Answer:
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
Explanation:
Given that rotor of gas turbine is decelerating at constant rate, it is required to obtained the value of angular acceleration as a function of time, as well as initial and final angular speeds. That is:
[tex]\dot n = \dot n_{o} + \ddot n \cdot t[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]t[/tex] - Time, measured in minutes.
[tex]\ddot n[/tex] - Angular acceleration, measured in revoiutions per square minute.
The angular acceleration is now cleared:
[tex]\ddot n = \frac{\dot n - \dot n_{o}}{t}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]t = 3.5\,min[/tex], the angular acceleration is:
[tex]\ddot n = \frac{0\,\frac{rev}{min}-7000\,\frac{rev}{min} }{3.5\,min}[/tex]
[tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex]
Now, the final angular speed as a function of initial angular speed, angular acceleration and the change in angular position is represented by this kinematic equation:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot (n-n_{o})[/tex]
Where [tex]n[/tex] and [tex]n_{o}[/tex] are the initial and final angular position, respectively.
The change in angular position is cleared herein:
[tex]n-n_{o} = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex], the change in angular position is:
[tex]n-n_{o} = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(7000\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-2000\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n-n_{o} = 12250\,rev[/tex]
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
A 5 kg block is sliding on a horizontal surface while being pulled by a child using a rope attached to the center of the block. The rope exerts a constant force of 28.2 N at an angle of \theta=θ = 30 degrees above the horizontal on the block. Friction exists between the block and supporting surface (with \mu_s=\:μ s = 0.25 and \mu_k=\:μ k = 0.12 ). What is the horizontal acceleration of the block?
Answer:
The horizontal acceleration of the block is 4.05 m/s².
Explanation:
The horizontal acceleration can be found as follows:
[tex] F = m \cdot a [/tex]
[tex] Fcos(\theta) - \mu_{k}N = m\cdot a [/tex]
[tex] Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)] = m\cdot a [/tex]
[tex] a = \frac{Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)]}{m} [/tex]
Where:
a: is the acceleration
F: is the force exerted by the rope = 28.2 N
θ: is the angle = 30°
[tex]\mu_{k}[/tex]: is the kinetic coefficient = 0.12
m: is the mass = 5 kg
g: is the gravity = 9.81 m/s²
[tex] a = \frac{28.2 N*cos(30) - 0.12[5 kg*9.81 m/s^{2} - 28.2 N*sen(30)]}{5 kg} = 4.05 m/s^{2} [/tex]
Therefore, the horizontal acceleration of the block is 4.05 m/s².
I hope it helps you!
A light ray in air strikes water at an angle of incidence equal to 40°. If the index of refraction for water is 1.33, what is the angle of refraction?
Answer:
The angle of refraction is 28.68°Explanation:
Given data
angle of incidence, i=40°
angle of refraction,r = ?
index of refraction u= 1.33
applying the formula
[tex]u=\frac{ sin i}{ sin r}[/tex]
According to Snell's law, the incident, normal and refracted rays all act on the same point.
Substituting and solving for r we have.
[tex]1.33= \frac{sin (40)}{sin (r)} \\\\sin(40)= 1.33* sin(r)\\\0.642= 1.33* sin(r)[/tex]
divide both sides by 1.33 we have
[tex]sin(r)= \frac{0.642}{1.33} \\\sin(r)= 0.48\\\r= sin^-^10.48\\\r= 28.68[/tex]
r= 28.68°
g Question 11 pts Consider two masses connected by a string hanging over a pulley. The pulley is a uniform cylinder of mass 3.0 kg. Initially m1 is on the ground and m2 rests 2.9 m above the ground. After the system is released, what is the speed of m2 just before it hits the ground? m1= 30 kg and m2= 35 kg Group of answer choices 2.1 m/s 1.4 m/s 9.8 m/s 4.3 m/s 1.9 m/s
Answer:
The speed of m2 just before it hits the ground is 2.1 m/s
Explanation:
mass on the ground m1 = 30 kg
mass oat rest at the above the ground m2 = 35 kg
height of m2 above the ground =2.9 m
Let the tension on the string be taken as T
for the mass m2 to reach the ground, its force equation is given as
[tex]m_{2} g - T = m_{2}a[/tex] ....equ 1
where g is acceleration due to gravity = 9.81 m/s^2
and a is the acceleration with which it moves down
For mass m1 to move up, its force equation is
[tex]T - m_{1} g = m_{1} a[/tex]
[tex]T = m_{1}a + m_{1}g[/tex]
[tex]T = m_{1}(a + g)[/tex] ....equ 2
substituting T in equ 1, we have
[tex]m_{2} g - m_{1}(a+g) = m_{2}a[/tex]
imputing values, we have
[tex](35*9.81) - 30(a+9.81) = 35a[/tex]
[tex]343.35 - 30a-294.3 = 35a[/tex]
[tex]343.35 -294.3 = 35a+ 30a[/tex]
[tex]49.05 = 65a[/tex]
a = 49.05/65 = 0.755 m/s^2
The initial velocity of mass m2 = u = 0
acceleration of mass m2 = a = 0.755 m/s^2
distance to the ground = d = 2.9 m
final velocity = v = ?
