Answer:
P₁ = 192 W and P₂ = 384 W
Explanation:
It is given that,
Resistor 1, [tex]R_1=12\ \Omega[/tex]
Resistor 2, [tex]R_1=6\ \Omega[/tex]
Voltage, V = 48 V
Power dissipated by resistor 1 is given by :
[tex]P_1=\dfrac{V^2}{R_1}[/tex]
[tex]P_1=\dfrac{(48)^2}{12}[/tex]
P₁ = 192 watts
Power dissipated by resistor 2 is given by :
[tex]P_2=\dfrac{V^2}{R_1}[/tex]
[tex]P_2=\dfrac{(48)^2}{6}[/tex]
P₂ = 384 watts
So, the power dissipated by both the resistors is 192 watts and 384 watts respectively. Hence, this is the required solution.
Frequency and velocity of a particle in simple harmonic motion Problem in that the amplitude is 0.24 in. and the maximum acceleration is 225 ft/s maximum velocity of the particle ist/s, and the frequency of its motion s z Print 20r
Answer:
velocity maximum = 21.6 ft /s
frequency = 16.88 Hz
Explanation:
Amplitude, A = 0.24 in = 0.02 ft
maximum acceleration, a = 225 ft/s\
The formula for maximum acceleration is
a = ω² A
225 = ω² x 0.02
ω² = 11250
ω = 106.06 rad/s
Maximum velocity, v = ω A
v = 106.06 x 0.02 = 2.1 ft/s
Let f be the frequency
ω = 2 x 3.14 x f
f = 106.06 / (2 x 3.14) = 16.88 Hz
velocity maximum = 21.6 ft /s
frequency = 16.88 Hz
Amplitude, A = 0.24 in = 0.02 ft
maximum acceleration, a = 225 ft/s
The formula for maximum acceleration is
a = ω² A
225 = ω² x 0.02
ω² = 11250
ω = 106.06 rad/s
Maximum velocity, v = ω A
v = 106.06 x 0.02 = 2.1 ft/s
Let f be the frequency
ω = 2 x 3.14 x f
f = 106.06 / (2 x 3.14) = 16.88 Hz
A battery with an emf of 12 V and an internal resistance of 1 Ω is used to charge a battery with an emfof 10 V and an internal resistance of 1 Ω. The current in the circuit is : A) 2 A B) 1 A C) 4A D) 11A E) 22 A
Answer:
Option B
Explanation:
The net emf in the circuit
E = 12 - 10 = 2 V
Total effective resistance,
r = 1 + 1 = 2 ohm
By using Ohm's law
E = I × R
I = 2 / 2 = 1 A
To find the current in the circuit, use Ohm's Law and solve two equations.
Explanation:The current in the circuit can be found using Ohm's Law: V = IR, where V is the voltage, I is the current, and R is the resistance.
In this case, the emf of the first battery is 12 V with an internal resistance of 1 Ω. Let's assume the current in the circuit is I1. So, the terminal voltage across the first battery can be calculated using Ohm's Law: 12 = I1 * (1 + 1).
Similarly, for the second battery with emf of 10 V and internal resistance of 1 Ω, the terminal voltage can be calculated as 10 = (I1 - I) * (1 + 1), where I is the current flowing in the second battery.
Now, we need to solve these two equations to find the value of I, which is the current in the circuit.
A length change - 0.18 m will occur for an object that is L- 80 m long, If the coeffcient of thermal expansion is12 x 106/C and if the original temperature is 83 C, find the final temperature.
Answer:
The final temperature is 270.5°C
Explanation:
Given that,
Change length = 0.18 m
Length = 80 m
Coefficient of thermal expansion [tex]\alpha=12\times10^{-6}\ C^{\circ}[/tex]
Temperature = 83°C
We need to calculate the final temperature
Using formula thermal expansion
[tex]\delta l=l\alpha\times\delta t[/tex]
Where, [tex]\Delta l[/tex] =change in length
[tex]\Delta t[/tex] = change in temperature
[tex]\alpha[/tex] = coefficient of thermal expansion
Put the value into the formula
[tex]0.18=80\times12\tmes10^{6}\times(T_{f}-T_{i})[/tex]
[tex]T_{f}-83=\dfrac{0.18}{80\times12\times10^{-6}}[/tex]
[tex]T_{f}=187.5+83=270.5^{\circ} C[/tex]
Hence, The final temperature is 270.5°C
what Is the gravitational force between a roast beef sandwich (m = .05 kg) and a hungry tarantula (m = 87 grams) if they are 17 cm apart?
