Answer:
Distance from the center of wire is 0.05 meters.
Explanation:
It is given that,
Current flowing in the wire, I = 2 A
Magnetic field, [tex]B=8\ \mu T=8\times 10^{-6}\ T[/tex]
Let d is the distance from the center of the wire. The magnetic field at a distance d from the wire is given by :
[tex]B=\dfrac{\mu_oI}{2\pi d}[/tex]
[tex]d=\dfrac{\mu_oI}{2\pi B}[/tex]
[tex]d=\dfrac{4\pi \times 10^{-7}\times 2\ A}{2\pi \times 8\times 10^{-6}\ T}[/tex]
d = 0.05 meters
So, the distance from the wire is 0.05 meters. Hence, this is the required solution.
To find the distance from the wire to point P where the magnetic field is 8 μT due to a 2-A current, the formula B = μ₀I/(2πR) is used, and the calculation reveals that R = 0.1 mm.
To, calculating the distance from the center of the wire to a point P where the magnetic field due to a 2-A current flowing in a long, straight wire is 8 μT.
To find this distance, we use the formula for the magnetic field around a long straight wire, given by B = μ₀I/(2πR), where B is the magnetic field strength, μ₀ is the permeability of free space (4π x 10⁻⁷ Tm/A), I is the current, and R is the distance from the wire.
Plugging the given values into this equation, we get:
8 x 10⁻⁶ T = (4π x 10⁻⁷ Tm/A)(2A) / (2πR)
Simplifying, we find that R = (4π x 10⁻⁷ Tm/A)(2A) / (2π x 8 x 10⁻⁶ T) = 0.0001 m or 0.1 mm.
A 300g marble is travelling at a speed of 3.00 m/s horizontally along a tabletop that is 1.00 m above the floor. The marble rolls up a ramp that launches it into the air. At the highest point in its flight the marble has a horizontal velocity of 1.10 m/s. If friction is negligible, how high above the floor is the marble at its highest point?
Answer:
The marble at its highest point are hmax= 1.39 meters above the floor.
Explanation:
hi= 1m
V= 3m/s
Vx= 1.1 m/s
Vy= ?
α=cos⁻¹(1.1/3)
α= 68.48º
Vy= V * sin(α)
Vy= 2.79 m/s
Vertical Velocity in highest point are Vy(hmax) = 0
Vy(hmax)= Vy - g*t
t= Vy/g
t= 0.28 sec
t : time to get the highest point
hmax= hi + Vy*t - g*t²/2
hmax= 1.39 m
What is the minimum uncertainty in the energy state of an atom if an electron remains in this state for 10 s
Answer:
The minimum uncertainty in the energy state of an atom is [tex]1.0557\times10^{-35}\ J[/tex]
Explanation:
Given that,
Time t = 10 s
We need to calculate the minimum uncertainty in the energy state
Using Hisen burg of uncertainty principle equation
[tex]\Delta E\Delta t=\dfrac{h}{2\pi}[/tex]
[tex]\Delta E=\dfrac{h}{2\pi}\times\dfrac{1}{\Delta t}[/tex]
Where, E = energy
t = time
Put the value into the formula
[tex]\Delta E=\dfrac{6.63\times10^{-34}}{2\times3.14}\times\dfrac{1}{10}[/tex]
[tex]\Delta E=1.0557\times10^{-35}\ J[/tex]
Hence, The minimum uncertainty in the energy state of an atom is [tex]1.0557\times10^{-35}\ J[/tex]
Final answer:
The minimum uncertainty in the energy state of an atom can be found using the Heisenberg Uncertainty Principle. If an electron remains in a state for 10 seconds, the minimum uncertainty in its energy is approximately 5.3x10^-25 J. The uncertainty principle states that there is a fundamental limit to how precisely we can simultaneously know certain pairs of physical properties, such as energy and time.
Explanation:
The minimum uncertainty in the energy state of an atom can be found using the Heisenberg Uncertainty Principle. The uncertainty in energy (AE) is given by the equation AEAt ≥ h/4π, where At is the uncertainty in time and h is Planck's constant. In this case, if the electron remains in the state for 10 s (At = 10 s), the minimum uncertainty in the energy can be calculated. Substituting the known values into the equation, we find the minimum uncertainty in energy to be approximately 5.3x10-25 J.
The uncertainty principle states that there is a fundamental limit to how precisely we can simultaneously know certain pairs of physical properties, such as energy and time. The uncertainty in energy is inversely proportional to the uncertainty in time. Therefore, the longer the electron remains in the energy state, the smaller the uncertainty in energy will be. This means that the more stable the state, the more accurately we can determine its energy value.
It is important to note that the uncertainty in energy is relatively small compared to typical excitation energies in atoms, which are on the order of 1 eV. Therefore, the uncertainty principle has a minimal effect on the accuracy with which we can measure the energy of such states.
On a snowy day, when the coefficient of friction μs between a car’s tires and the road is 0.50, the maximum speed that the car can go around a curve is 20 mph. What is the maximum speed at which the car can take the same curve on a sunny day when μs=1.0?