using Newton's equation of motion
[tex]v^{2}= u^{2} + 2ad[/tex]
substituting values, we have
[tex]v^{2}= 0^{2} + 2*0.755*2.9[/tex]
[tex]v^{2}= 2*0.755*2.9 = 4.379\\v = \sqrt{4.379}[/tex]
v = 2.1 m/s
At what frequency f, in hertz, would you have to move the comb up and down to produce red light, of wavelength 600 nm
Answer:
f = 500 x 10^12Hz
Explanation:
E=hc/wavelength
E=hf
hc/wavelength =hf
c/wavelength =f
f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz
Consider a block of mass equal to 10kg sliding on an inclined plane of 30°, as shown in the figure below. The coefficient of kinetic friction between the block and the plane surface is c = 0.4 (a) Determine the value of the horizontal and vertical acceleration of the block. (b) If the block starts from rest in t=0s and when it is in the X=0 and Y=5m position, calculate what its horizontal and vertical position will be at the instant t=1s. (C) How long does the LM block take to reach the base of the tilted plane?
Answer:
(a) aₓ = 1.33 m/s² and aᵧ = -0.770 m/s²
(b) x = 0.665 m and y = 4.62 m
(c) 3.61 s
Explanation:
(a) There are two ways we can solve this. The first way is to sum the forces in the x and y direction, then use the relation tan 30° = -aᵧ/aₓ, where aᵧ is the acceleration in the +y direction (up) and aₓ is the acceleration in the +x direction (right).
The second way is to sum the forces in the parallel and perpendicular directions to find the acceleration parallel to the incline, a. Then, use the relations aᵧ = -a sin 30° and aₓ = a cos 30°.
Let's try the first method. Sum of forces in the +y direction:
∑F = ma
N cos 30° + Nμ sin 30° − mg = maᵧ
N cos 30° + Nμ sin 30° − mg = -maₓ tan 30°
Sum of forces in the +x direction:
∑F = ma
N sin 30° − Nμ cos 30° = maₓ
Substituting:
N cos 30° + Nμ sin 30° − mg = -(N sin 30° − Nμ cos 30°) tan 30°
N cos 30° + Nμ sin 30° − mg = -N sin 30° tan 30° + Nμ sin 30°
N cos 30° − mg = -N sin 30° tan 30°
N (cos 30° + sin 30° tan 30°) = mg
N = mg / (cos 30° + sin 30° tan 30°)
N = (10 kg) (10 m/s²) / (cos 30° + sin 30° tan 30°)
N = 86.6 N
Now, solving for the accelerations:
N sin 30° − Nμ cos 30° = maₓ
aₓ = N (sin 30° − μ cos 30°) / m
aₓ = (86.6 N) (sin 30° − 0.4 cos 30°) / 10 kg
aₓ = 1.33 m/s²
N cos 30° + Nμ sin 30° − mg = maᵧ
aᵧ = N (cos 30° + μ sin 30°) / m − g
aᵧ = (86.6 N) (cos 30° + 0.4 sin 30°) / 10 kg − 10 m/s²
aᵧ = -0.770 m/s²
Now let's try the second method.