Answer:
Gravitational force, [tex]F=1.003\times 10^{-11}\ N[/tex]
Explanation:
It is given that,
Mass of roast beef sandwich, m₁ = 0.05 kg
Mass of hungry tarantula, m₂ = 87 gm = 0.087 kg
Distance between two objects, d = 17 cm = 0.17 m
We need to find the gravitational force between them. It is given by :
[tex]F=G\dfrac{m_1m_2}{d^2}[/tex]
[tex]F=6.67\times 10^{-11}\times \dfrac{0.05\ kg\times 0.087\ kg}{(0.17\ m)^2}[/tex]
[tex]F=1.003\times 10^{-11}\ N[/tex]
So, the gravitational force between two objects is [tex]F=1.003\times 10^{-11}\ N[/tex]. Hence, this is the required solution.
You are in downtown Chicago (where streets run N-S and E-W). You started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station. Length of one block is 100 [m]. What is the magnitude and direction of your displacement from the start position?
Answer:
Displacement is 565.69 m at 45° west of north
Explanation:
Let north represent positive y axis and east represent positive x axis.
We have journey started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station.
3 blocks toward north = 300 j m
4 blocks toward west = -400 i m
1 blocks toward north = 100 j m
Total displacement = -400 i + 400 j m
Magnitude
[tex]s=\sqrt{(-400)^2+400^2}=565.69m[/tex]
Direction,
[tex]\theta =tan^{-1}\left ( \frac{400}{-400}\right )=135^0[/tex]
Direction is 45° west of north.
Displacement is 565.69 m at 45° west of north
(a) If TH = 1100 K and TC = 400 K, what is the thermal efficiency?
Answer:
63.6 %
Explanation:
TH = 1100 K , Tc = 400 k
Efficiency is given by
η = 1 - Tc / TH
η = 1 - 400 / 1100
η = 1 - 0.36
η = 0.636
η = 63.6 %
For fully developed laminar pipe flow in a circular pipe, the velocity profile is u(r) = 2(1-r2 /R2 ) in m/s, where R is the inner radius of the pipe. Assuming the pipe diameter is 4 cm, find the maximum and average velocities in the pipe as well as the volume flow rate
Answer:
a) [tex]v_{max} = 2\ \textup{m/s}[/tex]
b) [tex]v_{avg} = 1\ \textup{m/s}[/tex]
c) Q = 1.256 × 10⁻³ m³/s
Explanation:
Given:
The velocity profile as:
[tex]u(r) = 2(1-\frac{r^2}{R^2} )[/tex]
Now, the maximum velocity of the flow is obtained at the center of the pipe
i.e r = 0
thus,
[tex]v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )[/tex]
or
[tex]v_{max} = 2\ \textup{m/s}[/tex]
Now,
[tex]v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}[/tex]
or
[tex]v_{avg} = \frac{2}}{2}\ \textup{m/s}[/tex]
or
[tex]v_{avg} = 1\ \textup{m/s}[/tex]
Now, the flow rate is given as:
Q = Area of cross-section of pipe × [tex]v_{avg}[/tex]
or
Q = [tex]\frac{\pi D^2}{4}\times v_{avg}[/tex]
or
Q = [tex]\frac{\pi 0.04^2}{4}\times 1[/tex]
or
Q = 1.256 × 10⁻³ m³/s
For fully developed laminar pipe flow in a circular pipe, the maximum velocity is 2 m/s, the average velocity is (4/3) m/s, and the volume flow rate is (16/3)π cm^3/s. The velocity profile equation and the formulas for maximum velocity, average velocity, and volume flow rate are explained in detail.
Explanation:For fully developed laminar pipe flow in a circular pipe, the velocity profile is given by the equation u(r) = 2(1-r^2/R^2) in m/s, where R is the inner radius of the pipe. To find the maximum velocity, we need to substitute r = R into the equation. The maximum velocity u_max is then equal to 2(1-1^2/R^2), which simplifies to 2 m/s.
The average velocity can be found by integrating the velocity profile equation over the entire cross-sectional area of the pipe. The cross-sectional area is given by A = πR^2, so the average velocity v_avg is equal to (1/A) times the integral of 2(1-r^2/R^2) over the range r = 0 to r = R. Simplifying the integral and dividing by A, we get v_avg = (4/3) m/s.