Answer:
28.1 mph
Explanation:
The force of friction acting on the car provides the centripetal force that keeps the car in circular motion around the curve, so we can write:
[tex]F=\mu mg = m \frac{v^2}{r}[/tex] (1)
where
[tex]\mu[/tex] is the coefficient of friction
m is the mass of the car
g = 9.8 m/s^2 is the acceleration due to gravity
v is the maximum speed of the car
r is the radius of the trajectory
On the snowy day,
[tex]\mu=0.50\\v = 20 mph = 8.9 m/s[/tex]
So the radius of the curve is
[tex]r=\frac{v^2}{\mu g}=\frac{(8.9)^2}{(0.50)(9.8)}=16.1 m[/tex]
Now we can use this value and re-arrange again the eq. (1) to find the maximum speed of the car on a sunny day, when [tex]\mu=1.0[/tex]. We find:
[tex]v=\sqrt{\mu g r}=\sqrt{(1.0)(9.8)(8.9)}=12.6 m/s=28.1 mph[/tex]
The maximum speed at which the car can take the same curve on a sunny day is about 28 mph
[tex]\texttt{ }[/tex]
Further explanationCentripetal Acceleration can be formulated as follows:
[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]
a = Centripetal Acceleration ( m/s² )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
[tex]\texttt{ }[/tex]
Centripetal Force can be formulated as follows:
[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]
F = Centripetal Force ( m/s² )
m = mass of Particle ( kg )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
coefficient of friction on a snowy day = μs₁ = 0.50
maximum speed of the car on a snowy day = v₁ = 20 mph
coefficient of friction on a sunny day = μs₂ = 1.0
Asked:
maximum speed of the car on a snowy day = v₂ = ?
Solution:
Firstly , we will derive the formula to calculate the maximum speed of the car:
[tex]\Sigma F = ma[/tex]
[tex]f = m \frac{v^2}{R}[/tex]
[tex]\mu N = m \frac{v^2}{R}[/tex]
[tex]\mu m g = m \frac{v^2}{R}[/tex]
[tex]\mu g = \frac{v^2}{R}[/tex]
[tex]v^2 = \mu g R[/tex]
[tex]\boxed {v = \sqrt { \mu g R } }[/tex]
[tex]\texttt{ }[/tex]
Next , we will compare the maximum speed of the car on a snowy day and on the sunny day:
[tex]v_1 : v_2 = \sqrt { \mu_1 g R } : \sqrt { \mu_2 g R }[/tex]
[tex]v_1 : v_2 = \sqrt { \mu_1 } : \sqrt { \mu_2 }[/tex]
[tex]20 : v_2 = \sqrt { 0.50 } : \sqrt { 1.0 }[/tex]
[tex]20 : v_2 = \frac{1}{2} \sqrt{2}[/tex]
[tex]v_2 = 20 \div \frac{1}{2} \sqrt{2}[/tex]
[tex]v_2 = 20 \sqrt{2} \texttt{ mph}[/tex]
[tex]\boxed{v_2 \approx 28 \texttt{ mph}}[/tex]
[tex]\texttt{ }[/tex]
Learn moreImpacts of Gravity : https://brainly.com/question/5330244Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454The Acceleration Due To Gravity : https://brainly.com/question/4189441[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Circular Motion
A camera of weight 8.9 newtons is dropped from a drone at a height of 18.6 meters and enters free fall. Assuming no air resistance, what is the final velocity of the camera a moment before it shatters on the ground?
Answer:
19.09 m/s
Explanation:
u = 0, h = 18.6 m
Use third equation of motion
v^2 = u^2 + 2 g h
v^2 = 0 + 2 x 9.8 x 18.6
v = 19.09 m/s
A space station of diameter 20.0 meters is turning about its axis to simulate gravity at its center rim. How fast must it rotate to produce an outer rim acceleration of 9.80 m/s^2 ?
Answer:
9.89 m/s
Explanation:
d = diameter of the space station = 20.0 m
r = radius of the space station
radius of the space station is given as
r = (0.5) d
r = (0.5) (20.0)
r = 10 m
a = acceleration produced at outer rim = 9.80 m/s²
v = speed at which it rotates
acceleration is given as
[tex]a = \frac{v^{2}}{r}[/tex]
[tex]9.80 = \frac{v^{2}}{10}[/tex]
v = 9.89 m/s
A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding?
Answer:
Coefficient of static friction between the road and the car's tire is 0.81
Explanation:
It is given that,
Velocity of the car, v = 40 m/s
Radius of the curve, r = 200 m
We need to find the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding. Let it is equal to μ.
The centripetal force of the car is balanced by the force of friction as :
[tex]\dfrac{mv^2}{r}=\mu mg[/tex]
[tex]\mu=\dfrac{v^2}{rg}[/tex]
[tex]\mu=\dfrac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}[/tex]
[tex]\mu=0.81[/tex]
So, the coefficient of static friction between the road and the car's tire is 0.81 Hence, this is the required solution.
The minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding is 0.816.
Given the data in the question;
Velocity of car; [tex]v = 40.0m/s[/tex]Radius of curve; [tex]r = 200 m[/tex]For a not car not to slide of the road, the frictional force and centripetal force should balance each other.
That is; Frictional force = Centripetal force
[tex]Frictional\ Force = uF = umg \\\\Centripetal\ Force = \frac{mv^2}{r}[/tex]
So,
[tex]u_{min}mg = \frac{mv^2}{r} \\\\u_{min}g =\frac{v^2}{r}\\\\u_{min} =\frac{v^2}{gr}\\[/tex]
Where [tex]u_{min}[/tex] is the minimum coefficient friction, v is the velocity, r is the radius and g is acceleration due to gravity( [tex]g = 9.8m/s^2\\[/tex])
We substitute our values into the equation
[tex]u_{min} = \frac{(40.0m/s)^2}{9.8m/s^2\ *\ 200m}\\\\ u_{min} = \frac{1600m^2/s^2}{1960m^2/s^2} \\\\u_{min} = 0.81[/tex]
Therefore, the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding is 0.816
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What is the pressure ratio for a sound that is 90 dB SPL?