Sum of forces in the perpendicular direction:
∑F = ma
N − mg cos 30° = 0
N = mg cos 30°
Sum of forces in the parallel direction:
∑F = ma
mg sin 30° − Nμ = ma
mg sin 30° − mgμ cos 30° = ma
a = g (sin 30° − μ cos 30°)
a = (10 m/s²) (sin 30° − 0.4 cos 30°)
a = 1.536 m/s²
Solving for the accelerations:
aₓ = a cos 30°
aₓ = 1.33 m/s²
aᵧ = -a sin 30°
aᵧ = -0.770 m/s²
As you can see, the second method is faster and easier, but both methods will give you the same answer.
(b) In the x direction:
Given:
x₀ = 0 m
v₀ = 0 m/s
aₓ = 1.33 m/s²
t = 1 s
Find: x
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (1 s) + ½ (1.33 m/s²) (1 s)²
x = 0.665 m
In the y direction:
Given:
y₀ = 5 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
t = 1 s
Find: y
y = y₀ + v₀ t + ½ at²
y = 5 m + (0 m/s) (1 s) + ½ (-0.770 m/s²) (1 s)²
y = 4.62 m
(c) In the y direction:
Given:
y₀ = 5 m
y = 0 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 m = 5 m + (0 m/s) t + ½ (-0.770 m/s²) t²
t = 3.61 s
An astronaut out on a spacewalk to construct a new section of the International Space Station walks with a constant velocity of 2.30 m/s on a flat sheet of metal placed on a flat, frictionless, horizontal honeycomb surface linking the two parts of the station. The mass of the astronaut is 71.0 kg, and the mass of the sheet of metal is 230 kg. (Assume that the given velocity is relative to the flat sheet.)
Required:
a. What is the velocity of the metal sheet relative to the honeycomb surface?
b. What is the speed of the astronaut relative to the honeycomb surface?
Answer:
Explanation:
Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction , so their relative velocity
= u + v = 2.3 m /s ( given )
We shall apply conservation of momentum law for the movement of astronaut and metal plate
mu = M v where m is mass of astronaut , M is mass of metal plate
71 u = 230 x v
71 ( 2.3 - v ) = 230 v
163.3 = 301 v
v = .54 m / s
u = 1.76 m / s
honeycomb will be at rest because honeycomb surface is frictionless . Plate will slip over it . Over plate astronaut is walking .
a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s
b ) velocity of astronaut relative to honeycomb will be + .54 m /s
Here + ve direction is assumed to be the direction of astronaut .
if the current in the circuit decreases, what does that mean about the rate at which the charge(and voltage) change in a capacitor?
2. the exponent of the exponential function contains RC for the given circuit. who's is a constant. use units R and C to find units of RC. write ohms in terms of volts and amps and write farads in terms of volts and coulombs. Simplify
units of RC are__________
Answer:
`1. charge Q, on the capacitor increases, while the current will decrease
2. τ = t = secs
Explanation:
1. consider RC of a circuit to be am external source
voltage across the circuit is given as
v =v₀(1 - [tex]e^{\frac{t}{τ} }[/tex])
where v = voltage
v₀ = peak voltage
t = time taken
τ= time constant
as the charge across the capacitor increases, current decreases
the charge across the circuit is given as
Q= Q₀(1 - [tex]e^{\frac{t}{τ} }[/tex])
charge Q is inversely proportional to the current I
hence the charge across the circuit increases
2. τ = RC
unit of time constant, τ,
= Ω × F
=[tex]\frac{V}{I}[/tex] ˣ [tex]\frac{C}{V}[/tex]
=[tex]\frac{C}{A}[/tex]
=[tex]\frac{C}{C/t}[/tex]
τ = t = secs
Four equal masses m are located at the corners of a square of side L, connected by essentially massless rods. Find the rotational inertia of this system about an axis (a) that coincides with one side and (b) that bisects two opposite sides.
Answer:
Explanation:
a )
Moment of inertial of four masses about axis that coincides with one side :
Out of four masses . location of two masses will lie on the axis so their moment of inertia will be zero .
Moment of inertia of the two remaining masses
= m L² + m L²
= 2 mL²
b )
Axis that bisects two opposite sides
Each of the four masses will lie at a distance of L / 2 from this axis so moment of inertia of the four masses
= 4 x m x ( L/2 )²
= 4 x mL² / 4
= m L² .