The volume flow rate Q can be calculated by multiplying the cross-sectional area A by the average velocity v_avg. Using the given diameter of the pipe, we can find the radius R = 2 cm. Substituting the values into the equation Q = (πR^2)(4/3), we get Q = (16/3)π cm^3/s.
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Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite each other. Between the electrodes but not near their edges, the electric field strength is 2.5 × 106 N/C. What is the magnitude of the charge on each electrode? (ε0 = 8.85 × 10-12 C2/N ∙ m2)
Answer:
1.77 x 10^-8 C
Explanation:
Let the surface charge density of each of the plate is σ.
A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2
d = 2 mm
E = 2.5 x 10^6 N/C
ε0 = 8.85 × 10-12 C2/N ∙ m2
Electric filed between the plates (two oppositively charged)
E = σ / ε0
σ = ε0 x E
σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2
The surface charge density of each plate is ± σ / 2
So, the surface charge density on each = ± 22.125 x 10^-6 / 2
= ± 11.0625 x 10^-6 C/m^2
Charge on each plate = Surface charge density on each plate x area of each plate
Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C
A parallel-plate capacitor initially has a potential difference of 400 V and is then disconnected from the charging battery. If the plate spacing is now doubled (without changing Q), what is the new value of the voltage?
The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold welghs 198 N and is 3.2 m long. what is the tension in each rope whien the 6oo-N worker stands 1.12 m from one end? smaller tension arger tension
Answer:
[tex]T_2 = 309 N[/tex]
[tex]T_1 = 489 N[/tex]
Explanation:
As we know that total tension in both the ropes is counter balancing the weight of scaffold and worker both
so here we will have
[tex]T_1 + T_2 = (m + M)g[/tex]
now we have
[tex]T_1 + T_2 = 198 + 600 = 798 N[/tex]
now we also know that net torque due to both tension force in the string with respect to the position of worker must be zero so that platform will remain in equilibrium and horizontal in position
so here we will have
[tex]T_1(1.12) + (198)(1.6 - 1.12) = T_2(3.2 - 1.12)[/tex]
[tex]T_1+ 84.86 = 1.86 T_2[/tex]
now from above two equations we will have
[tex](1.86 T_2 - 84.86) + T_2 = 798 [/tex]
[tex]T_2 = 309 N[/tex]
also we have
[tex]T_1 = 489 N[/tex]
Final answer:
To calculate the tension in each rope supporting a scaffold with a window washer, the principles of torques and equilibrium are applied. By considering the system's equilibrium, equations based on the torques created by the worker's weight, the scaffold's weight, and the tensions in the ropes can be set up and solved.
Explanation:
To solve for the tension in each rope when a window washer stands on a scaffold, we use the concept of torques and equilibrium. The entire system is in equilibrium, meaning the sum of torques around any pivot point is zero. Considering the window washer weighing 600 N and standing 1.12 m from one end of a 3.2 m scaffold that weighs 198 N, we choose the pivot point at one end of the scaffold for easier calculations.
Let T1 be the tension in the rope closest to the worker and T2 be the tension in the other rope. The force due to the worker creates a clockwise torque, and the force due to the scaffold creates a clockwise torque as well, whereas the tensions in the ropes create counterclockwise torques. The torques are calculated by multiplying the force by the distance from the pivot point.
To find T1 and T2, we set up the equilibrium condition for torques and solve the equations considering the weight of the worker and the scaffold acting at their respective centers of mass. The specific calculations require numerical values that depend on the distances given and the gravitational force.
Ultimately, by solving these equations, we find the tensions T1 and T2 that support the system in equilibrium.
Laura, Jeff, and Amy are in the same work group; they work well with other people in the work group but they don’t get along. This type of conflict can best be described as ________.
Chameleons can rapidly project their very long tongues to catch nearby insects. The tongue of the tiny Rosette-nosed chameleon has the highest acceleration of a body part of any amniote (reptile, bird, or mammal) ever measured. In a somewhat simplified model of its tongue motion, the tongue, starting from rest, first undergoes a constant-acceleration phase with an astounding magnitude of 2500 m/s^2. This acceleration brings the tongue up to a final speed of 5.0 m/s. It continues at this speed for 22 m/s until it hits its target. What is the total distance traveled by the chameleon's tongue?
