The tip of one prong of a tuning fork undergoes SHM of frequency 1000 Hz and amplitude 0.40 mm. For this tip, what is the magnitude of the
(a) maximum acceleration,
(b) maximum velocity,
(c) acceleration at tip displacement 0.20 mm, and
(d) velocity at tip displacement 0.20 mm?
Answer:(a) 15795.5m/s²
(b) 2.5m/s
(c) 7897.7 m/s²
(d) 2.2m/s
Explanation:The displacement, y, of a body undergoing simple harmonic motion (SHM) is given by
y = A sin (ωt + φ) ------------------(i)
Where;
A = maximum displacement or amplitude of the body
ω = angular frequency of the body
t = time taken for the displacement
φ = phase constant
The velocity, v, of the body can be found by differentiating equation (i) as follows;
v = Aω cos (ωt + φ) ------------------(ii)
Where;
Aω = maximum velocity or amplitude of the velocity of the body
Also, the acceleration of the body can be found by differentiating equation (ii) as follows;
a = -Aω² sin(ωt + φ) --------------------(iii)
Where;
-Aω² = maximum acceleration or amplitude of the acceleration of the body
(a) From equation (iii), the magnitude of the maximum acceleration [tex]a_{max}[/tex] is given by;
[tex]a_{max}[/tex] = Aω² ----------------(iv)
Where;
A = amplitude = 0.40mm = 0.00040m
ω = 2 π f [Take π = 3.142. Also, f = frequency of the motion = 1000Hz]
=> ω = 2 x 3.142 x 1000 = 6284 rad/s
Substitute these values into equation (iv) as follows;
[tex]a_{max}[/tex] = 0.00040 x 6284² = 15795.5m/s²
Therefore, the magnitude of the maximum acceleration is 15795.5m/s²
========================================================
(b) From equation (ii), the magnitude of the maximum velocity [tex]v_{max}[/tex], is given by;
[tex]v_{max}[/tex] = Aω ----------------(v)
Where;
A = amplitude = 0.40mm = 0.00040m
ω = 2 π f [Take π = 3.142. Also, f = frequency of the motion = 1000Hz]
=> ω = 2 x 3.142 x 1000 = 6284 rad/s
Substitute these values into equation (v) as follows;
[tex]v_{max}[/tex] = 0.00040 x 6284 = 15795.5m/s²
Therefore, the magnitude of the maximum velocity is 2.5m/s
========================================================
(c) Comparing equations (i) and (iii), equation (iii) can be written as;
a = -ω² y -------------------(vi)
Therefore, to get the acceleration at tip displacement of 0.20mm, substitute y = 0.20mm = 0.00020m and ω = 6284rad/s into equation (vi) as follows;
a = - 6284² x 0.00020
a = - 7897.7 m/s²
Therefore, the magnitude of the acceleration at the tip displacement is 7897.7 m/s²
========================================================
(d) Recall that;
sin²θ + cos²θ = 1
=> cos²θ = 1 - sin²θ
=> cosθ = √(1 - sin²θ )
=> cos (ωt + φ) = √(1 - sin² (ωt + φ))
Substitute this value into equation (ii) as follows;
v = Aω √(1 - sin² (ωt + φ))
v = ω√(A² - A²sin² (ωt + φ))
Now, comparing the equation above and equation (i), the equation above can be written as;
v = ω√(A² - y²) -------------(vii)
Therefore, to get the velocity at tip displacement of 0.20mm, substitute y = 0.20mm = 0.00020m, ω = 6284rad/s and A = 0.00040m into equation (vii) as follows;
v = 6284√(0.00040² - 0.00020²)
v = 6284√(0.00000060)
v = 2.2m/s
Therefore, the magnitude of the velocity at the tip displacement is 2.2 m/s
Two bodies are falling with negligible air resistance, side byside, above a horizontal plane near Earth's surface. If one of thebodies is given an additional horizontal acceleration during itsdescent, it ........................
(a) strikes the plane at the same time as the other body
(b) has the vertical component of its velocity altered
(c) has the vertical component of its accelerationaltered
(d) follows a hyperbolic path
(e) follows a straight line path along the resultantacceleration vector
Answer:
(a) strikes the plane at the same time as the other body
Explanation:
Two bodies are falling with negligible air resistance, side by side, above a horizontal plane near Earth's surface. If one of the bodies is given an additional horizontal acceleration during its descent, it strikes the plane at the same time as the other body
The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons transported from the positive electrode to the negative electrode from the time the battery is first used until it is essentially dead?
a. 3.7 x 10^18
b. 5.3 x 10^23
c. 4.4 x 10^22
d. 1.6 x 10^19
e. 1.2 x 10^23
Answer:
e. 1.2 x 10²³
Explanation:
According to the problem, The current equation is given by:
[tex]I(t)=0.88e^{-t/6\times3600s}[/tex]
Here time is in seconds.
Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.
The relation between current and number of charge carriers is:
[tex]q=\int\limits {I} \, dt[/tex]
Here the limits of integration is from 0 to infinite. So,
[tex]q=\int\limits {0.88e^{-t/6\times3600s}}\, dt[/tex]
[tex]q=0.88\times(-6\times3600)(0-1)[/tex]
q = 1.90 x 10⁴ C
Consider N be the total number of charge carriers. So,
q = N e
Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.