Solution:
To calculate the pressure ratio for a sound for measuring sound, we use the following logarithmic formula of Sound Pressure Level (SPL):
SPL = [tex]20\log_{10}\frac{p}{p_{ref}} dB[/tex]
where,
p = pressure to be measured
[tex]p_{ref} [/tex] = reference pressure
[tex]\frac{90}{20}[/tex] = [tex]\log_{10}\frac{p}{p_{ref}} dB[/tex]
[tex]10^{4.5}[/tex] = [tex]\frac{p}{p_{ref}}[/tex]
pressure ratio is :
[tex]\frac{p}{p_{ref}}[/tex] = [tex]3.16\times 10^{5}[/tex]
Final answer:
A 90 dB sound has a pressure ratio [tex]10^{4.5[/tex] times greater than the threshold of hearing (0 dB). Every 10 dB increase corresponds to doubling the sound pressure level, so a 90 dB sound is exponentially more intense than the reference level.
Explanation:
To compute the pressure ratio for a sound that is 90 dB SPL (Sound Pressure Level), we must first understand the relationship between decibels and pressure ratios. The reference pressure for 0 dB is 20 micropascals, which is equivalent to 10⁻⁹ atm. A change of 20 dB represents a tenfold increase in the pressure amplitude.
Since every 10 dB increase corresponds to the sound pressure level doubling, a 90 dB sound is 9 factors of 10 or 10⁹ times more intense than the threshold of hearing (0 dB). Therefore, the pressure ratio of a 90 dB sound compared to the reference sound (0 dB) is 10⁹ times greater.
To confirm, every 20 dB increase multiplies the pressure ratio by 10, so:
0 dB = 1 (reference level)
20 dB = 10
40 dB = 10²
60 dB = 10³
80 dB = 10
90 dB =[tex]10^{4.5[/tex]
Thus, a 90 dB sound signal is 1[tex]10^{4.5[/tex] times the pressure of the reference signal.
(b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?
Answer:
Explanation:
a) using the energy conservation equation
mgh = 0.5mv^2 + 0.5Iω^2
I(moment of inertia) (basket ball) = (2/3)mr^2
mgh = 0.5mv^2 + 0.5( 2/3mr^2) ( v^2/r^2)
gh = 1/2v^2 + 1/3v^2
gh = v^2( 5/6)
v = [tex]\sqrt{\frac{6gh}{5} }[/tex]
putting the values we get
[tex]6.6 ^{2} = \frac{6\times9.8h}{5}[/tex]
solving for h( height)
h = 3.704 m apprx
b) velocity of solid cylinder
mgh = 0.5mv^2 + 0.5( mr^2/2)( v^2/r^2) where ( I ofcylinder = mr^2/2)
g*h = 1/2v^2 + 1/4v^2
g*h = 3/4v^2
putting the value of h and g we get
v= = 6.957 m/s apprx
A projectile is fired into the air at a nonzero angle with the horizontal. When the projectile reaches its maximum height, the speed of the projectile is 50.8% of its original speed. What angle was the projectile originally fired at?
Answer:
59.5 deg
Explanation:
[tex]v[/tex] = original speed at which the projectile is launched
θ = angle of launch of projectile
[tex]v_{x}[/tex] = component of speed along the horizontal direction = [tex]v [/tex] Cosθ
At the highest position, the vertical component of velocity becomes zero and there is only horizontal component of velocity, hence
[tex]v_{highest}[/tex] = velocity at the highest point = [tex]v_{x}[/tex] = [tex]v [/tex] Cosθ
it is given that
[tex]v_{highest}[/tex] = 0.508 [tex]v[/tex]
so
[tex]v[/tex] Cosθ = 0.508 [tex]v[/tex]
Cosθ = 0.508
θ = 59.5 deg
A 250-lb block is subjected to a horizontal force P. The coefficient of friction between the block and surface is µs = 0.2. Determine the force P required to start moving the block up the incline
Answer:
force required to push the block = 219.714 lb
Explanation:
GIVEN DATA:
weight W of block = 250 lb
coefficient of friction = 0.2
consider equilibrium condition in x direction
[tex]P*cos(30)-W*sin(30)-\mu _{s}N = 0[/tex]
[tex]P*0.866-0.2N = 125[/tex].........................(1)
consider equilibrium condition in Y direction
[tex]N-Wcos(30)-Psin(30)= 0[/tex]
[tex]N-0.5P=216.503[/tex].....................(2)
SOLVING 1 and 2 equation we get N value
N = 326.36 lb
putting N value in either equation we get force required to push the block = 219.714 lb
Kevin has a mass of 87- kg and is skating with in-line skates. He sees his 22-kg younger brother up ahead standing on the sidewalk, with his back tuned. Coming up, grabs his just before he brother and rolls off at a speed of 2.4 m/s. Ignoring friction, find Kevin speed just before he grabbed his brother.
Answer:
3 m/s
Explanation:
M1 = 87 kg, u1 = ?
M2 = 22 kg, u = 0
After grabbing, let the velocity is V.
V = 2.4 m/s
By using the conservation of momentum
Momentum before grabbing = momentum after grabbing
M1 x u1 + M2 x u2 = (M1 + M2) x V
87 x u1 + 0 = (87 + 22) x 2.4
87 u1 = 261.6
u1 = 3 m/s
Components of some computers communicate with each other through optical fibers having an index of refraction n =1.55. What time in nanoseconds is required for a signal to travel 0.35 m through such a fiber? Your answer should be a number with two decimal places, do not include the unit.