Answer:
0.115 m
Explanation:
Consider the motion of tongue during acceleration :
v₀ = initial velocity of the tongue = 0 m/s
v = final velocity of the tongue = 5 m/s
a = acceleration = 2500 m/s²
d = distance traveled during acceleration phase
Using the equation
v² = v₀² + 2 a d
5² = 0² + 2 (2500) d
d = 0.005 m
Consider the motion of tongue after it attains constant speed
d' = distance traveled during constant velocity
v = constant velocity = 5 m/s
t = time of travel = 22 ms = 0.022 s
using the equation
d' = v t
d' = 5 x 0.022
d' = 0.11 m
D = Total distance traveled by tongue
Total distance traveled by tongue is given as
D = d + d'
D = 0.005 + 0.11
D = 0.115 m
A spring of equilibrium length L1 and spring constant k1 hangs from the ceiling. Mass m1 is suspended from its lower end. Then a second spring, with equilibrium length L2 and spring constant k2, is hung from the bottom of m1. Mass m2 is suspended from this second spring. How far is m2 below the ceiling?
Answer:
[tex]d = L_1 + L_2 + \frac{(m_1 + m_2)g}{K_1} + \frac{m_2g}{K_2}[/tex]
Explanation:
For first spring the total extension is given as
[tex]F_{net} = K_1 x_1[/tex]
here net force is due to weight of both masses
[tex](m_1 + m_2)g = K_1 x_1[/tex]
so extension of first spring is
[tex]x_1 = \frac{(m_1 + m_2)g}{K_1}[/tex]
now similarly the extension in second spring is only due to the weight of mass m2
so here we will have
[tex]m_2g = K_2 x_2[/tex]
[tex]x_2 =\frac{m_2g}{K_2}[/tex]
so the total distance from the ceiling for mass m2 is given as
[tex]d = L_1 + L_2 + \frac{(m_1 + m_2)g}{K_1} + \frac{m_2g}{K_2}[/tex]
The total distance of m₂ from the ceiling is [tex]d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]
The given parameters;
equilibrium of length of the spring, = L₁spring constant, = K₁mass on the spring = m₁length of the first spring, = L₂spring constant = K₂mass on the second spring, = m₂The extension of the first spring due to mass m₁ and m₂ is calculated as;
[tex]F = kx\\\\ g(m_1+m_2) = k_1x_1\\\\x_1 = \frac{g(m_1+m_2)}{k_1}[/tex]
The extension of the second spring due to mass (m₂) is calculated as;
[tex]m_2g = k_2x_2\\\\x_2 = \frac{m_2g}{k_2}[/tex]
The total distance of m₂ from the ceiling is calculated as follows;
[tex]d = L_1 + L_2 + x_1 + x_2\\\\d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]
Thus, the total distance of m₂ from the ceiling is [tex]d = L_1 + L_2 + \frac{g(m_1 + m_2)}{k_1} + \frac{m_2g}{k_2}[/tex]
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A powerful grasshopper launches itself at an angle of 45° above the horizontal and rises to a maximum height of 1.01 m during the leap. With what speed v did it leave the ground? Neglect air resistance. Hint: Use conservation of energy. A. 6.29 m/s B. 7.15 m/s C. 5.98 m/s D. 6.72 m/s E. 5.37 m/s
Answer:
6.29 m/s option (A)
Explanation:
theta = 45 degree, H = 1.01 m
let v be the launch speed
Use the formula for the maximum height for the projectile
H = v^2 Sin^θ / 2g
1.01 = v^2 x Sin^2(45) / (2 x 9.8)
1.01 = 0.0255 v^2
v^2 = 39.59
v = 6.29 m/s
The initial velocity of the grasshopper is 6.29 m/s.
Initial velocity of the grasshopper
The Initial velocity of the grasshopper is calculated from the following kinematic equation.
[tex]H = \frac{v_0^2 sin^2 \theta}{2g}[/tex]
where;
H is the maximum heightv is the initial velocity[tex]v_0^2 = \frac{2gH}{sin^2\theta} \\\\v_0^2 = \frac{2 \times 9.8 \times 1.01 }{(sin45)^2} \\\\v_0^2 = 39.6\\\\v_0 = 6.29 \ m/s[/tex]
Thus, the initial velocity of the grasshopper is 6.29 m/s.