N = q/e
Substitute the suitable values in the above equation.
[tex]N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}[/tex]
N = 1.2 x 10²³
A spaceship of proper length ? = 100 m travels in the positive x direction at a speed of 0.700 c0 relative to Earth. An identical spaceship travels in the negative x direction along a parallel course at the same speed relative to Earth. At t = 0, an observer on Earth measures a distance d = 58,000 km separating the two ships.
Part A
At what instant does this observer see the leading edges of the two ships pass each other?
Answer:
observer see the leading edges of the two ships pass each other at time 0.136 s
Explanation:
given data
spaceship length = 100 m
speed of 0.700 Co = [tex]0.700\times 3 \times 10^8[/tex] m/s
distance d = 58,000 km = 58000 × 10³ m
solution
as here distance will be half because both spaceship travel with same velocity
so they meet at half of distance
Distance Da = [tex]\frac{d}{2}[/tex] ............1
Distance Da = [tex]\frac{58000\times 10^3}{2}[/tex]
Distance Da = 29 ×[tex]10^{6}[/tex] m
and
time at which observer see leading edge of 2 spaceship pass
Δ time = [tex]\frac{Da}{v}[/tex] ..........2
Δ time = [tex]\frac{29 \times 10^6}{0.700\times 3 \times 10^8}[/tex]
Δ time = 0.136 s
Answer:
The time at which the observer see the leading edges of the two ships pass each other is 0.138 sec.
Explanation:
Given that,
Length = 100 m
Speed = 0.700 c
Distance = 58000 km
The distance should be halved because the spaceships both travel the same speed.
So they will meet at the middle of the distance
We need to calculate the distance
Using formula for distance
[tex]d'=\dfrac{d}{2}[/tex]
Put the value into the formula
[tex]d'=\dfrac{58000}{2}[/tex]
[tex]d'=29000\ km[/tex]
We need to calculate the time at which the observer see the leading edges of the two ships pass each other
Using formula of time
[tex]\Delta t=\dfrac{d}{V}[/tex]
Put the value into the formula
[tex]\Delta t=\dfrac{29\times10^{6}}{0.700\times3\times10^{8}}[/tex]
[tex]\Delta t=0.138\ sec[/tex]
Hence, The time at which the observer see the leading edges of the two ships pass each other is 0.138 sec.
A string is stretched to a length of 212 cm and both ends are fixed. If the density of the string is 0.02 g/cm, and its tension is 357 N, what is the fundamental frequency? Course hero
The fundamental frequency of a string fixed at both ends is f₁ = c/2L, where 'c' is calculated from the tension and linear mass density of the string. Using the given tension and density, one can find the wave speed 'c' first, then substitute into the formula to get the fundamental frequency.
To calculate the fundamental frequency of a string fixed at both ends, you can use the formula for the fundamental frequency of a string, which is given by f1 = c/2L, where f1 is the fundamental frequency, c is the speed of the wave on the string, and L is the length of the string.
To find the speed of the wave on the string, we use the formula c = \/(T/μ), where c is the speed of the wave, T is the tension in the string, and μ is the linear mass density of the string. Substituting the given values, c = \/(357 N / (0.02 g/cm * 100 cm/m)) = \/(357 / 0.0002 kg/m), we can calculate c and then use the result to find f1.
car leaves a stop sign and exhibits a constant acceleration of 0.300 m/s2 parallel to the roadway. The car passes over a rise in the roadway such that the top of the rise is shaped like an arc of a circle of radius 500 m. Now the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What are the magnitude and direction of the total acceleration vector for the car at this instant
Answer:
0.308 m/s2 at an angle of 13.5° below the horizontal
Explanation:
The parallel acceleration to the roadway is the tangential acceleration on the rise.
The normal acceleration is the centripetal acceleration due to the arc. This is given by
[tex]a_N = \dfrac{v^2}{r} = \dfrac{36^2}{500}=0.072[/tex]
The tangential acceleration, from the question, is
[tex]a_T = 0.300[/tex]
The magnitude of the total acceleration is the resultant of the two accelerations. Because these are perpendicular to each other, the resultant is given by
[tex]a^2 =a_T^2 + a_N^2 = 0.300^2 + 0.072^2[/tex]
[tex]a = 0.308[/tex]
The angle the resultant makes with the horizontal is given by
[tex]\tan\theta=\dfrac{a_N}{a_T}=\dfrac{0.072}{0.300}=0.2400[/tex]
[tex]\theta=13.5[/tex]
Note that this angle is measured from the horizontal downwards because the centripetal acceleration is directed towards the centre of the arc
. An electron moving at 4.00×103m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40×10−16N . What angle does the velocity of the electron make with the magnetic field? There are two answers.
Answer: 9.59° and 350.41°
Explanation: The formulae that relates the force F exerted on a moving charge q with velocity v in a magnetic field of strength B is given as
F =qvB sin x
Where x is the angle between the strength of magnetic field and velocity of the charge.
q = 1.609×10^-19 C
v = 4×10³ m/s
B = 1.25 T
F = 1.40×10^-16 N
By substituting the parameters, we have that
1.40×10^-16 = 1.609×10^-19 × 4×10³ × 1.25 × sinx
sin x = 1.40×10^-16/ 1.609×10^-19 × 4×10³ × 1.25
sin x = 1.40×10^-16 /8.045*10^(-16)
sin x = 0.1666
x = 9.59°
The value of sin x is positive in first and fourth quadrant.
