Explanation:
N = 8, A = 0.0775 m^2, R = 8.53 ohm, B = 0.222 T, f = 51 Hz
e0 = N B A w
e0 = 8 x 0.222 x 0.0775 x 2 x 3.14 x 51
e0 = 44.08 Volt
The question pertains to the calculation of the maximum induced emf (voltage) in an AC generator. This can be calculated using the generator's specifications and the formula ε_max = NBAω.
Explanation:The subject of your question pertains to electromagnetic induction in physics. The induction of emf in an AC generator is described by the equation ε = NBAω sin ωt, where ε is the induced emf, N is the number of turns of wire, B is the magnetic field strength, A is the cross-sectional area of the coil, and ω is the angular velocity of the generator. However, considering you are asking specifically for the maximum induced emf, we calculate this using the equation ε_max = NBAω, as sin ωt=1 at its peak. In your case, the generator consists of 8 turns of wire (N=8), an area of 0.0775 m^2 (A=0.0775), the strength of the magnetic field is 0.222 T (B=0.222), and a frequency of 51 Hz (f=51) which converts to angular velocity (ω) using the formula ω = 2πf. Substituting these values into the equation will give you the maximum induced emf.
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When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (Round your answers to two decimal places.) (a) What is the temperature of the drink after 45 minutes? °C (b) When will its temperature be 16°C? min
(a) The temperature of the drink after 45 minutes is 13.85°C.
(b) The temperature of the drink will be ( 16°C ) after ( 67.75) minutes.
To solve this problem, we can use Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature. The formula can be written as:
[tex]\[ \frac{dT}{dt} = -k(T - T_a) \][/tex]
(a) To find the temperature of the drink after 45 minutes, we first need to determine the constant (k). We can do this using the given data points:
1. At ( t = 0 ), ( T = 5°C).
2. At ( t = 25) minutes, ( T = 10°C), and [tex]\( T_a[/tex] = 20°C.
Using these values, we can write the equation as:
[tex]\[ 10 = 20 - (20 - 5)e^{-25k} \][/tex]
Solving for (k):
[tex]\[ e^{-25k} = \frac{20 - 10}{20 - 5} \] \[ e^{-25k} = \frac{10}{15} \] \[ e^{-25k} = \frac{2}{3} \] \[ -25k = \ln\left(\frac{2}{3}\right) \] \[ k = -\frac{\ln\left(\frac{2}{3}\right)}{25} \][/tex]
Now that we have (k), we can find the temperature after 45 minutes:
[tex]\[ T = 20 - (20 - 5)e^{-45k} \] \[ T = 20 - 15e^{45\left(\frac{\ln\left(\frac{2}{3}\right)}{25}\right)} \] \[ T = 20 - 15e^{1.8\ln\left(\frac{2}{3}\right)} \] \[ T = 20 - 15\left(\frac{2}{3}\right)^{1.8} \] \[ T \approx 20 - 15(0.41) \] \[ T \approx 20 - 6.15 \] \[ T \approx 13.85C \][/tex]
So, the temperature of the drink after 45 minutes is approximately ( 13.85°C ).
(b) To find when the temperature of the drink will be (16°C), we use the same formula and solve for ( t ):
[tex]\[ 16 = 20 - (20 - 5)e^{-kt} \] \[ e^{-kt} = \frac{20 - 16}{20 - 5} \] \[ e^{-kt} = \frac{4}{15} \] \[ -kt = \ln\left(\frac{4}{15}\right) \] \[ t = -\frac{\ln\left(\frac{4}{15}\right)}{k} \] \[ t = -\frac{\ln\left(\frac{4}{15}\right)}{-\frac{\ln\left(\frac{2}{3}\right)}{25}} \] \[ t = 25\frac{\ln\left(\frac{4}{15}\right)}{\ln\left(\frac{2}{3}\right)} \][/tex]
Using a calculator, we find:
[tex]\[ t \approx 25\frac{\ln(0.2667)}{\ln(0.6667)} \] \[ t \approx 25 \times 2.71 \] \[ t \approx 67.75 \][/tex]
So, the temperature of the drink will be ( 16°C ) after ( 67.75) minutes.
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm . The explorer finds that the pendulum completes 99.0 full swing cycles in a time of 128 s . What is the magnitude of the gravitational acceleration on this planet? Express your answer in meters per second per second.