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What accelerating potential is needed to produce electrons of wavelength 5.20 nm? V- 10.5563 Previous Answers Reguest Answer Submit Incorrect; Try Again; 5 attempts remaining ▼ Part B What would be the energy of photons having the same wavelength as these electrons? eV Submit Request Answer
The accelerating potential needed for electrons of a particular wavelength can be found by rearranging the de Broglie wavelength equation. The energy of photons of the same wavelength can be found using Planck's equation, by substituting the given wavelength.
Explanation:The accelerating potential needed to produce electrons of a wavelength (de Broglie wavelength) can be calculated using the equation: λ = h / √(2mVq), where h is Planck's constant, m is the mass of the electron, V is the accelerating voltage, and q is the charge of the electron. For the wavelength of 5.20 nm, one can rearrange and solve for V to get V = h² / (2mqλ²).
Now, for the energy of photons having the same wavelength as the electrons, we can use Planck's equation, E = hv = hc/λ. Here, 'h' is Planck's constant, 'v' is the frequency of light, 'c' is the speed of light and 'λ' is the wavelength. Substituting λ = 5.20 nm, we get the energy of photons in terms of electron-volts (eV).
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A flywheel in the form of a uniformly thick disk of radius 1.23 m has a mass of 93.6 kg and spins counterclockwise at 369 rpm . Calculate the constant torque required to stop it in 2.25 min .
Answer:
20.26 Nm
Explanation:
r = 1.23 m , m = 93.6 kg, w = 0, f0 = 369 rpm = 369 / 60 = 6.15 rps
w0 = 2 x 3.14 x 6.15 = 38.622 rad/s
t = 2.25 min = 2.25 x 60 = 135 second
Moment of inertia = 1/2 m r^2 = 0.5 x 93.6 x 1.23 x 1.23 = 70.8 kg m^2
use first equation of motion for rotational motion
w = w0 + α t
0 = 38.622 - α x 135
α = 0.286 rad/s^2
torque = moment of inertia x angular acceleration
Torque = 70.8 x 0.286 = 20.26 Nm
The crankshaft in a race car goes from rest to 3000 rpm in 3.0 s . (a) What is the crankshaft's angular acceleration?
(b) How many revolutions does it make while reaching 3000 rpm ?
Answer:
75 rotations
Explanation:
f0 = 0, f = 3000 rpm = 50 rps, t = 3 s
(a) use first equation of motion for rotational motion
w = w0 + α t
2 x 3.14 x 50 = 0 + α x 3
α = 104.67 rad/s^2
(b) Let θ be the angular displacement
use second equation of motion for rotational motion
θ = w0 t + 1/2 α t^2
θ = 0 + 0.5 x 104.67 x 3 x 3
θ = 471.015 rad
The angle turn in one rotation is 2 π radian.
Number of rotation = 471.015 / (2 x 3.14) = 75 rotations
A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times greater is the sound intensity of the train whistle than that of the library?
Answer:
The sound intensity of train is 1000 times greater than that of the library.
Explanation:
We have expression for sound intensity level,
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )[/tex]
A train whistle has a sound intensity level of 70 dB
We have
[tex]70=10log_{10}\left ( \frac{I_1}{I_0}\right )[/tex]
A library has a sound intensity level of about 40 dB
We also have
[tex]40=10log_{10}\left ( \frac{I_2}{I_0}\right )[/tex]
Dividing both equations
[tex]\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000[/tex]
The sound intensity of train is 1000 times greater than that of the library.
The electric field of a charge is defined by the force on: An electron A probe charge A proton. A source charge.
Answer:
A probe charge
Explanation:
As we know that electric field intensity is the force experienced by the probe charge which is placed in the electric field region
Here we can say it as
[tex]E = \frac{F}{q}[/tex]
so here that probe charge should be very small so that it will not disturb the electric field in the space.
If the probe charge is of large magnitude then the field will get disturbed and the intensity which is to be measured is different from its actual value.
Also the sign of the probe charge is taken to be positive.
so correct answer here will be
A probe charge
Final answer:
The electric field of a charge is defined by the force experienced by a probe charge placed within the electric field, so the correct option will be D probe charge
Explanation:
The electric field of a charge is defined by the force on a probe charge. An electric field is a vector field that affects the space surrounding a source charge and exerts a force on any test charge (probe charge) within that region. The direction of the electric field represents the direction a positive probe charge would move if placed within the field.
It's calculated as the force per unit charge exerted on a test charge and can be represented mathematically by the equation F = qE, where F is the force experienced by the test charge, q represents the charge of the test charge, and E is the electric field strength.