Hence to get the second value of x, we move to the 4th quadrant of the trigonometric quadrant which is 360 - x
Hence = 360 - 9.59 = 350.41°
Final answer:
The angle between the velocity of the electron and the magnetic field is approximately 6.27° or 173.73°.
Explanation:
To find the angle between the velocity of the electron and the magnetic field, we can use the formula:
F = q(vsinθ)B
Where F is the magnetic force, q is the charge of the electron, v is the velocity of the electron, θ is the angle between the velocity and the magnetic field, and B is the magnetic field strength.
In this case, the magnetic force is given as 1.40 × 10^-16 N, the charge of an electron is 1.6 × 10^-19 C, the velocity of the electron is 4.00 × 10³ m/s, and the magnetic field strength is 1.25 T.
Plugging in these values, we can solve for θ:
1.40 × 10^-16 N = (1.6 × 10^-19 C)(4.00 × 10³ m/s)(sinθ)(1.25 T)
Solving for sinθ:
sinθ = (1.40 × 10^-16 N) / [(1.6 × 10^-19 C)(4.00 × 10³ m/s)(1.25 T)]
sinθ ≈ 0.109
Taking the inverse sine of 0.109, we find that θ ≈ 6.27° or θ ≈ 173.73°.
A solenoid of length 2.50 cm and radius 0.750 cm has 25 turns. If the wire of the solenoid has 1.85 amps of current, what is the magnitude of the magnetic field inside the solenoid
Answer:
13.875 T
Explanation:
Parameters given:
Length of solenoid, L = 2.5 cm = 0.025 m
Radius of solenoid, r = 0.75 cm = 0.0075 m
Number of turns, N = 25 turns
Current, I = 1.85 A
Magnetic field, B, is given as:
B = (N*r*I) /L
B = (25 * 0.0075 * 1.85)/0.025
B = 13.875 T
(a) Given a 48.0-V battery and 24.0-Ω and 96.0-Ω resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.
Answer with Explanation:w
a.We are given that
Potential difference, V=48 V
[tex]R_1=24\Omega[/tex]
[tex]R_2=96\Omega[/tex]
Equivalent resistance when R1 and R2 are connected in series
[tex]R_{eq}=R_1+R_2[/tex]
Using the formula
[tex]R_{eq}=24+96=120\Omega[/tex]
We know that
[tex]I=\frac{V}{R_{eq}}=\frac{48}{120}=0.4 A[/tex]
In series combination, current passing through each resistor is same and potential difference across each resistor is different.
Power, P=[tex]I^2 R[/tex]
Using the formula
Power,[tex]P_1=I^2R_1=(0.4)^2\times 24=3.84 W[/tex]
Power, [tex]P_2=I^2 R_2=(0.4)^2(96)=15.36 W[/tex]
b.
In parallel combination, potential difference remains same across each resistor and current passing through each resistor is different..
Current,[tex]I=\frac{V}{R}[/tex]
Using the formula
[tex]I_1=\frac{V}{R_1}=\frac{48}{24}=2 A[/tex]
[tex]I_2=\frac{V}{R_2}=\frac{48}{96}=0.5 A[/tex]
[tex]P_1=\frac{V^2}{R_1}=\frac{(48)^2}{24}=96 W[/tex]
[tex]P_2=\frac{V^2}{R_2}=\frac{(48)^2}{96}=24 W[/tex]
When two resistors are in series, the current through each is the same and is calculated by dividing the total voltage by the total resistance. The power through each resistor is the current squared times the resistance of each resistor. When resistors are in parallel, the total resistance decreases and the power increases. The current remains the same.
Explanation:When the 24-ohm and 96-ohm resistors are connected in series, the total resistance can be calculated using the formula R = R1 + R2. Here, R1 is the resistance of the first resistor (24 ohms) and R2 is the resistance of the second resistor (96 ohms). R equals 120 ohms. The current, I, is calculated using Ohm's Law, I = V / R. Therefore, the current flowing through the circuit is I = 48.0 V / 120 Ω = 0.4 A. The power, P, through each resistor is calculated using the formula P = I^2 * R. Therefore, for the 24-ohm resistor, P is (0.4 A)^2 * 24 Ω = 3.84 W, and for the 96-ohm resistor, P is (0.4 A)^2 * 96 Ω = 15.36 W. When the resistors are connected in parallel, the total resistance is calculated using the formula 1/R = 1/R1 + 1/R2. The current through each resistor remains the same (0.4 A) and the power for each resistor can be calculated in the same way as described above.
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The vertical normal stress increase caused by a point load of 10 kN acting on the ground surface at a point 1m vertically below its point of application is: (a) 0 (b) 4.775 kN (c) 5 kN (d) 10 kN
Answer:
(b) 4.775 kN
Explanation:
see the attached file
A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+138 determines the height of the rock above the surface of the water (in feet) in terms of the number of seconds t since the rock was thrown. g
Answer:
a) 138 ft
b) 4.62 s
c) 1.375 s
d) 168.25 ft
Explanation:
The height of a rock (thrown from the top of a bridge) above the level of water surface as it varies with time when thrown is given in the question as
h = f(t) = -16t² + 44t + 138
with t in seconds, and h in feet
a) The bridge's height above the water.