The magnitude of the gravitational acceleration (g) on this planet is equal to 12.52 [tex]m/s^2[/tex].
Given the following data:
Length = 53.0 cm to m = 0.53 m.
Number of cycle, n = 99.0 cycles.
Time = 128 seconds.
How to determine the magnitude of the gravitational acceleration.First of all, we would calculate the period for a full swing cycle as follows:
[tex]T=\frac{time}{n} \\\\T=\frac{128}{99.0}[/tex]
Period, T = 1.293 seconds.
Mathematically, the time taken (period) by a pendulum is given by this formula:
[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]
Making g the subject of formula, we have:
[tex]g=\frac{4\pi^2L}{T^2} \\\\g=\frac{4(3.142)^2 \times 0.53}{1.293^2} \\\\g=\frac{20.929}{1.672}[/tex]
g = 12.52 [tex]m/s^2[/tex].
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To determine the gravitational acceleration on an unfamiliar planet, the period of a pendulum was used. After calculating the period per swing, the pendulum formula T = 2π√(l/g) was rearranged to solve for g, yielding approximately 9.826 m/s² as the gravitational acceleration.
Explanation:The student's question involves calculating the magnitude of the gravitational acceleration on an unfamiliar planet using the period of a simple pendulum. The pendulum's length is 53.0 cm and it completes 99.0 swings in 128 seconds. The formula to calculate the period of a pendulum (T) is T = 2π√(l/g), where l is the length of the pendulum and g is the acceleration due to gravity.
First, calculate the period (T) of one full swing by dividing the total time by the number of cycles: T = 128 s / 99.0 = 1.29292929 s approximately.Next, use the formula T = 2π√(l/g) to solve for g, where l = 0.53 m. Rearrange the formula to solve for g: g = 4π²(l/T²).Plug in the values: g = 4π²(0.53 m / (1.29292929 s)²) = 9.826 m/s² approximately.Therefore, the gravitational acceleration on the planet is around 9.826 m/s².
An electron is confined in a harmonic oscillator potential well. What is the longest wavelength of light that the electron can absorb if the net force on the electron behaves as though it has a spring constant of 74 N/m? (m el = 9.11 × 10-31 kg, c = 3.00 × 108 m/s, 1 eV = 1.60 × 10-19 J, ℏ = 1.055 × 10-34 J · s, h = 6.626 × 10-34 J · s)
Answer:
The longest wavelength of light is 209 nm.
Explanation:
Given that,
Spring constant = 74 N/m
Mass of electron [tex]m= 9.11\times10^{-31}\ kg[/tex]
Speed of light [tex]c= 3\times10^{8}\ m/s[/tex]
We need to calculate the frequency
Using formula of frequency
[tex]f =\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
Where, k= spring constant
m = mass of the particle
Put the value into the formula
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{74}{9.11\times10^{-31}}}[/tex]
[tex]f=1.434\times10^{15}\ Hz[/tex]
We need to calculate the longest wavelength that the electron can absorb
[tex]\lambda=\dfrac{c}{f}[/tex]
Where, c = speed of light
f = frequency
Put the value into the formula
[tex]\lambda =\dfrac{3\times10^{8}}{1.434\times10^{15}}[/tex]
[tex]\lambda=2.092\times10^{-7}\ m[/tex]
[tex]\lambda=209\ nm[/tex]
Hence, The longest wavelength of light is 209 nm.
Three forces act on an object. If the object is in translational equilibrium, which of the following must be true? I. The vector sum of the three forces must equal zero; II. The magnitude of the three forces must be equal; III. The three forces must be parallel.
Answer:
Option I
Explanation:
When ever the system is in equilibrium, it means the net force on the system is zero.
If the number of forces acting on a system and then net force on the system is zero, it only shows that the vector sum of all the forces is zero.
A long straight wire runs up and down vertically and carries current downward. What is the direction of the magnetic field produced by this current at a location nearby and directly east of the wire? A) southward
B) downward
C) upward
D) northward
E) westward
Answer:
A) southward
Explanation:
Applying the rule of the right hand for the determination of the magnetic field produced by the current flowing through the cable yields the following results: between the wires it is to the north, the west of the wires is to the south and east of the wires , which asks the problem is to the south.