Torque is dependent on the angle between the force applied and the length of the level arm. When is the torque maximum? Not enough information. θ = 90 θ = 0 θ = 45
Answer:
90 degree
Explanation:
According to the formula of torque
torque = force x displacement x Sine of angle between force and displacement
So, for the maximum torque, the value of Sin theta should be maximum.
the maximum value of Sin theta is 1.
that means the value of theta is 90 degree.
When ________ is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system? pressure and volume temperature temperature and volume pressure volume
Answer:
temperature
Explanation:
When temperature is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system.
When pressure is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system
What is enthalpy?Enthalpy, the sum of the internal energy and the product of the pressure and volume of a thermodynamic system. Enthalpy is an energy-like property or state function—it has the dimensions of energy (and is thus measured in units of joules or ergs),
and its value is determined entirely by the temperature, pressure, and composition of the system and not by its history. In symbols, the enthalpy, H, equals the sum of the internal energy, E, and the product of the pressure, P, and volume, V, of the system: H = E + PV.
According to the law of energy conservation, the change in internal energy is equal to the heat transferred to, less the work done by, the system.
If the only work done is a change of volume at constant pressure, the enthalpy change is exactly equal to the heat transferred to the system
Hence when pressure is constant, the enthalpy change of a process equal to the amount of heat transferred into or out of the system
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A motorcycle moving at 13.2 m/s increases in speed to 25.7 m/s over a period of 8.6 s. If the motorcycle's mass is 352 kg, what net force in Newtons acts on the motorcycle?
Answer:
510.4 N
Explanation:
u = 13.2 m /s, v = 25.7 m/s, t = 8.6 s, m = 352 kg
Use first equation of motion
v = u + a t
a = (25.7 - 13.2) / 8.6 = 1.45 m/s^2
Use Newton's second law
F = m a = 352 x 1.45 = 510.4 N
A 230 g air-track glider is attached to a spring. The glider is pushed in 8.2 cm and released. A student with a stopwatch finds that 13 oscillations take 19.0 s . What is the spring constant? Express your answer using two significant figures.
Answer:
4.3 N/m
Explanation:
m = 230 g = 0.230 kg, x = 8.2 cm
in 13 oscillations, time taken = 19 s
In 1 oscillation, time taken = 19 / 13 = 1.46 s
By the use of formula of time period , Let k be the spring constant.
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]1.46 = 2\pi \sqrt{\frac{0.230}{k}}[/tex]
0.054 = 0.230 / k
k = 4.26 N/m
k = 4.3 N/m
One billiard ball is shot east at 2.00 m/s. A second, identical billiard ball is shot west at 1.00 m/s. The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90° and sending it north at 1.41 m/s. What are the speed and direction of the first ball after the collision?
Answer:
Velocity is 1.73 m/s along 54.65° south of east.
Explanation:
Let unknown velocity be v, mass of billiard ball be m and east direction be positive x axis.
Here momentum is conserved.
Initial momentum = Final momentum
Initial momentum = m x 2i + m x (-1)i = m i
Final momentum = m x v + m x 1.41 j = mv + 1.41 m j
Comparing
mi = mv + 1.41 m j
v = i - 1.41 j
Magnitude of velocity
[tex]v=\sqrt{1^2+(-1.41)^2}=1.73m/s[/tex]
Direction,
[tex]\theta =tan^{-1}\left ( \frac{-1.41}{1}\right )=-54.65^0[/tex]
Velocity is 1.73 m/s along 54.65° south of east.
Using the law of conservation of momentum, one can deduce the speed and direction of the first ball after the collision. It's found to be traveling east at 1 m/s.
Explanation:The scenario described is an example of a two-dimensional collision. In such collision, the law of conservation of momentum applies both in the east-west direction (x-axis) and the north-south direction (y-axis).
With the given velocities before the collision, the total momentum in the x-axis before the collision is: momentum_east = mass*velocity = m*2.00 m/s, and momentum_west = m*(-1.00) m/s. Therefore, total momentum in the x-axis before the collision = m*2.00 m/s + m*(-1.00) m/s = m m/s.
After the collision, the first ball keeps moving in the east-west direction (since the second ball is deflected north), but we don't know its velocity, let's call it v1. Applying the conservation of momentum after the collision in the x-direction, we get: total momentum = m*v1 + 0 (since the second ball no longer moves in the east-west direction) = m m/s. From this, we can solve for v1 and find that v1 = 1 m/s east. Thus, the first billiard ball is traveling east at 1 m/s after the collision.