At t=0 s, the rock is at the level of the Bridge's height.
At t = 0,
h = 0 + 0 + 138 = 138 ft
b) How many seconds after being thrown does the rock hit the water?
The rock hits the water surface when h = 0 ft. Solving,
h = f(t) = -16t² + 44t + 138 = 0
-16t² + 44t + 138 = 0
Solving this quadratic equation,
t = 4.62 s or t = -1.87 s
Since time cannot be negative,
t = 4.62 s
c) How many seconds after being thrown does the rock reach its maximum height above the water?
At maximum height or at the maximum of any function, the derivative of that function with respect to the independent variable is equal to 0.
At maximum height,
(dh/dt) = f'(t) = (df/dt) = 0
h = f(t) = -16t² + 44t + 138
(dh/dt) = (df/dt) = -32t + 44 = 0
32t = 44
t = (44/32)
t = 1.375 s
d) What is the rock's maximum height above the water?
The maximum height occurs at t = 1.375 s,
Substituting this for t in the height equation,
h = f(t) = -16t² + 44t + 138
At t = 1.375 s, h = maximum height = H
H = f(1.375) = -16(1.375²) + 44(1.375) + 138
H = 168.25 ft
Hope this Helps!!!
A typical cell membrane is 8.0 nm thick and has an electrical resistivity of 1.3 107 Ohms · m.
(a) If the potential difference between the inner and outer surfaces of a cell membrane is 80 mV, how much current flows through a square area of membrane 1.2 µm on a side?
_____________A
(b) Suppose the thickness of the membrane is doubled, but the resistivity and potential difference remain the same. Does the current increase or decrease or remain the same?
By what factor?
Explanation:
Below is an attachment containing the solution.
Considering the form of the Gibb's Free Energy and the condition about it that some process be spontaneous, if some system has the following properties, what would be the result?
Answer:
A) Always Spontaneous.
B) Spontaneous at lower temperatures.
C) Spontaneous at higher temperatures.
D) Never Spontaneous.
Explanation:
The change in Gibb's free energy for a process, ΔG, is given by
ΔG = ΔH - TΔS
where ΔH is the change in enthalpy for the process
T = absolute temperature in Kelvin of the process
ΔS = change in enthalpy of the process.
And for a process to be spontaneous, its change in Gibb's free energy must be negative.
If it is positive, then the reaction isn't spontaneous.
So, we consider the conditions given one by one.
A) The enthalpy change is negative and the entropy change is positive.
ΔH = -ve, ΔS = +ve
ΔG = ΔH - TΔS = (negative number) - T(positive number) = (negative number - negative number) = negative number (since T isn't negative for these processes.)
Hence, the process with these conditions is always spontaneous.
B) The enthalpy change is negative and the entropy change is negative.
ΔH = negative, ΔS = negative
ΔG = ΔH - TΔS
For this relation, ΔG can only be positive if the numerical value of the ΔH (without the sign) is greater than TΔS.
This will happen mostly at low temperatures as low T, helps to reduce the numerical value of TΔS, thereby making ΔH (without the negative sign) the bigger number and subsequently makethe overall expression negative and the process, spontaneous.
C) The enthalpy change is positive and the entropy change is positive.
ΔH = positive, ΔS = positive.
ΔG = ΔH - TΔS
For this to be spontaneous,
TΔS > ΔH
And the one thing that favours this is high temperatures for the process.
At high temperatures, TΔS gives a much larger number which would drive the overall expression towards the negative sign, thereby making the process spontaneous.
D) The enthalpy change is positive and the entropy change is negative.
ΔH = positive, ΔS = negative.
ΔG = ΔH - TΔS
The signs on these state functions mean that the change in Gibb's free energy will always be positive for this set of conditions (since the temperature can't go as low as being negative).
A process with these conditions, is never spontaneous.
The spring of constant k = 170 N/m is attached to both the support and the 1.5-kg cylinder, which slides freely on the horizontal guide. If a constant 14-N force is applied to the cylinder at time t = 0 when the spring is undeformed and the system is at rest, determine the velocity of the cylinder when x = 65 mm. Also determine the maximum displacement of the cylinder.
Answer:
Velocity at 64 mm is 0.532 m/s
Maximum displacement = 0.082 m or 82 mm
Explanation:
The maximum displacement or amplitude is determined by the applied force from Hooke's law
[tex]F = kA[/tex]
[tex]A=\dfrac{F}{k}=\dfrac{14 \text{ N}}{170 \text{ N/m}}=0.082 \text{ m}[/tex]
The velocity at at any point, x, is given by
[tex]v=\sqrt{\dfrac{k}{m}(A^2 - x^2)}[/tex]
m is the mass of the load, here the cylinder.
In fact, the expression [tex]\sqrt{\dfrac{k}{m}}[/tex] represents the angular velocity, [tex]\omega[/tex].