What is the power of the man who pushes the box 8m with a force of 15N in a 6seconds? A. 20 Watts B. 30 Watts C. 40 Watts D. 40 Watts
Answer:
A. 20 Watts
Explanation:
Power is the rate of doing work.
P = Fd/t = (15 N)(8 m)/(6 s) = 20 N·m/s = 20 W
20 watts. If the power of the man who pushes the box 8m with a force of 15n is 6 seconds, it will force 20 watts in 6 seconds.
First, you do is multiply and then divide to give us the final answer.
[tex]\displaystyle 8m\times15n\div6s[/tex]
[tex]\displaystyle 8\times15=120\div6=20[/tex]
[tex]\Large\boxed{20}[/tex], which is our answer.
A rigid tank contains 1 kg of air (ideal gas) at 15 °C and 210 kPa. A paddle wheel supplies work input to the air such that final temperature is 97 "C. The specific heats are Cp 1.005 kJ(kg.K) and C-0.718 kJ/(kg.K). Determine the total work done in kJ:
Answer:
-58.876 kJ
Explanation:
m = mass of air = 1 kg
T₁ = Initial temperature = 15°C
T₂ = Final temperature = 97°C
Cp = Specific heat at constant pressure = 1.005 kJ/kgk
Cv = Specific heat at constant volume = 0.718 kJ/kgk
W = Work done
Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)
ΔU = Change in internal energy
Q = W+ΔU
⇒Q = W+mCvΔT
⇒0 = W+mCvΔT
⇒W = -mCvΔT
⇒Q = -1×0.718×(97-15)
⇒Q = -58.716 kJ
An electric dipole consists of two opposite charges of magnitude q = 1uC separated by a distance of 2 cm. The dipole is placed in an external field of 1.2 M N/C. What is the maximum torqued exerted on the dipole by the electric field?
Answer:
[tex]\tau_{max} = 0.024 Nm[/tex]
Explanation:
Dipole moment of the electric dipole is given by the equation
[tex]P = qd[/tex]
here we have
[tex]q = 1\mu C[/tex]
d = 2 cm[/tex]
[tex]P = (1\mu C)(0.02)[/tex]
[tex]P = 2\times 10^{-8} C-m[/tex]
now the maximum torque due to electric field is given as
[tex]\tau = \vec P \times \vec E[/tex]
[tex]\tau_{max} = PE sin90[/tex]
[tex]\tau_{max} = (2\times 10^{-8})(1.2\times 10^6)[/tex]
[tex]\tau_{max} = 0.024 Nm[/tex]
The maximum torque exerted on an electric dipole with a charge magnitude of 1µC and separated by 2 cm in an external electric field of 1.2 MN/C is 0.024 N·m.
The dipole moment (p) of an electric dipole is defined as the product of the charge q and the distance d between the charges. Calculating the dipole moment for a dipole with charges of magnitude q = 1µC (micro Coulombs) separated by a distance of d = 2 cm (0.02 meters), we have p = q * d. To find the maximum torque, we use the equation T = pE, where E is the external electric field magnitude. Given that E = 1.2 MN/C (Mega Newtons per Coulomb), the equation becomes T = (1µC * 0.02m) * 1.2MN/C, which gives us T = (1e-6 C * 0.02 m) * 1.2e6 N/C, leading to a maximum torque of 0.024 N·m.
A proton moves perpendicular to a uniform magnetic field B S at a speed of 1.00 3 107 m/s and experiences an acceleration of 2.00 3 1013 m/s2 in the positive x direction when its velocity is in the positive z direction. Determine the magnitude and direction of the field.
Explanation:
It is given that,
Speed of proton, [tex]v=10^7\ m/s[/tex]
Acceleration of the proton, [tex]a=2\times 10^{13}\ \ m/s^2[/tex]
The force acting on the proton is balanced by the magnetic force. So,
[tex]ma=qvB\ sin(90)[/tex]
[tex]B=\dfrac{ma}{qv}[/tex]
m is the mass of proton
[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 2\times 10^{13}\ \ m/s^2}{1.6\times 10^{-19}\times 10^7\ m/s}[/tex]
B = 0.020875
or
B = 0.021 T
So, the magnitude of magnetic field is 0.021 T. As the acceleration in +x direction, velocity in +z direction. So, using right hand rule, the magnitude of B must be in -y direction.