Answer: 1.00
Explanation:
The index of refraction [tex]n[/tex] is a number that describes how fast light propagates through a medium or material.
Being its equation as follows:
[tex]n=\frac{c}{v}[/tex] (1)
Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum and [tex]v[/tex] its speed in the other medium and [tex]n=1.55[/tex].
So, from (1) we can find the velocity at which the signal travels and then the time it requires to travel:
[tex]v=\frac{c}{n}[/tex] (2)
[tex]v=\frac{3(10)^{8}m/s}{1.55}[/tex] (3)
[tex]v=193548387.1m/s[/tex] (4)
Now, knowing the velocity [tex]v[/tex] is the distance [tex]d=0.35m[/tex] traveled in a time [tex]t[/tex]:
[tex]v=\frac{d}{t}[/tex] (5)
We can isolate [tex]t[/tex] from (5) and find the value of the required time:
[tex]t=\frac{d}{v}[/tex] (6)
[tex]t=\frac{0.35m}{193548387.1m/s}[/tex] (7)
[tex]t=0.000000001s=1(10)^{-9}s=1ns[/tex] (8) This is the time it takes the signal to travel through the optical fiber: 1 nanosecond.
Final answer:
The time for a signal to travel 0.35 m through an optical fiber with an index of refraction of 1.55 is approximately 1.81 nanoseconds.
Explanation:
To calculate the time required for a signal to travel through an optical fiber with an index of refraction of 1.55, we use the equation that relates the speed of light in a vacuum (c) to the speed of light in a material (v) using the index of refraction (n): v = c/n. Given that the speed of light in a vacuum is approximately 3.00×108 m/s, we can find the speed of light in the optical fiber.
The speed of light in the fiber (v) is:
v = (3.00 × 108 m/s) / 1.55 = 1.935×108 m/s.
Now, we calculate the time (t) it takes for light to travel 0.35 m in the fiber, using the formula
t = distance/speed:
t = 0.35 m / (1.935 × 108 m/s).
The calculated time t is found to be approximately 1.81 nanoseconds, after converting from seconds by multiplying by 109.
If the absolute pressure of a gas is 550.280 kPa, its gage pressure is A. 101.325 kPa. B. 651.605 kPa. C. 448.955 kPa. D. 277.280 kPa.
Answer:
Option C is the correct answer.
Explanation:
Absolute pressure is sum of gauge pressure and atmospheric pressure.
That is
[tex]P_{abs}=P_{gauge}+P_{atm}[/tex]
We have
[tex]P_{abs}=550.280 kPa\\\\P_{atm}=1atm=101325Pa=101.325kPa[/tex]
Substituting
[tex]P_{abs}=P_{gauge}+P_{atm}\\\\550.280=P_{gauge}+101.325\\\\P_{gauge}=448.955kPa[/tex]
Option C is the correct answer.
Professional Application: A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal into it. (a) What is the final velocity of the loaded freight car?
Answer:
0.182 m/s
Explanation:
m1 = 30,000 kg, m2 = 110,000 kg, u1 = 0.85 m/s
let the velocity of loaded freight car is v
Use the conservation of momentum
m1 x u1 + m2 x 0 = (m1 + m2) x v
30,000 x 0.85 = (30,000 + 110,000) x v
v = 0.182 m/s
The potential difference between the accelerating plates of a TV set is about 25 kV. If the distance between the plates is 1.0 cm, find the magnitude of the uniform electric field in the region between the plates.
Answer:
E = 2.5 x 10⁶ N/C
Explanation:
V = Potential difference between the plates of a TV set = 25 kV = 25000 Volts
d = Distance between the plates of TV set = 1.0 cm = 0.01 m
E = Electric field in the region between the plates
Electric field between the plates is given as
[tex]E= \frac{V}{d}[/tex]
Inserting the values
[tex]E= \frac{25000}{0.01}[/tex]
E = 2.5 x 10⁶ N/C
The uniform electric field is the field in which the value of the electric field strength remain same at each point.
The magnitude of the uniform electric field in the region between the plates [tex]2.5\times10^6 \rm N/C[/tex].
What is uniform electric field?The uniform electric field is the field in which the value of the electric field strength remain same at each point.
It can be given as,
[tex]E=\dfrac{V}{d}[/tex]
Here, [tex]V[/tex] is the potential difference between two points and [tex]d[/tex] is the distance between two points.
Given information-
The potential difference between the accelerating plates of a TV set is about 25 k-V or 25000 V.
The distance between the plates is 1.0 cm or 0.01 meters.
Use the above formula to find the magnitude of the uniform electric field in the region between the plates as,
[tex]E=\dfrac{25000}{0.01}\\E=2.5\times10^6 \rm N/C[/tex]
Thus the magnitude of the uniform electric field in the region between the plates [tex]2.5\times10^6 \rm N/C[/tex].
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An object moving at a constant speed requires 4.0 s to go once around a circle with a diameter of 5.0 m. What is the magnitude of the instantaneous acceleration of the particle during this time? O2.20 m/s2 3.93 m/s2 6.17 m/s2 12.3m/2 15.4 m/g2
The magnitude of the instantaneous acceleration of the object is 12.3 m/s².