Substituting given values,
[tex]v=\sqrt{\dfrac{170}{1.5}(0.082^2 - 0.065^2)} = 0.532 \text{ m/s}[/tex]
Three charged particles form a triangle: particle 1 with charge Q1 = 80.0 nC is at xy coordinates (0, 3.00 mm), particle 2 with charge Q2 is at (0, −3.00 mm), and particle 3 with charge q = 18.0 nC is at (4.00 mm, 0). In unit-vector notation, what is the electrostatic force on particle 3 due to the other two particles if Q2 is equal to (a) 80.0 nC and (b) −80.0 nC?
The net electrostatic force on particle 3 due to the other two particles depends on the charges of these particles. When Q2 = 80.0 nC, the forces from the other two particles cancel out and when Q2 = -80.0 nC, the forces add up.
Explanation:The electrostatic force on a charge due to other charges can be determined using Coulomb’s law. For the setup in the question, the deletion force on particle 3 because of particles 1 and 2 can be obtained by vectorially adding the forces it experiences due to each of these particles separately. When Q2 = -80.0 nC, the forces that particle 3 experiences due to particle 1 and particle 2 are in the same direction this time, therefore they add up to give the net force on particle 3. We can determine the exact value by substituting the given values in the equation for Coulomb's Law.
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An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on of possible sites for adsorption (see sketch at right). Calculate the entropy of this system. Round your answer to significant digits, and be sure it has the correct unit symbol.
The given question is incomplete. The complete question is as follows.
An oxygen molecule is adsorbed onto a small patch of the surface of a catalyst. It's known that the molecule is adsorbed on 1 of 36 possible sites for adsorption. Calculate the entropy of this system.
Explanation:
It is known that Boltzmann formula of entropy is as follows.
s = k ln W
where, k = Boltzmann constant
W = number of energetically equivalent possible microstates or configuration of the system
In the given case, W = 36. Now, we will put the given values into the above formula as follows.
s = k ln W
= [tex]1.38 \times 10^{-23} ln (36)[/tex]
= [tex]4.945 \times 10^{-23} J/K[/tex]
Thus, we can conclude that the entropy of this system is [tex]4.945 \times 10^{-23} J/K[/tex].
The entropy of the given system is [tex]4.954 \times 10^{-23 } \rm\; J/K[/tex] in which an oxygen molecule is absorbed in one of the 36 possible states.
From the Boltzmann formula of entropy,
[tex]s = k \times ln W[/tex]
Where,
[tex]k[/tex]= Boltzmann constant = [tex]1.38\times 10^{-23} \rm\; J/K[/tex]
[tex]W[/tex] = Number of energetically equivalent possible microstates of the system = 36
Put the values in the formula
[tex]s =1.38\times 10^{-23} \times ln (36)\\\\s = 4.954 \times 10^{-23 } \rm\; J/K[/tex]
Therefore, the entropy of the given system is [tex]4.954 \times 10^{-23 } \rm\; J/K[/tex]
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A simple circuit consists only of of a 1.0-μF capacitor and a 15-mH coil in series. At what frequency does the inductive reactance equal the capacitive reactance? 67 Hz 0.77 kHz 15 kHz 1.3 kHz
Answer:
The frequency is 1.3 kHz.
Explanation:
Given that,
Capacitance of the capacitor, [tex]C=1\ \mu F=10^{-6}\ C[/tex]
Inductance of the inductor, [tex]L=15\ mH=15\times 10^{-3}\ H[/tex]
We need to find the frequency when the inductive reactance equal the capacitive reactance such that :
[tex]X_c=X_L[/tex]
[tex]2\pi fL=\dfrac{1}{2\pi fC}[/tex]
[tex]f=\dfrac{1}{2\pi \sqrt{LC} }[/tex]
[tex]f=\dfrac{1}{2\pi \sqrt{15\times 10^{-3}\times 10^{-6}} }[/tex]
f = 1299.49 Hz
[tex]f=1.29\times 10^3\ Hz[/tex]
or
[tex]f=1.3\ kHz[/tex]
So, the frequency is 1.3 kHz. Therefore, the correct option is (d).
Suppose an Atwood machine has a mass of m1 = 6.0 kg and another mass of m2 = 2.0 kg hanging on opposite sides of the pulley. Assume the pulley is massless and frictionless, and the cord is ideal. Determine the magnitude of the acceleration of the two objects and the tension in the cord.
Answer:
Acceleration=[tex]4.9 /s^2[/tex]
Tension=29.4 N
Explanation:
We are given that
[tex]m_1=6 kg[/tex]
[tex]m_2=2 kg[/tex]
We have to find the magnitude of the acceleration of the two objects and the tension in the cord.
Tension, [tex]T=m_1(a+g)[/tex]
[tex]m_2g-T=m_2a[/tex]
Substitute the values
[tex]m_2g-m_1(a+g)=m_2a[/tex]
[tex]m_2g-m_1a-m_1g=m_2a[/tex]
[tex]g(m_2-m_1)=m_2a+m_1a=a(m_1+m_2)[/tex]
[tex]a=\frac{(m_2-m_1)g}{m_1+m_2}[/tex]
Substitute the values
[tex]a=\frac{(2-6)\times 9.8}{2+6}=-4.9m/s^2[/tex]
Where [tex]g=9.8m/s^2[/tex]
Hence, the magnitude of the acceleration of the two objects =[tex]4.9 m/s^2[/tex]
Substitute the values of a
[tex]T=m_1(a+g)=6(-4.9+9.8)=29.4 N[/tex]
Two cars are driving towards an intersection from perpendicular directions. The first car's velocity is 2 meters per second and the second car's velocity is 9 meters per second. At a certain instant, the first car is 8 meters from the intersection and the second car is 6 meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)? Choose 1 answer:
a -10
b -7
© -8.4
Answer:
Explanation:.