The velocity of an object moving along a straight line is given by – v(t) = t 2 − 3t + 2 (a) Find the displacement of the object from t = 0 to t = 3. (b) Find the distance traveled from t = 0 to t = 3.
Answer:
a) Displacement of the object from t = 0 to t = 3 is 1.5 m
b) Distance of the object from t = 0 to t = 3 is 1.83 m
Explanation:
Velocity, v(t) = t² - 3t + 2
a) Displacement is given by integral of v(t) from 0 to 3.
[tex]s=\int_{0}^{3}(t^2-3t+2)dt=\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_0^3=\frac{3^3}{3}-\frac{3^3}{2}+6=1.5m[/tex]
b) t² - 3t + 2 = (t-1)(t-2)
Between 1 and 2, t² - 3t + 2 is negative
So we can write t² - 3t + 2 as -(t² - 3t + 2)
Distance traveled
[tex]s=\int_{0}^{1}(t^2-3t+2)dt+\int_{1}^{2}-(t^2-3t+2)dt+\int_{2}^{3}(t^2-3t+2)dt\\\\s=\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_0^1-\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_1^2+\left [ \frac{t^3}{3}-\frac{3t^2}{2}+2t\right ]_2^3\\\\s=\frac{1^3}{3}-\frac{3\times 1^2}{2}+2-\left ( \frac{2^3}{3}-\frac{3\times 2^2}{2}+4\right )+\frac{1^3}{3}-\frac{3\times 1^2}{2}+2+\frac{3^3}{3}-\frac{3\times 3^2}{2}+6-\left ( \frac{2^3}{3}-\frac{3\times 2^2}{2}+4\right )=1.83m[/tex]
The displacement of the object from t = 0 to t = 3 is -4.5 m, indicating the object moved 4.5 meters in the opposite direction from its initial position. The distance the object traveled during the same period can be found by taking the integral of the absolute value of the velocity function from 0 to 3, and adding the magnitudes for time intervals when the velocity was positive and when it was negative.
Explanation:The given function indicates the velocity of an object moving along a straight line as a function of time - v(t) = t^2 - 3t + 2. It's a quadratic function so one way to find the displacement from t = 0 to t = 3, is to integrate the velocity function. The integral of v(t) from 0 to 3 gives the total change in position, or displacement, which would be the integral ∫ from 0 to 3 of (t^2 - 3t + 2) dt = [t^3/3 - 1.5t^2 + 2t] from 0 to 3 = 3^3/3 - 1.5 * 3^2 + 2*3 - (0 - 0 + 0) = 3 - 13.5 + 6 = -4.5 m.
On the other hand, distance is a scalar quantity and does not account for the direction, only the magnitude of movement. As such the object's total distance travelled from t = 0 to t = 3 may be calculated by finding the integral of the absolute value of the velocity from 0 to 3. In this case, v(t) is positive from t = 0 to t = 1 and negative from t = 1 to t = 3 (as substantiated by equating the velocity function to 0). Thus, the total distance traveled by the object is the sum of distances in segments (0,1) and (1,3), obtained as the sum of integrals ∫ from 0 to 1 of (t^2 - 3t + 2) dt + ∫ from 1 to 3 of (-t^2 + 3t - 2) dt.
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Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m.
Final answer:
To bring a 950-kg car to rest from 90 km/h over 120 m requires an average force of 2473.96 N. If the car hits a concrete abutment and stops within 2 m, the force exerted is much higher, at 148,437.50 N. This illustrates the impact of stopping distance on the force experienced by a vehicle.
Explanation:
Work-Energy Theorem Application
We'll first convert the speed from km/h to m/s by multiplying by 1000/3600. Therefore, 90.0 km/h is 25 m/s. Using the work-energy theorem, we know the work done to stop the car is equal to the change in kinetic energy.
Plugging in the values: W = KE = 1/2mv²
W = 1/2(950 kg)(25 m/s)²
W = 1/2(950 kg)(625 m²/s²)
W = 296,875 J
Since work is also equal to force times distance (W = Fd), the force needed can be found by dividing the work by the distance.
F = W/d = 296,875 J/120 m = 2473.96 N
This is the average force required to stop the car over 120 m. Now let's calculate the force if the car hits a concrete abutment and stops in 2.00 m:
F = 296,875 J/2 m = 148,437.50 N
The force exerted on the car in this case is significantly higher, showing the importance of cushioning distance in reducing impact forces.