The metal bar on the conducting rails is moving to the left. The magnitude of the uniform magnetic field is 0.80 Tesla. The magnetic field is directed out of the page. The length L is 0.95 m. The resistance R of the circuit is 3.60 Ω. (a) If the induced current needs to be 0.54 A, what should be the induced emf
Answer:
Explanation:
induced emf = ?
induced emf = induced current x resistance of the circuit
= .54 x 3.6
= 1.944 V
induced emf required = 1.944 V .
The intensity of sunlight at the Earth's distance from the Sun is 1370 W/m2. (a) Assume the Earth absorbs all the sunlight incident upon it. Find the total force the Sun exerts on the Earth due to radiation pressure. N (b) Explain how this force compares with the Sun's gravitational attraction.
Answer:
F= 3.56e22N
Explanation:
Using the force of radiation acting on the earth which is
force = radiation pressure x area = (intensity/c)xpi R^2
force = 1370W/m^2 x pi x( 6.37x10^6m)^2/3x10^8m/s
force = 5.82x10^8 N
But the sun's gravitational attraction means the magnitude of the solar gravitational force on earth: If that's the case, the answer is approx 10^22 N:
F=GMm/r^2
G=6.67x10^(-11)=6.67e-11
M=mass sun = 2x10^30kg=2e30
m=mass earth = 6x10^24kg
r=earth sun distance = 1.5x10^11m
F=(6.6e-11)(2e30)(6e24)/(1.5e11)^2 =
F= 3.56e22N
In the circuit shown, the galvanometer shows zero current. The value of resistance R is :
A) 1 W
B) 2 W
C) 4 W
D) 9 W
Answer:
its supposed to be (a) 1W
A parallel-plate capacitor in air has a plate separation of 1.31 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator.
(a) Determine the charge on the plates before and after immersion.
before pC
after pC
(b) Determine the capacitance and potential difference after immersion.
Cf = F
ΔVf = V
(c) Determine the change in energy of the capacitor.
[ ] nJ
Answer:
a) before immersion
C = εA/d = (8.85e-12)(25e-4)/(1.31e-2) = 1.68e-12 F
q = CV = (1.68e-12)(255) = 4.28e-10 C
b) after immersion
q = 4.28e-10 C
Because the capacitor was disconnected before it was immersed, the charge remains the same.
c)*at 20° C
C = κεA/d = (80.4*)(8.85e-12)(25e-4)/(1.31e-2) = 5.62e-10 F
V = q/C = 4.28e-10 C/5.62e-10 C = 0.76 V
e)
U(i) = (1/2)CV^2 = (1/2)(1.68e-12)(255)^2 = 5.46e-8 J
U(f) = (1/2)(5.62e-10)(0.76)^2 = 1.62e-10 J
ΔU = 1.62e-10 J - 5.46e-8 J = -3.84e-8 J
Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
The object with the twice the area of the other object, will have the larger drag coefficient.
Explanation:
The equation for drag force is given as
[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]
where [tex]F_{D}[/tex] IS the drag force on the object
p = density of the fluid through which the object moves
u = relative velocity of the object through the fluid
p = density of the fluid
[tex]C_{D}[/tex] = coefficient of drag
A = area of the object
Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid
The above equation can also be broken down as
[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A
where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A
Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]
which also clarifies that the drag force is approximately proportional to the abject's area.
In this case, the object with the twice the area of the other object, will have the larger drag coefficient.
The headlights of a car are 1.4 m apart. What is the maximum distance (in km) at which the eye can resolve these two headlights? Take the pupil diameter to be 0.30 cm. (Assume the average wavelength of visible light is 555 nm.)
Answer:
5.4x10^4km
Explanation:
See attached file
Two 15-Ω and three 25-Ω light bulbs and a 24 V battery are connected in a series circuit. What is the current that passes through each bulb?
Answer:
0.229AExplanation:
Before we determine the amount of current in each bulb, we must first know that the same current flows in a series connected resistors. Since the Two 15-Ω and three 25-Ω light bulbs are connected in series, same current will flow in all of them.
According to Ohm's law, E = IRt where;
E is the supply voltage = 24V
I is the total current flowing in the circuit
Rt is the total equivalent resistance.
First, we need to calculate Rt.