Given that
First car velocity is 2m/s
Second car velocity is 9m/s.
At a certain time the first car is 8m from intersection
And at the same time second car is 6m from intersection.
The rate of change of distance, i.e dx/dt, which is the speed of the car.
Using Pythagoras theorem, their distance apart is given as
Z²=X²+Y²
Z²=6²+8²
Z²=36+64
Z²=100
Z=√100
Z=10m
Let assume the direction,
Let assume the first car is moving in positive x direction, then
dx/dt=2m/s
And also second car will be moving in negative y direction
dy/dt=-9m/s
Now, to know dz/dt, let use the Pythagoras formulae above
x²+y²=z²
differentiate with respect to t
2dx/dt+2dy/dt=2dz/dt
Divide through by 2
dz/dt=dx/dt+dy/dt
dz/dt=2-9
dz/dt=-7m/s
The rate of change of distance between the two body is -7m/s
Option B is correct
Answer:
Rate of change of the distance between the cars = -7 m/s
Explanation:
Let the distance between the first car and the intersection be p = 8 meters
Let the distance between the second car and the intersection be q = 6 meters
velocity of the first car, dp/dt = -2 m/s
velocity of the second car, dq/dt = -9 m/s
We can get the distance between the two cars using pythagora's theorem
s² = p² + q²...................................(1)
s² = 8² + 6²
s = √(8² + 6²)
s = 10 m
Differentiating equation (1) through with respect to t
2s ds/dt = 2p dp/dt + 2q dq/dt
(2*10*ds/dt) = (2*8*(-2)) + (2*6*(-9))
20 ds/dt = -32 - 108
20 ds/dt = -140
ds/dt = -140/20
ds/dt = -7 m/s
In a dentist's office, an X-ray of a tooth is taken using X-rays that have a frequency of 9.81 1018 Hz. What is the wavelength in vacuum of these X-rays?
Answer:
[tex]\lambda=3.056\times10^{-11}m[/tex]
Explanation:
The equation that involves frequency (f) and wavelength ([tex]\lambda[/tex]) of a wave is [tex]v=\lambda f[/tex], where v is the speed of the wave. X-rays are a type of electromagnetic wave, so in vacuum it moves at the speed of light c, which means that our wavelength will be:
[tex]\lambda=\frac{c}{f}=\frac{299792458m/s}{9.81\times10^{18}Hz}=3.056\times10^{-11}m[/tex]
To calculate the wavelength of an X-ray with a frequency of 9.81 x 1018 Hz, we substitute the given frequency into the formula λ = c/f. The wavelength of these X-rays is approximately 3.06 x 10-11 meters.
Explanation:In Physics, we can find the wavelength of an X-ray given its frequency using the formula: λ=c/f, where 'λ' is the wavelength, 'c' is the speed of light (approximately 3.0 x 108 m/s in a vacuum), and 'f' is the frequency. Here, the frequency of the X-rays is given as 9.81 x 1018 Hz.
Substituting the values into the formula, we have: λ = (3.0 x 108) / (9.81 x 1018) = ~3.06 x 10-11 meters. The result indicates that the wavelength of these X-rays is incredibly small, which is consistent with our understanding of X-rays as high energy radiation with very short wavelengths and high frequencies.
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A solid conducting sphere has net positive charge and radiusR = 0.600 m . At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 18.0 V . Assume that V = 0 at an infinite distance from the sphere.
What is the electric potential at the center of the sphere?
Express your answer with the appropriate units.
Answer:
V_inside = 36 V
Explanation:
Given
We are given a sphere with a positive charge q with radius R = 0.400 m Also, the potential due to this charge at distance r = 1.20 m is V = 24.0 V.
Required
We are asked to calculate the potential at the centre of the sphere
Solution
The potential energy due to the sphere is given by equation
V = (1/4*π*∈o) × (q/r) (1)
Where r is the distance where the potential is measured, it may be inside the sphere or outside the sphere. As shown by equation (1) the potential inversely proportional to the distance V
V ∝ 1/r
The potential at the centre of the sphere depends on the radius R where the potential is the same for the entire sphere. As the charge q is the same and the term (1/4*π*∈o) is constant we could express a relation between the states , e inside the sphere and outside the sphere as next
V_1/V_2=r_2/r_1
V_inside/V_outside = r/R
V_inside = (r/R)*V_outside (2)
Now we can plug our values for r, R and V_outside into equation (2) to get V_inside
V_inside = (1.2 m )/(0.600)*18
= 36 V
V_inside = 36 V
Answer:
36 V
Explanation:
The solid conducting sphere is a positive charge
and has radius R₁ = 0.6m
at a point R₂ = 1.20 m, the electric potential V = 18.0 V
V, electric potential = K q/R where k = 1/4 πε₀
V is inversely proportional to R
V₁ = electric potential at the center
V₂ = electric potential at 1.2 m
then
V₁ /V₂ = R₂ / R₁
V₁ = V₂ ( R₂ / R₁) = 18.0 V ( 1.2 / 0.6 ) = 36 V
A 5 meter ladder leans against a wall. The bottom of the ladder is 1 meter from the wall at time t = 0 and slides away from the wall at 1 meters per second. Find the velocity of the top of the ladder at time t = 2.