A car tries to tow a stranded van out of a ditch, and the car applies 300 N and pulls the van for 3 meters in less than a minute before getting it safely back onto the road. How much work was done by the car on the car? The van on the car?
Answer:
Work done, W = 900 Joules
Explanation:
It is given that,
Force applies by the car on the van, F = 300 N
The van pulls a distance of, d = 3 m
We need to find the work was done by the car on the car. We know that the product of force and distance is equal to the work done by the object i.e.
[tex]W=F\times d[/tex]
[tex]W=300\ N\times 3\ m[/tex]
W = 900 Joules
So, the work done by the car on the van is 900 Joules. Hence, this is the required solution.
An aluminum wire has a resistance of 7.00 Ω at 30.0°C. Determine its resistance (in Ω) at 430°C. The temperature coefficient of resistivity for aluminum wire is 3.90 ✕ 10−3 (°C)−1. (Assume that the temperature coefficient of resistivity was measured using the reference temperature 20°C.)
Answer:
17.92 Ω
Explanation:
R₀ = Initial resistance of the aluminum wire at 30.0°C = 7.00 Ω
R = resistance of the aluminum wire at 430.0°C = ?
α = temperature coefficient of resistivity for aluminum = 3.90 x 10⁻³ °C⁻¹
ΔT = Change in temperature = 430 - 30 = 400 °C
Resistance of the wire is given as
R = R₀ (1 + α ΔT)
R = (7) (1 + (3.90 x 10⁻³) (400))
R = 17.92 Ω
A copper wire is 1.6 m long and its diameter is 1.1 mm. If the wire hangs vertically, how much weight (in N) must be added to its free end in order to stretch it 2.8 mm?
Answer:
Weight required = 194.51 N
Explanation:
The elongation is given by
[tex]\Delta L=\frac{PL}{AE}[/tex]
Length , L= 1.6 m
Diameter, d = 1.1 mm
Area
[tex]A=\frac{\pi d^2}{4}=\frac{\pi \times (1.1\times 10^{-3})^2}{4}=9.50\times 10^{-7}m^2[/tex]
Change in length, ΔL = 2.8 mm = 0.0028 m
Young's modulus of copper, E = 117 GPa = 117 x 10⁹ Pa
Substituting,
[tex]\Delta L=\frac{PL}{AE}\\\\0.0028=\frac{P\times 1.6}{9.50\times 10^{-7}\times 117\times 10^9}\\\\P=194.51N[/tex]
Weight required = 194.51 N
What is the equation used to calculate the density of UO2?
Answer:
D = [tex]\frac{ZM}{a^{3}N_A}[/tex]
Explanation:
the equation used for calculation of density of [tex]UO_2[/tex]
where D = density of [tex]UO_2[/tex]
M=molar mass of [tex]UO_2[/tex]
a= lattice constant here [tex]UO_2[/tex] is a body centered lattice and for body centered lattice a=0.547 nm
[tex]N_A[/tex]= Avogadro number which is equal to [tex]6.023\times 10^{23}[/tex]
At the 18th green of the U.S. Open you need to make a 23.5ft putt to win the tournament. When you hit the ball, giving it an initial speed of 1.45 m/s, it stops 5.10 ft short of the hole. Assuming the deceleration caused by the grass is constant, what should the initial speed have been to just make the putt? 2.What initial speed do you need to make the remaining 5.10 ft putt?
Answer:
1.63 m/s
Explanation:
a = acceleration of the ball
d = stopping distance = 23.5 - 5.10 = 18.4 ft = 5.61 m
v₀ = initial velocity of the car = 1.45 m/s
v = final velocity of the car = 0 m/s
Using the equation
v² = v₀² + 2 a d
0² = 1.45² + 2 a (5.61)
a = - 0.187 m/s²
To win the tournament :
a = acceleration of the ball = - 0.187 m/s²
d = stopping distance = 23.5 ft = 7.1 m
v₀ = initial velocity of the car = ?
v = final velocity of the car = 0 m/s
Using the equation
v² = v₀² + 2 a d
0² = v₀² + 2 (- 0.187) (7.1)
v₀ = 1.63 m/s
The initial speed required can be calculated using the formulas of motion under constant deceleration. Here, we first calculated deceleration from the given initial velocity and distance and then applied that deceleration to find the initial speed required for the remaining distance.