Rt = 15Ω+15Ω+25Ω+25Ω+25Ω (Two 15-Ω and three 25-Ω light bulb in series)
Rt = 105Ω
From ohms law formula, I = E/Rt
I = 24/105
I = 0.229Amp
Since the total current in the circuit is 0.229A, therefore the amount of current that passes through each bulb is the same as the total current i.e 0.229A
The current that passes via each bulb is 0.229A
Ohm law:According to the above law,
E = IRt
Here
E should be the supply voltage = 24V
I should be the total current flowing in the circuit
Rt should be the total equivalent resistance.
Now Rt should be
Rt = 15Ω+15Ω+25Ω+25Ω+25Ω
Rt = 105Ω
Now the current is
I = E/Rt
I = 24/105
I = 0.229Amp
Therefore, The current that passes via each bulb is 0.229A.
Learn more about current here: https://brainly.com/question/21075693
Four 50-g point masses are at the corners of a square with 20-cm sides. What is the moment of inertia of this system about an axis perpendicular to the plane of the square and passing through its center
Answer:
moment of inertia I ≈ 4.0 x 10⁻³ kg.m²
Explanation:
given
point masses = 50g = 0.050kg
note: m₁=m₂=m₃=m₄=50g = 0.050kg
distance, r, from masses to eachother = 20cm = 0.20m
the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by
= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m
moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center
mathematically,
I = ∑m×d²
remember, a square will have 4 equal points
I = ∑m×d² = 4(m×d²)
I = 4 × 0.050 × (14.12 x 10⁻² m)²
I = 0.20 × 1.96 × 10⁻²
I = 3.92 x 10⁻³ kg.m²
I ≈ 4.0 x 10⁻³ kg.m²
attached is the diagram of the equation
An air-filled parallel-plate capacitor has plates of area 2.30 cm2 2 separated by 1.50 mm. The capacitor is connected to a 12.0-V battery. Find the value of its capacitance.
Answer:
[tex]1.357\times 10^{-12}[/tex]
Explanation:
Relevant Data provided
Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2
Distance which indicates d = 1.50 x 10^-3 m
Voltage which indicates V = 12 V
According to the requirement, the computation of value of its capacitance is shown below:-
[tex]Capacitance, C = \frac{\epsilon oA}{D}[/tex]
[tex]= \frac{= 8.854\times 10^{-12}\times 2.3\times 10^{-4}}{(1.5 \times 10^{-3})}[/tex]
= [tex]1.357\times 10^{-12}[/tex]
Therefore for computing the capacitance we simply applied the above formula.
A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g and a volume of 4,064 cm3. The density of seawater is 1.03 g/mL.
a. What is the weight of the statuette?
b. What is the mass of displaced water?
c. What is the weight of displaced water?
d. What is the buoyant force on the statuette?
e. What is the net force on the statuette?
f. How much force would be required to lift the statuette?
Answer:
A) W = 103.55 N
B) mass of displaced water = 4186 g
C) W_displaced water = 41.06 N
D) Buoyant force = 41.06 N.
E) ZERO
F) 62.54 N
Explanation:
We are given;
mass of statuette;m = 10,566 g = 10.566 kg
volume = 4,064 cm³
Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³
A) The dry weight of the statuette can be calculated as;
W = mg
So;
W = 10.556 × 9.81
W = 103.55 N
B) Mass of displaced water is calculated from;
Density = mass/volume
So, mass = Density × Volume
m = 1.03 × 4,064 = 4186 g
C) Weight of displaced water is given by;
W_displaced water = (m_displaced water) × g
W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N
D) The buoyant force is the same as the weight of the displaced water.
Thus, Buoyant force = 41.06 N.
E) The apparent weight of the statuette is calculated from;
Apparent weight = Dry weight - Weight of displaced water
Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.
F. From E above, The Force required to lift the statuette = 62.54 N
When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink. To do this, the separation of the dots must be less than Raleigh's criterion. Take the pupil of the eye to be 3.2 mm and the distance from the paper to the eye of 42 cm; find the maximum separation (in cm) of two dots such that they cannot be resolved. (Assume the average wavelength of visible light is 550 nm.)
Answer:
y <8 10⁻⁶ m
Explanation:
For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.
Therefore the diffraction equation for slits with m = 1 remains
a sin θ = λ
in general these experiments occur for oblique angles so
sin θ = θ
θ = λ / a
in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant
θ = 1.22 λ / a
The angles in these measurements are taken in radians, therefore
θ = s / R
as the angle is small the arc approaches the distance s = y
y / R = 1.22 λ / s
y = 1.22 λ R / a
let's calculate
y = 1.22 500 10⁻⁹ 0.42 / 0.032
y = 8 10⁻⁶ m
with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)
y <8 10⁻⁶ m
A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?