Explanation:The magnitude of the instantaneous acceleration of the object can be calculated using the formula:
a = v² / r
Where v is the speed and r is the radius of the circle. In this case, the speed is the distance traveled divided by the time taken, which is equal to the circumference of the circle divided by the time taken:
v = (2πr) / t
Substituting the values given in the question, we can calculate the acceleration:
a = [(2πr) / t]² / r = (4π²r²) / t² = (4π² * (2.5)²) / (4)² = 12.3 m/s²
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The instantaneous acceleration of a particle moving in a circular path of diameter 5.0 m at a constant speed, completing the path in 4.0 s, is approximately 6.17 m/s².
Explanation:The subject of this question is the instantaneous acceleration of a particle moving in a circle, which is a topic within physics, specifically, kinematics. This is a case of uniform circular motion, which implies a constant speed, but changing direction, thereby resulting in centripetal (towards the center) acceleration.
The radius of the circle is half of its diameter: r = 5.0 m/2 = 2.5 m. As the particle completes a full circle in 4.0 s, the speed can be calculated using circumference (2πr) and time: v = 2πr/t = 2π * 2.5 m/4.0 s ≈ 3.92 m/s. The centripetal acceleration (the instantaneous acceleration for a particle in a uniform circular motion) can be calculated as: a = v²/r = (3.92 m/s)²/2.5 m ≈ 6.17 m/s².
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Physicist S. A. Goudsmit devised a method for measuring the mass of heavy ions by timing their period of revolution in a known magnetic field. A singly charged ion makes 5.00 revolutions in a 50.0 mT field in 1.11 ms. Calculate its mass, in atomic mass units.
Answer:
170.36 amu
Explanation:
Ion makes 5 revolutions in 1.11 ms
Frequency = 5 / (1.11 x 10^-3) = 4504.5 rps
B = 50 mT = 0.05 T
q = 1.6 x 10^-19 C
Let m be the mass in kg
Time period is given by
T = (2 π m) / (B q)
Frequency is the reciprocal of time period.
f = 1 / T = B q / 2πm
So,
m = B q / 2 π f
m = (0.05 x 1.6 x 10^-19) / ( 2 x 3.14 x 4504.5) = 2.828 x 10^-25 kg
as we know that
1 amu = 1.66 x 10^-27 kg
So, m = (2.828 x 10^-25 ) / (1.66 x 10^-27) = 170.36 amu
The person has an upward acceleration of 0.70 m/s^2 and is lifted from rest through a distance of 13 m. A rescue helicopter lifts a 80-kg person straight up by means the cable? (a) What is the tension in the cable? (b) How much work is done by the tension in the cable? (c) How much work is done by the person's weight? (d) Use the work-energy theorem and find the final speed of the person.
Answer:
a)
840 N
b)
10920 J
c)
- 10192 J
d)
4.3 m/s
Explanation:
a)
T = tension force in the cable in upward direction = ?
a = acceleration of the person in upward direction = 0.70 m/s²
m = mass of the person being lifted = 80 kg
Force equation for the motion of person in upward direction is given as
T - mg = ma
T = m (g + a)
T = (80) (9.8 + 0.70)
T = 840 N
b)
d = distance traveled in upward direction = 13 m
[tex]W_{t}[/tex] = Work done by tension force
Work done by tension force is given as
[tex]W_{t}[/tex] = T d
[tex]W_{t}[/tex] = (840) (13)
[tex]W_{t}[/tex] = 10920 J
c)
d = distance traveled in upward direction = 13 m
[tex]W_{g}[/tex] = Work done by person's weight
Work done by person's weight is given as
[tex]W_{g}[/tex] = - mg d
[tex]W_{g}[/tex] = - (80 x 9.8) (13)
[tex]W_{g}[/tex] = - 10192 J
d)
[tex]F_{net}[/tex] = Net force on the person = ma = 80 x 0.70 = 56 N
v₀ = initial speed of the person = 0 m/s
v = final speed
Using work-energy theorem
[tex]F_{net}[/tex] d = (0.5) m (v² - v₀²)
(56) (13) = (0.5) (80) (v² - 0²)
v = 4.3 m/s
In SI units, the electric field in an electromagnetic wave is described by Ey = 112 sin(1.40 107x − ωt). (a) Find the amplitude of the corresponding magnetic field oscillations. µT (b) Find the wavelength λ. µm (c) Find the frequency f. Hz
Explanation:
The electric field in an electromagnetic wave is described by:
[tex]E_y=112\ sin(1.4\times 10^7x-\omega t)[/tex]
The general equation is given by :
[tex]E=E_0\ sin(kx-\omega t)[/tex]
Here,
The amplitude in electric field is [tex]E_0=112[/tex]
Propagation constant, [tex]k=1.4\times 10^7[/tex]
[tex]\omega[/tex] is the angular frequency
(a) The amplitude of the corresponding magnetic field oscillations is given by :
[tex]B_0=\dfrac{E_0}{c}[/tex]
[tex]B_0=\dfrac{112}{3\times 10^8}[/tex]
[tex]B_0=3.73\times 10^{-7}\ T[/tex]
[tex]B_0=0.37\ \mu T[/tex]
(b) We know that the propagation constant is given by :
[tex]k=\dfrac{2\pi}{\lambda}[/tex]
[tex]\lambda=\dfrac{2\pi}{k}[/tex]
[tex]\lambda=\dfrac{2\pi}{1.4\times 10^7}[/tex]
[tex]\lambda=4.48\times 10^{-7}\ m[/tex]
[tex]\lambda=0.44\ \mu T[/tex]
(c) [tex]f=\dfrac{c}{\lambda}[/tex]
[tex]f=\dfrac{3\times 10^8\ m/s}{4.48\times 10^{-7}\ m}[/tex]
[tex]f=6.69\times 10^{14}\ Hz[/tex]
Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.070m2, and the magnitude of the fluid velocity is 3.50m/s.
part A
What is the fluid speed at point in the pipe where the cross-sectional area is 0.105m2?
part b
What is the fluid speed at point in the pipe where the cross-sectional area is 0.047m2?
part c
Calculate the volume of water discharged from the open end of the pipe in 1.00hour.