Explanation:In the given problem, the ball falls short of the target. It indicates that the ball decelerated while it was in motion. The distance covered and the initial speed are given, so the deceleration can be calculated using the formula for motion under constant acceleration (v^2 = u^2 + 2as). In this case, the final velocity (v) is 0, the initial velocity (u) is 1.45m/s, and the distance covered (s) is the total - the short distance, which in meters turns out to be (23.5ft - 5.10ft) * 0.3048 = 5.608m. By substituting these values, we can solve for acceleration (a). For the remaining 5.10ft, which is approximately 1.554m, we can use the found deceleration and the physics equation v = sqrt(u^2 + 2as) to find the initial speed that is required for the remaining distance.
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A 60-watt light bulb carries a current of 0.5 ampere. The total charge passing through it in one hour is: A) 3600 C B) 3000C C) 2400C D) 120 C E) 1800 C
Answer:
Total charge, q = 1800 C
Explanation:
It is given that,
Power of light bulb, P = 60 watts
It carries a current, I = 0.5 A
Time, t = 1 hour = 3600 seconds
We need to find the total charge passing through it in one hour. We know that current through an electrical appliance is defined as the charge flowing per unit time i.e.
[tex]I=\dfrac{q}{t}[/tex]
[tex]q=I\times t[/tex]
[tex]q=0.5\ A\times 3600\ s[/tex]
q = 1800 C
So, the total charge passing through it is 1800 C. Hence, this is the required solution.
The current of 0.5 amperes means that 0.5 coulombs of charge pass every second. Therefore, over an hour (3600 seconds), a total charge of 1800 Coulombs would pass. So, the answer to the question is E) 1800 C.
Explanation:The question is about understanding the relationship between electric current, time and charge. The electric current is the charge passing per unit time. In this case, we know that our current is 0.5 amperes, which conveys that 0.5 coulombs of electric charge is passing every second.
Therefore, when you want to find out how much charge passes in an hour, you would multiply by the number of seconds in an hour which is equal to 3600 (60 seconds per minute x 60 minutes).
Hence, the total charge passing for an hour is 0.5 amperes x 3600 seconds = 1800 Coulombs, so the answer is (E) 1800 C.
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Two vehicles, a car and a truck, leave an intersection at the same time. The car heads east at an average speed of 20 miles per hour, while the truck heads south at an average speed of 50 miles per hour. Find an expression for their distance apart d (in miles) at the end of t hours.
Answer:
[tex]d = 10\times t\sqrt{29}miles[/tex]
Explanation:
Given:
't' hour be the time taken for travel by both the vehicles and 'd' be the distance between then
then
Distance traveled by the car = 20 × t miles
and
Distance traveled by the truck = 50 × t miles
now, using the Pythagoras theorem
[tex]d = \sqrt{(20t)^2+(50t)^2}[/tex]
or
[tex]d = \sqrt{400t^2+2500t^2}[/tex]
or
[tex]d = \sqrt{2900t^2}[/tex]
or
[tex]d = 10\times t\sqrt{29}[/tex]
thus, the equation relating the distance 'd' with the time 't' comes as
[tex]d = 10\times t\sqrt{29}miles[/tex]
Final answer:
To find the distance between the car and truck after t hours, apply the Pythagorean theorem. Calculate the hypotenuse of the right triangle formed by their paths. The distance apart in miles is represented by the expression 50√1.16t.
Explanation:
The question involves finding an expression for the distance d between two vehicles traveling perpendicularly away from an intersection, with one vehicle going east and another going south at different speeds. To solve this, we can apply the Pythagorean theorem which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. If the car travels east at 20 miles per hour and the truck travels south at 50 miles per hour, after t hours, the car will have traveled 20t miles and the truck 50t miles. These distances represent the two legs of a right triangle, and the distance d between the vehicles is the hypotenuse.
So, the distance d (in miles) at the end of t hours is given by:
d = √{(20t)² + (50t)²}
You can simplify this further to find:
d = t * √{20² + 50²}
d = t *√{400 + 2500}
d = t *√2900
d = t *50√1.16
d = 50√1.16t
Therefore, the distance apart in miles at the end of t hours is given by the expression 50√1.16t.