Answer:
2.9tons
Explanation:
Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:
parallel ("tangential"): F_t = F sin a
perpendicular ("normal"): F_n = F cos a
At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:
F sin a ~~ (46 tons)*0.064 ~~ 2.9tons
Also see attached file
The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".
Angle and ForceAccording to the question,
Angle, a = 3.7 degrees or,
Sin a = 0.064
Force, F = 46 tons
We know the relation,
Parallel (tangential), [tex]F_t[/tex] = F Sin a
By substituting the values,
= 46 × 0.064
= 2.9 tons
Thus the response above is appropriate answer.
Find out more information about Force here:
https://brainly.com/question/25239010
g When attempting to determine the coefficient of kinetic friction, why is it necessary to move the block with constant velocity
Answer:
This is because motion is intended to occur but at zero acceleration. It means at a constant velocity, henceFor that to happen the pulling force F must exactly equal the frictional force Fk .
Two forces act at a point in the plane. The angle between the two forces is given. Find the magnitude of the resultant force. forces of and newtons, forming an angle of
Answer:
F = 44.22 N
Explanation:
Let force 1, [tex]F_1=19\ N[/tex]
Force 2, [tex]F_2=32\ N[/tex]
The angle between forces, [tex]\theta=118^{\circ}[/tex]
We need to find the magnitude of the resultant force. It is based on the law of cosines. The formula is given by :
[tex]F^2=F_1^2+F_2^2-2AB\cos\theta\\\\F^2=(19)^2+(32)^2-2\times 19\times 32\times \cos(118)\\\\F=\sqrt{(19)^{2}+(32)^{2}-2\times19\times32\times\cos(118)}\\\\F=44.22\ N[/tex]
So, the magnitude of resultant force is 44.22 N.
Correct question: Two forces act on a point on the plane. The angle between the two forces is given. Find the magnitude of the resultant force. forces of 19 and 32 newtons, forming an angle of 118 degrees.
the magnitude of the resultant force is 28.53 N.
The resultant of the two vectors can be calculated using parallelogram theorem.
parallelogram theorem states that if two vectors are represented by the adjacent side of a parallelogram, the resultant of the vectors is the diagonal of the parallelogram drawn from the point of intersection of the vectors.
This can be expressed mathematically as
R² = P²+Q²-2PQcos(180-∅).............. Equation 1
Where R = resultant of the vectors, P and Q = the two vectors respectively, ∅ = angle between the vectors.
From the question,
Given: P = 19 N, Q = 32 N, ∅ = 118°
Substitute these values into equation 2
R² = 19²+32²-2×19×32cos(180-118)
R² = 361+1024-1216cos62°
R² = 1385-1216(0.4695)
R² = 1385-570.878
R² = 814.122
R = √(814.122)
R = 28.53 N
Hence, the magnitude of the resultant force is 28.53 N
Learn more about resultant force here:https://brainly.com/question/21852571
(a) Find the speed of waves on a violin string of mass 717 mg and length 24.3 cm if the fundamental frequency is 980 Hz. (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string? (Take the speed of sound in air to be 343 m/s.)
Answer:
a)v = 476.28 m / s , b) T = 6.69 10⁵ N , c) λ = 0.486 m , d) λ = 0.35 m
Explanation:
a) The speed of a wave on a string is
v = √T /μ
also all the waves fulfill the relationship
v = λ f
they indicate that the fundamental frequency is f = 980 Hz.
The wavelength that is fixed at its ends and has a maximum in the center
L = λ / 2
λ = 2L
we substitute
v = 2 L f
let's calculate
v = 2 0.243 980
v = 476.28 m / s
b) The tension of the rope
T = v² μ
the density of the string is
μ = m / L
T = v² m / L
T = 476.28² 0.717 / 0.243
T = 6.69 10⁵ N
c) λ = 2L
λ = 2 0.243
λ = 0.486 m
d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air
v = λ f
λ= v / f
λ = 343/980
λ = 0.35 m
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will
Which has more mass electron or ion?