Answer:
V = –0.89m/s
Explanation:
Please see attachments below.
The dean of a university located near the ocean (who was not a science major in college) proposes building an infrared telescope right on campus and operating it in a nice heated dome so that astronomers will be comfortable on cold winter nights. Criticize this proposal, giving your reasoning.
Answer:
The proposal is bad and has many mistakes in itself.
Explanation:
-Telescopes are preferably to be placed far from cities, ideally a remote place on a mountain or raised land. It should not be erected at sea level.
-It should not be erected at sea level. High humidity present at sea level clouds observations made as infrared observations are not possible at sea level.
-Dome should not be heated at all. Strong air currents are generated during heating which inturn ruins observations. Also, a heated dome will emit infrared radiation which ultimately swamps astronomical signals.
Final answer:
Coastal location and heated dome are unsuitable for infrared telescope due to atmospheric interference, light pollution, and dome-related issues. Consider remote location, dry climate, and remote operation for optimal observations. Consult experts and consider environmental impact.
Explanation:
Criticisms of Building an Infrared Telescope Near the Ocean in a Heated Dome:
While the intention might seem well-meaning, building an infrared telescope near the ocean in a heated dome has several drawbacks:
Unsuitable Location:
Atmospheric interference: Coastal areas have higher humidity and turbulent air, which negatively affects infrared observations due to water vapor absorption and image distortion.
Light pollution: City lights and nearby human activity create significant light pollution, impacting observations of faint infrared sources.
Dome Issues:
Cost: Building and maintaining a heated dome adds significant expense compared to an open-air observatory.
Heat distortion: Heating the dome creates air currents that can distort telescope observations.
Ventilation challenges: Maintaining controlled airflow and humidity within the dome can be complex and costly.
Alternatives:
Remote location: Building the telescope at a high-altitude, dry site with minimal light pollution would be more suitable for infrared observations.
Remote access and automation: Modern telescopes can be operated remotely, eliminating the need for astronomers to be physically present during observations.
Additional considerations:
Expertise: Consult with professional astronomers for advice on telescope placement and operation.
Environmental impact: Consider the potential ecological impact of the telescope and dome on the coastal environment.
Overall, while the concern for astronomer comfort is understandable, a coastal location with a heated dome is not optimal for an infrared telescope. Exploring alternative locations and remote operation technologies would be more effective and cost-effective for achieving high-quality scientific observations.
( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration would the force give to an object with a mass of (m2-m1)
Answer:
[tex]4.725 m/s^{2}[/tex]
Explanation:
We know that from Newton's second law of motion, F=ma hence making acceleration the subject then [tex]a=\frac {F}{m}[/tex] where a is acceleration, F is force and m is mass
Also making mass the subject of the formula [tex]m=\frac {F}{a}[/tex]
For [tex]m1= \frac {F}{13.5}[/tex] and [tex]m2=\frac {F}{3.5}[/tex] hence [tex]F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}[/tex]
a bag of ice cubes absorbs 149,000 J of heat, which causes its temperature to increase by 5.23 degrees celsius. what is the mass of the ice in the bag?
Ice=c=2000J/(kg*c) (unit=kg)
Answer:
14.3kg
Explanation:
Given parameters:
Quantity of heat = 149000J
Change in temperature = 5.23°C
specific heat of the ice = 2000J/kg°C
Unknown:
Mass of the ice in the bag = ?
Solution:
The heat capacity of a substance is given as:
H = m c Ф
H is the heat capacity
m is the mass
c is the specific heat
Ф is the temperature change;
since m is the unknown, we make it the subject of the expression;
m = H/ mФ
m = [tex]\frac{149000}{2000 x 5.23}[/tex] = 14.3kg
Answer: 14.2447
Explanation:
When a car drives through the Earth's magnetic field, an emf is induced in its vertical 60-cm-long radio antenna. Part A If the Earth's field (5.0×10−5 T) points north with a dip angle of 38∘, which direction(s) will the car be moving to produce the maximum emf induced in the antenna?
Answer:
Explanation:
The magnetic field is in north south direction . In order to cut lines of forces to the maximum extent , car has to move in east- west direction ie towards east or towards west to produce maximum emf.
emf produced = B L V
where B is horizontal component of earth magnetic field
L is length of rod
v is velocity of car carrying antenna rod .
The car should be moving east or west to produce the maximum emf induced in the antenna.
Explanation:When a car drives through the Earth's magnetic field, the maximum emf induced in its vertical radio antenna occurs when the car is moving in a direction perpendicular to the magnetic field. In this case, the Earth's magnetic field points north with a dip angle of 38°. To produce the maximum emf, the car should be moving east or west.
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A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.
When the beads reach the ends of the rod, the angular speed of the system is approximately 5.76 rad/s.