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A spring-loaded gun, fired vertically, shoots a marble 9.0 m straight up in the air. What is the marble's range if it is fired horizontally from 1.8 m above the ground?
The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.
Explanation:To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.
First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.
Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.
So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.
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The marble's range if it is fired horizontally from 1.8 m above the ground is 7.968 m.
At the highest point, the velocity of the gun will be zero. The initial velocity of the shot can be determined using the kinematic equation,
[tex]v^2 = u^2 + 2gs[/tex]
here, v = 0 m/s
g = 9.8 m/s² (downward, hence taken negative)
s = 9.0 m (upwards, hence taken positive.)
Using proper sign convention, we get:
[tex](0 \hspace{0.8 mm} m/s)^2 = u^2 - 2 \times (9.8 \hspace{0.8 mm} m/s^2) \times (9.0 \hspace{0.8 mm} m )[/tex]
u = 13.28 m/s
Now, when the marble is fired at a height of 1.8 m above the ground, the time taken by the marble to reach the ground can be determined by the kinematic equation:
[tex]h = ut + \frac{1}{2}gt^2[/tex],
using u = 0 m/s (as we will consider the time of fall of the marble from highest point), g = 9.8 m/s² (downward, hence taken to be negative), h = 1.8 m (downwards, hence taken positive), we get:
[tex]- 1.8 \hspace{0.8 mm} m = (0 \hspace{0.8 mm} m/s)t - \frac{1}{2}(9.8 \hspace{0.8 mm} m/s^2)t^2[/tex]
or, [tex]1.8 \hspace{0.8 mm} m = (4.9 \hspace{0.8 mm} m/s^2)t^2[/tex]
or, t = 0.60 s
Range of the marble will be determined by the horizontal velocity of the marble. It will be the maximum horizontal distnace covered by the marble as it falls down in time t = 0.60 s.
R = [tex]v_x[/tex]t
Since, the horizontal velocity is not influenced by any acceleration and will remain constant.
∴ R = [tex]v_x[/tex]t = 13.28 m/s × 0.60 s = 7.968 m
A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 39.5 s. The distance between two successive crests is 37.9 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed?
(a) 7.9 s
The period of a wave is time that passes between two consecutive crests (or two consecutive troughs).
In this case, we are told that five crests pass in a time of 39.5 s. Therefore we can find the period by using the proportion:
[tex]\frac{5}{39.5 s}=\frac{1}{T}[/tex]
Where T is the period. Re-arranging the equation, we find
[tex]T=\frac{(39.5)(1)}{5}=7.9 s[/tex]
(b) 0.127 Hz
The frequency of a wave is equal to the reciprocal of the period:
[tex]f=\frac{1}{T}[/tex]
where
f is the frequency
T is the period
For this wave, we have T = 7.9 s, so its frequency is
[tex]f=\frac{1}{7.9 s}=0.127 Hz[/tex]
(c) 37.9 m
The wavelength of a wave is the distance between two consecutive crests (or two consecutive troughs). For this wave, the distance between two successive crests is 37.9 m, so the wavelength of the wave is
[tex]\lambda=37.9 m[/tex]
(d) 4.81 m/s
The speed of a wave is given by
[tex]v=\lambda f[/tex]
where
[tex]\lambda[/tex] is the wavelength
f is the frequency
For the wave in the problem, we have
[tex]\lambda=37.9 m\\f=0.127 Hz[/tex]
Therefore, the speed of the wave is
[tex]v=(37.9)(0.127)=4.81 m/s[/tex]
What was the speed of a space shuttle that orbited Earth at an altitude of 1482 km?
Answer:
v = 7121.3 m/s
Explanation:
As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed
so here we will have
[tex]\frac{mv^2}{r} = \frac{GMm}{r^2}[/tex]
here we know that
r = orbital radius = 6370 km + 1482 km
[tex]r = 7.852 \times 10^6 m[/tex]
also we know that
[tex]M = 5.97 \times 10^{24} kg[/tex]
now we will have
[tex]v^2 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{7.852 \times 10^6}[/tex]
[tex]v^2 = 5.07 \times 10^7[/tex]
[tex]v = 7121.3 m/s[/tex]
The speed of the space shuttle that orbited the earth at an altitude of 1482km will be [tex]V=7121.3\dfrac{m}{s}[/tex]
What will be the speed of a space shuttle that orbited Earth at an altitude of 1482 km?