Calculate the wavelengths of the first five members of the Lyman series of spectral lines, providing the result in units Angstrom with precision one digit after the decimal point.
Answer:
Explanation:
The formula for hydrogen atomic spectrum is as follows
energy of photon due to transition from higher orbit n₂ to n₁
[tex]E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV[/tex]
For layman series n₁ = 1 and n₂ = 2 , 3 , 4 , ... etc
energy of first line
[tex]E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})[/tex]
10.2 eV
wavelength of photon = 12375 / 10.2 = 1213.2 A
energy of 2 nd line
[tex]E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})[/tex]
= 12.08 eV
wavelength of photon = 12375 / 12.08 = 1024.4 A
energy of third line
[tex]E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})[/tex]
12.75 e V
wavelength of photon = 12375 / 12.75 = 970.6 A
energy of fourth line
[tex]E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})[/tex]
= 13.056 eV
wavelength of photon = 12375 / 13.05 = 948.3 A
energy of fifth line
[tex]E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})[/tex]
13.22 eV
wavelength of photon = 12375 / 13.22 = 936.1 A
Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from a point 6 965 km from the Earth's center to the Moon, 385 000 km from the Earth's center.
Answer:
0.0665 days
Explanation:
We are given;
The mean distance from the Earth's center to the moon;a1 = 385000 km
The mean distance from the Earth's center to the space craft;a2 = 6965 km
Formula for kepplers third law is;
T² = 4π²a³/GM
However, the proportion of both distances would be;
(T1)²/(T2)² = (a1)³/(a2)³
Where;
T1 is the period of orbit of the moon around the earth. T1 has a standard value of 27.322 days
T2 is the period of the space craft orbit.
Making T2 the subject, we have;
T2 = √((T1)²×(a2)³)/(a1)³)
Thus, plugging in the relevant values;
T2 = √(27.322² × 6965³)/(385000)³
T2 = 0.0665 days
The tune-up specifications of a car call for the spark plugs to be tightened to a torque of 38N⋅m38N⋅m. You plan to tighten the plugs by pulling on the end of a 25-cm-long wrench. Because of the cramped space under the hood, you'll need to put at an angle of 120∘with respect to the wrench shaft. With what force must you pull?
Answer:
F= 175.5N
Explanation:
Given:
Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance
torque of 38N⋅m
angle of 120∘
Torque(τ): 38Nm
position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m
Angle: 120°
To find the Force, the torque equation will be required which is expressed below
τ = Frsinθ
We need to solve for F, if we rearrange the equation, we have the expression below
F= τ/rsinθ
Note: the torque is maximum when the angle is 90 degrees
But θ= 180-120=60
F= 38/0.25( sin(60) )
F= 175.5N
A plane progressive
the expression
in time, ys
where you
progressivo ware is no presented by
(At + A
y- 5 sin
in metre, t es in time the doplicensel
Calculate
the amplitude of the wave.
Answer:
Amplitude, A = 5 m
Explanation:
Let a progressive wave is given by equation :
[tex]y=5\sin (100\pi t-0.4\pi x)[/tex] .....(1)
The general equation of a progressive wave is given by :
[tex]y=A\sin (\omega t-kx)[/tex] ....(2)
Here,
A is the amplitude of the wave
[tex]\omega[/tex] is the angular frequency
k is propagation constant
We need to find the amplitude of the wave.
If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.
A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev
Answer:
a
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
b
[tex]\theta = 9124.5 \ rev[/tex]
Explanation:
From the question we are told that
The maximum angular speed is [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]
The time taken is [tex]t = 2.8 \ s[/tex]
The minimum angular speed is [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest
Apply the first equation of motion to solve for acceleration we have that
[tex]w_{max} = w_{mini} + \alpha * t[/tex]
=> [tex]\alpha = \frac{ w_{max}}{t}[/tex]
substituting values
[tex]\alpha = \frac{40950.73}{2.8}[/tex]
[tex]\alpha = 14625 .3 \ rad/s^2[/tex]
converting to [tex]rev/s^2[/tex]
We have
[tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
According to the first equation of motion the angular displacement is mathematically represented as
[tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]
substituting values
[tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]
[tex]\theta = 57331.2 \ radian[/tex]
converting to revolutions
[tex]revolution = 57331.2 * 0.159155[/tex]
[tex]\theta = 9124.5 \ rev[/tex]