Explanation:To find the angular speed of the system at the instant the beads reach the ends of the rod, we can use the principle of conservation of angular momentum. Initially, the angular momentum of the system is given by Li = Irodωi + 2(Ibead)ωi, where Irod is the moment of inertia of the rod about its center, Ibead is the moment of inertia of each bead about the center, and ωi is the initial angular speed of the system. When the beads reach the ends of the rod, the moment of inertia of the system changes, but the angular momentum remains constant. So, we have Li = Irodωf + 2(Ibead)ωf, where ωf is the final angular speed of the system. We can rearrange this equation to solve for ωf. Given the values of Irod, Ibead, and ωi, we can substitute them into the equation to find ωf.
Using the given values: mass of the rod = 190 g = 0.19 kg, length of the rod = 43 cm = 0.43 m, mass of each bead = 38 g = 0.038 kg, distance of the beads from the center = 10 cm = 0.1 m, and initial angular speed = 12 rad/s, we can calculate the moments of inertia as follows:
Irod = (1/12)mrodLrod2 = (1/12)(0.19)(0.43)2 = 0.002196 kg.m2
Ibead = mbeadR2 = (0.038)(0.1)2 = 0.000038 kg.m2
Now, substituting the values into the equation Li = Irodωi + 2(Ibead)ωi, we have (0.002196)(12) + 2(0.000038)(12) = (0.002196 + 2(0.000038))ωf. Solving for ωf, we get ωf ≈ 5.76 rad/s.
Three identical 4.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up from left to right and a force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 4.0 m/s2 . What is the magnitude of the force exerted on the right cube by the middle cube in this case
Final answer:
The magnitude of the force exerted on the right cube by the middle cube, when three identical cubes accelerate together on a frictionless surface, is calculated using Newton's second law (F = ma) and is found to be 16.0 Newtons.
Explanation:
The student's question pertains to Newton's second law of motion and the concept of force and acceleration. The problem involves three identical cubes on a frictionless surface accelerating due to a force applied to the first cube. To find the magnitude of the force exerted on the right cube by the middle cube, we use the formula F = ma, where F is the force, m is the mass, and a is the acceleration.
Since the cubes are identical and accelerate together, each 4.0-kg cube has an acceleration of 4.0 m/s². Therefore, the force exerted by one cube on another can be found by multiplying one cube's mass by the given acceleration:
F = m × aF = 4.0 kg × 4.0 m/s²F = 16.0 NThe force exerted on the right cube by the middle cube is 16.0 Newtons.
Before entering the cyclotron, the particles are accelerated by a potential difference V. Find the speed v with which the particles enter the cyclotron.
31. Three long, straight, parallel wires all lie in the yz plane and each carries a current of 20 A in the positive z direction. The two outer wires are each 4.0 cm from the center wire. What is the magnitude of the magnetic force on a 50-cm length of either of the outer wires
Final answer:
The magnitude of the magnetic force on a wire can be calculated using the formula F = B * I * L. In this case, the magnetic field created by the two outer wires can be calculated using Ampere's Law. Plugging in the values will give you the magnitude of the magnetic force on the wire.
Explanation:
The magnitude of the magnetic force on a wire can be calculated using the formula:
F = B * I * L
Where F is the magnetic force, B is the magnetic field, I is the current, and L is the length of the wire.
In this case, the magnetic field created by the two outer wires can be calculated using Ampere's Law:
B = (μ0 * I) / (2π * r)
Where μ0 is the permeability of free space, I is the current, and r is the distance from the wire.
Plugging in the values, the magnetic force per meter on either of the outer wires is:
F = (2 * 10-7 Tm/A * 20 A * 0.5 m) / (2π * 0.04 m)
Calculating this will give you the magnitude of the magnetic force on the wire.
Before the experiment, the total momentum of the system is 2.5 kg m/s to the right and the kinetic energy is 5J. After the experiment, the total momentum of the system is 2.5 kg m/s to the right and the kinetic energy is 4J.
a)This describes an elastic collision (and it could NOT be inelastic).
b)This describes an inelastic collision (and it could NOT be elastic).
c)This is NEITHER an elastic collision nor an inelastic collision
d)This describes a collision that is EITHER elastic or inelastic, but more information is required to determine which.
Answer:
B
Explanation:
Newton's third law of motion states that to every action there is equal an opposite reaction. Momentum is always conserved provided there is no net force on the body. Considering the experiment: the momentum is conserved as expected but the kinetic energy is not conserved meaning that the collision is not elastic; some energy is converted into internal energy. When the collision is elastic the total kinetic energy before will equal total kinetic energy after and when the body stick together, they move with a common velocity and that described a perfectly inelastic collision. 5J ≠ 4J
Final answer:
The scenario describes an inelastic collision because while the momentum is conserved, the kinetic energy decreases from 5J to 4J, indicating that not all kinetic energy is conserved in the collision.
Explanation:
When examining collisions in physics, an elastic collision is one in which both momentum and kinetic energy are conserved, whereas an inelastic collision is one where momentum is conserved but kinetic energy is not. Given that in the scenario the total momentum of the system remains the same before and after the collision, that part of the conservation law is satisfied in both cases. However, since the kinetic energy of the system decreases from 5J to 4J, this means that some of the kinetic energy has been transformed into other forms of energy, like heat or sound, which typically happens during an inelastic collision.
Therefore, the correct answer is (b): This describes an inelastic collision (and it could NOT be elastic) since the total kinetic energy of the system is not the same before and after the collision. An elastic collision would have required that the kinetic energy also remain constant.