Answer:
a) 2.33 m/s
b) 5.21 m/s
c) 882 m³
Explanation:
Using the concept of continuity equation
for flow through pipes
[tex]A_{1}\times V_{1} = A_{2}\times V_{2}[/tex]
Where,
A = Area of cross-section
V = Velocity of fluid at the particular cross-section
given:
[tex]A_{1} = 0.070 m^{2}[/tex]
[tex]V_{1} = 3.50 m/s[/tex]
a) [tex]A_{2} = 0.105 m^{2}[/tex]
substituting the values in the continuity equation, we get
[tex]0.070\times 3.50 = 0.105\times V_{2}[/tex]
or
[tex]V_{2} = \frac{0.070\times 3.5}{0.150}m/s[/tex]
or
[tex]V_{2} = 2.33m/s[/tex]
b) [tex]A_{2} = 0.047 m^{2}[/tex]
substituting the values in the continuity equation, we get
[tex]0.070\times 3.50 = 0.047\times V_{2}[/tex]
or
[tex]V_{2} = \frac{0.070\times 3.5}{0.047}m/s[/tex]
or
[tex]V_{2} = 5.21m/s[/tex]
c) we have,
Discharge[tex]Q = Area (A)\times Velocity(V)[/tex]
thus from the given value, we get
[tex]Q = 0.070m^{2}\times 3.5m/s\[/tex]
[tex]Q = 0.245 m^{3}/s[/tex]
Also,
Discharge[tex]Q = \frac{volume}{time}[/tex]
given time = 1 hour = 1 ×3600 seconds
substituting the value of discharge and time in the above equation, we get
[tex]0.245m^{3}/s = \frac{volume}{3600s}[/tex]
or
[tex]0.245m^{3}/s\times 3600 = Volume[/tex]
volume of flow = [tex]882 m^{3}[/tex]
Using the continuity equation for incompressible fluids, we calculated the velocity of water at points in a pipe with different cross-sectional areas and also determined the volume of water discharged from the pipe in one hour.
Explanation:To answer the fluid dynamics question, we will use the principle of conservation of mass, specifically the continuity equation for incompressible fluids, which states that the product of the cross-sectional area (A) and the velocity (v) of the fluid must remain constant at all points in the flow. This can be written as A1 * v1 = A2 * v2, where A1 and v1 are the area and velocity at point 1, and A2 and v2 are the corresponding values at point 2.
Part ATo find the fluid speed at the point where the cross-sectional area is 0.105 m2, we use the provided information:
A1 = 0.070 m2v1 = 3.50 m/sA2 = 0.105 m2Thus, v2 = (A1 * v1) / A2 = (0.070 m2 * 3.50 m/s) / 0.105 m2 = 2.33 m/s
Part BNext, to find the fluid speed at the point where the cross-sectional area is 0.047 m2, we have:
A3 = 0.047 m2The velocity at this point can be calculated as v3 = (A1 * v1) / A3 = (0.070 m2 * 3.50 m/s) / 0.047 m2 = 5.21 m/s
Part CFor the volume discharged from the open end of the pipe in 1.00 hour, we use the flow rate at point 1, with A1 and v1. The flow rate Q1 = A1 * v1 = 0.070 m2 * 3.50 m/s = 0.245 m3/s. To get the volume for one hour, we need to convert seconds to hours, knowing that there are 3600 seconds in one hour:
Volume = Q1 * time = 0.245 m3/s * 3600 s = 882 m3
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A balloon having an initial temperature of 17.8°C is heated so that the volume doubles while the pressure is kept fixed. What is the new value of the temperature?
Answer:
[tex]T = 308.6 ^0 C[/tex]
Explanation:
Here by ideal gas equation we can say
[tex]PV = nRT[/tex]
now we know that pressure is kept constant here
so we will have
[tex]V = \frac{nR}{P} T[/tex]
since we know that number of moles and pressure is constant here
so we have
[tex]\frac{V_2}{V_1} = \frac{T_2}{T_1}[/tex]
now we know that initial temperature is 17.8 degree C
and finally volume is doubled
So we have
[tex]\frac{2V}{V} = \frac{T_2}{(273 + 17.8)}[/tex]
so final temperature will be
[tex]T_2 = 581.6 k[/tex]
[tex]T_2 = 308.6 ^o C[/tex]
A 30-μF capacitor is charged to 80 V and then connected across an initially uncharged capacitor of unknown capacitance C. If the final potential difference across the 30-μF capacitor is 20 V, determine C.
The unknown capacitance C connected parallelly with a 30-μF capacitor initially charged at 80 V is calculated to be 90 μF. After being connected, the total potential difference across both capacitors reduces to 20 V.
Explanation:The question deals with capacitors and how they function in a circuit. A capacitor is an electronic component that stores electrical energy and releases it in the circuit when necessary. When a charged 30-μF capacitor is connected across an initially uncharged capacitor, the voltage or potential difference across the connected capacitors will equalize. In this case, the final potential difference is noted as 20 V.