As we know that the centripetal force for the space shuttle is due to the gravitational force of the earth due to which it will rotate in a circular path with constant speed
so here we will have
[tex]\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}[/tex]
[tex]V^2=\dfrac{GM}{r}[/tex]
here we know that
[tex]\rm r= orbital \ radius =6370+1482=7852 \ km[/tex]
[tex]r=7.852\times10^6\ m[/tex]
mass of earth [tex]M=5.97\times10^{24}\ kg[/tex]
Gravitational constant [tex]G=6.67\times10^{-11}[/tex]
By putting all the values we get
[tex]V^2=\dfrac{(6.67\times10^{-11} )(5.97\times10^{24})}{7.852\times10^{6}}[/tex]
[tex]V^2=5.07\times10^7[/tex]
[tex]V=7121.3 \dfrac{m}{s}[/tex]
Thus the speed of the space shuttle that orbited the earth at an altitude of 1482km will be [tex]V=7121.3\dfrac{m}{s}[/tex]
Two particles, one with charge -5.45 uC and one with charge 3.55 uC, are 4.34 cm apart. What is the magnitude of the force that one particle exerts on the other?
Answer:
[tex]F = 92.45 N[/tex]
Explanation:
As we know that the force between two charge particles is given by
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we know that
[tex]q_1 = 3.55 \mu C[/tex]
[tex]q_2 = 5.45 \mu C[/tex]
now the distance between the two charges is
r = 4.34 cm
now from the formula of electrostatic force we will have
[tex]F = \frac{(9\times 10^9)(3.55 \mu C)(5.45 \mu C)}{0.0434^2}[/tex]
[tex]F = 92.45 N[/tex]
A motorboat is moving at 4.0 m/s when it begins to accelerate at 1.0 m/s2. To the nearest tenth of a second, how long does it take for the boat to reach a speed of 17.0 m/s?
Answer:
13 sec
Explanation:
Hello
by definition the acceleration is a vector derived magnitude that indicates the speed variation per unit of time
[tex]a=\frac{V_{f}-V_{i}}{t_{f}-t_{i}}\\a(t_{f}-t_{i})={V_{f}-V_{i} }}\\t_{f}-t_{i}=\frac{V_{f}-V_{i}}{a} \\\\t_{f}=\frac{V_{f}-V_{i}}{a}+t_{i} \\\\[/tex]
Let
Vi=4.0 m/s
a=1.0 m/s2
Vf=17.0 m/s
ti= (0)sec
t2= unknown
[tex]t_{f}=\frac{V_{f}-V_{i} }{a}+t_{i}\\t_{f}=\frac{17-4 }{1}+0\\t_{f}=\ 13\ sec\\[/tex]
Answer: 13 sec
I hope it helps
Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 100 cm2, how many turns do you need?
Answer:
The number of turns is 64.
Explanation:
Given that,
Magnetic field = 0.050 T
Area of coil = 100 cm²
Frequency = 60 Hz
Output voltage emf= 12 V
We need to calculate the number of turns
Using formula of induced emf
[tex]emf =NAB\omega[/tex]
[tex]N=\dfrac{emf}{A\times B\times2\pi f}[/tex]
[tex]N=\dfrac{12}{0.01\times0.050\times2\times3.14\times60}[/tex]
[tex]N =63.6 = 64\ turns[/tex]
Hence, The number of turns is 64.
Answer:
You need 63.66 turns.
Explanation:
The number of turns of a magnetic field is given by the following formula:
[tex]N = \frac{V}{S*T*2\pi f}[/tex]
In which N is the number of turns, V is the maximum output voltage, S is the area of the rotating coil, in square meters and T is the measure of the magnetic field and f is the frequency.
In this problem, we have that:
Suppose that you wish to construct a simple ac generator having an output of 12 V maximum when rotated at 60 Hz. This means that [tex]V = 12[/tex] and [tex]f = 60[/tex].
A uniform magnetic field of 0.050 T is available. This means that [tex]T = 0.050[/tex].
If the area of the rotating coil is 100 cm2, how many turns do you need?
This means that [tex]S = 0.01[/tex]m². So:
[tex]N = \frac{V}{S*T*2\pi f}[/tex]
[tex]N = \frac{12}{0.01*0.05*120\pi}[/tex]
[tex]N = 63.66[/tex]
You need 63.66 turns.