We know that for capacitors connected in parallel, the total charge stored is the sum of the charges stored in each capacitor, i.e., Q = Q₁ + Q₂, where Q is the total charge, and Q₁ and Q₂ are the charges stored in capacitor 1 (the 30-μF capacitor) and 2 (unknown capacitance C), respectively.
By using the formula Q = CV (charge = capacitance × voltage), where C is the capacitance and V is the voltage, we understand that Q₁ (charge on the 30 uF capacitor) is initially 30 μF × 80 V = 2400 μC (micro coulombs). After connecting the uncharged capacitor, the voltage drops to 20V, thus the final charge on the 30 uF capacitor becomes 30 μF × 20 V = 600 μC. The remaining charge must then be stored in the previously uncharged capacitor.
So, the charge on the unknown capacitor C would be the difference i.e., 2400 μC - 600 μC = 1800 μC. Now using the Q = CV formula, we can calculate the unknown capacitance C. V here is 20V as capacitors connected in parallel have the same voltage. So, C = Q / V = 1800 μC / 20 V = 90 μF. Thus, the unknown capacitance is 90 μF.
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A person who weighs 846 N steps onto a spring scale in the bathroom, and the spring compresses by 0.574 cm. (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by 0.314 cm?
Explanation:
It is given that,
Weight of the person, W = F = 846 N
When the person steps onto a spring scale in the bathroom, the spring compresses by 0.574 cm, x = 0.00574 m
(a) The force acting on the spring is is given by Hooke's law as :
[tex]F=-kx[/tex]
[tex]k=\dfrac{F}{x}[/tex]
[tex]k=\dfrac{846\ N}{0.00574\ m}[/tex]
k = 147386.7 N/m
(b) If the spring is compressed by 0.314 cm or 0.00314 m, weight of the person is given by again Hooke's law as :
[tex]F=kx[/tex]
[tex]F=147386.7\ N/m\times 0.00314\ m[/tex]
F = 462.7 N
Hence, this is the required solution.
A 81 cm long brass rod has a diameter of 3 mm. The temperature of one end is 50 degrees higher than the other end. How much heat is conducted in 1.8 mins?
Answer:
5.14 J
Explanation:
l = 81 cm = 0.81 m
diameter = 3 mm
Radius, r = 1.5 mm = 1.5 x 10^-3 m
Change in temperature, T2 - T1 = 50 degree
t = 1.8 minutes = 1.8 x 60 = 108 s
k = 109 W/mk
Heat transfer = K A (T2 - T1) t / l
H = 109 x 3.14 x 1.5 x 10^-3 x 1.5 x 10^-3 x 50 x 108 / 0.81
H = 5.14 J
Compute the binomial expansion for (1+x)^5
Answer:
1+5x+10x^2+10x^3+5x^4+x^5
Explanation:
You just make it (1+x)(1+x)(1+x)(1+x)(1+x) and multiply it out until it's all one big term.
A 32.1-g ice cube at 0 °C is added to 120 g of water in a 64.2-g aluminum cup. The cup and the water have an initial temperature of 23.5 °C. Calculate the equilibrium temperature of the cup and its contents. Do not enter unit.
Answer:
[tex]T = 3.5^0 C[/tex]
Explanation:
Heat given by water + cup = Heat absorbed by the ice
here we can say that let the final temperature of the system is "T"
so we will have heat absorbed by the ice given as
[tex]Q = mL + ms\Delta T[/tex]
[tex]Q = (32.1)(335) + (32.1)(4.186)(T - 0)[/tex]
[tex]Q_{in} = 10753.5 + 134.4 T[/tex]
now we will have heat given by cup + water as
[tex]Q = m_w s_w(23.5 - T) + m_c s_c(23.5 - T)[/tex]
[tex]Q = 120(4.186)(23.5 - T) + 64.2(0.900)(23.5 - T)[/tex]
[tex]Q_{out} = 560.1(23.5 - T)[/tex]
now we have
[tex]Q_{in} = Q_{out}[/tex]
[tex]10753.5 + 134.4T = 560.1(23.5 - T)[/tex]
[tex]10753.5 + 134.4T = 13162.4 - 560.1 T[/tex]
[tex]694.5T = 2409[/tex]
[tex]T = 3.5^0 C[/tex]
A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant of 0.25 lb·s/ft and is acted on by an external force of 4 cos 2t lb. (a) Determine the steady state response of this system.
Answer:
y= 240/901 cos 2t+ 8/901 sin 2t
Explanation:
To find mass m=weighs/g
m=8/32=0.25
To find the spring constant
Kx=mg (given that c=6 in and mg=8 lb)
K(0.5)=8 (6 in=0.5 ft)
K=16 lb/ft
We know that equation for spring mass system
my''+Cy'+Ky=F
now by putting the values
0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given that C=0.25 lb.s/ft)
Lets assume that at steady state the equation of y will be
y=A cos 2t+ B sin 2t
To find the constant A and B we have to compare this equation with equation 1.
Now find y' and y" (by differentiate with respect to t)
y'= -2A sin 2t+2B cos 2t
y"=-4A cos 2t-4B sin 2t
Now put the values of y" , y' and y in equation 1
0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t
So by comparing the coefficient both sides
30 A+ B=8
A-30 B=0
So we get
A=240/901 and B=8/901
So the steady state response
y= 240/901 cos 2t+ 8/901 sin 2t
A steady-state response is the behavior of a circuit over a lengthy period of time when stable conditions have been achieved. The steady-state response of this system will be y= 240/901 cos 2t+ 8/901 sin 2t.
What is a steady-state response?A steady-state response is the behavior of a circuit over a lengthy period of time when stable conditions have been achieved following an external stimulus.
The given data in the problem will bge;
C=0.25 lb.s/ft
Weight is defined as the product of mass and gravity.
[tex]\rm{m=\frac{W}{g} }\\\\\rm{m=\frac{8}{32}[/tex]
[tex]\rm m=0.25[/tex]
Spring constant is defined as the ratio of force per unit displaced length.
The spring force is balanced by the weight;
[tex]\rm Kx=mg\\\\ \rm x= \frac{mg}{K} \\\\ \rm x=\frac{8}{0.5} lb/ft[/tex]
The equation for the spring-mass system is given by;
[tex]\rm {my''+Cy'+Ky=F }[/tex]
[tex]\rm 0.25 y"+0.25 y'+16 y=4 cos 20 t[/tex]
Steady-state equation;
[tex]\rm y=A cos 2t+ B sin 2t[/tex]
For finding the value of A and B
[tex]\rm y'= -2A sin 2t+2B cos 2ty"=-4A cos 2t-4B sin 2t[/tex]
By putting the value we got
[tex]\rm 0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t[/tex]
The value of cofficient obtained from the equation
[tex]30 A+ B=8[/tex]
Getting the value as
[tex]A= \frac{240/901}\\\\ B=\frac{8}{901}[/tex]
The steady-state response got
[tex]y= 240/901 cos 2t+ 8/901 sin 2t[/tex]
Hence the steady-state response of this system.y= 240/901 cos 2t+ 8/901 sin 2t
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A fisherman has caught a very large, 5.0 kg fish from a dock that is 2.0 m above the water. He is using lightweight fishing line that will break under a tension of 54 N or more. He is eager to get the fish to the dock in the shortest possible time.If the fish is at rest at the water's surface, what's the least amount of time in which the fisherman can raise the fish to the dock without losing it?
Answer:
The least amount of time in which the fisherman can raise the fish to the dock without losing it is t= 2 seconds.
Explanation:
m= 5 kg
h= 2m
Fmax= 54 N
g= 9.8 m/s²
W= m * g
W= 49 N
F= Fmax - W
F= 5 N
F=m*a
a= F/m
a= 1 m/s²
h= a * t²/2
t= √(2*h/a)
t= 2 seconds
The least amount of time in which the fisherman can raise the fish to the dock without losing it depends on the tension in the fishing line. The tension can be calculated using the equation T = mg, where T is the tension, m is the mass of the fish, and g is the acceleration due to gravity. As long as the tension in the line is below the breaking point, the fisherman can safely raise the fish to the dock.
Explanation:In order to determine the least amount of time in which the fisherman can raise the fish to the dock without losing it, we need to calculate the tension in the fishing line. We know that the fishing line will break under a tension of 54 N or more. The tension in the line can be calculated using the equation T = mg, where T is the tension, m is the mass of the fish, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Given that the mass of the fish is 5.0 kg, we can calculate the tension in the line as follows:
T = (5.0 kg) (9.8 m/s²) = 49.0 N
Therefore, the least amount of time in which the fisherman can raise the fish to the dock without losing it is determined by the tension in the line. As long as the tension in the line is below 54 N, the fisherman can safely raise the fish to the dock without the line breaking.
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A fish appears to be 3 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish? (n water = 1.33) A fish appears to be 3 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish? (n water = 1.33) 0.44 m 0.25 m 3.99 m 2.26 m
Answer:
3.99 m option c
Explanation:
n = 1.33
Real depth / Apparent depth = n
Real depth = n × apparent depth
Real depth = 1.33 × 3
Real depth = 3.99 m
The apparent depth of the fish is not the same as its actual depth due to refraction. The actual depth can be calculated using the equation for apparent depth and the refractive index of water. In this case, the actual depth of the fish is 2.26 m.
Explanation:The apparent depth of an object submerged in water can be different from its actual depth due to refraction. In this case, the fish appears to be 3 m below the surface when viewed from above by a fisherman. To find the actual depth, we can use the equation for apparent depth, which states that the apparent depth is equal to the actual depth divided by the refractive index of the medium.
Using this equation, we can calculate the actual depth as follows:
Actual depth = Apparent depth ÷ Refractive index of water
Given that the refractive index of water is 1.33, we can substitute the values and calculate the actual depth.
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Jupiter’s Great Red Spot rotates completely every six days. If the spot is circular (not quite true, but a reasonable approximation for this matter) and 26,000 km in diameter, what are the wind speeds (i.e. how fast the gas is moving) at the outer edges ofthe storm (in km/hr)? (Hints: Recall the definition of speed. Think about how far a cloud on the edge of the storm must travel.)
Answer:
[tex]v = 567.2 km/h[/tex]
Explanation:
As we know that if Jupiter Rotate one complete rotation then it means that the it will turn by 360 degree angle
so here the distance covered by the point on its surface in one complete rotation is given by
[tex]distance = 2\pi r[/tex]
[tex]distance = \pi D[/tex]
now we will have the time to complete the rotation given as
[tex]t = 6 days[/tex]
[tex]t = 6 (24 h) = 144 h[/tex]
now the speed is given by
[tex]speed = \frac{distance}{time}[/tex]
[tex]speed = \frac{\pi D}{t}[/tex]
[tex]speed = \frac{\pi(26000 km)}{144}[/tex]
[tex]v = 567.2 km/h[/tex]