Answer:
L₀ = 15.12 fermi
Explanation:
From Einstein's special theory of relativity's consequence of length contraction, we know that:
L₀ = L √(1 - v²/c²)
where,
L₀ = Length of the object (electron) measured by observer at rest = ?
L = Length of the object (electron) measured by observer in relative motion
L = 10 fermi
v = relative speed = 0.75 c
c = speed of light
Therefore,
10 fermi = L₀√(1 - 0.75²c²/c²)
10 fermi = L₀√(1 - 0.5625)
10 fermi = L₀√0.4375
10 fermi/0.66 = L₀
L₀ = 15.12 fermi
Passengers in a carnival ride move at constant speed in a circle of radius 5.0 m, making a complete revolution in 4.0 s. As they spin, they feel their backs pressing against the wall holding them in the ride. A. What is the direction of the passengers' acceleration? a. No direction (zero acceleration) b. Directed towards center c. Directed away from center d. Directed tangentially B. What is the passengers' linear speed in m/s? C. What is the magnitude of their acceleration in m/s^2? D. What is their angular speed in rad/s?
Answer:
A. b) Directed towards center
B. [tex]v = 7.854\ m/s[/tex]
C. [tex]a_c = 12.337\ m/s^2[/tex]
D. [tex]w = 1.57\ rad/s[/tex]
Explanation:
The "force" that they feel pressing their backs against the wall is because the reaction to the centripetal acceleration .
A.
This acceleration has its direction towards the center of the circle. (option b)
B.
Their linear speed can be calculated with the equation:
[tex]v = (\theta/t)*r[/tex]
Where [tex]\theta[/tex] is the total angular position moved in radians ([tex]1\ rev = 2\pi\ radians[/tex]), 't' is the time elapsed for the angular position moved and 'r' is the radius. So we have that:
[tex]v = (2\pi/4)*5 = 7.854\ m/s[/tex]
C.
The centripetal acceleration is given by the equation:
[tex]a_c = v^2/r[/tex]
[tex]a_c = 7.854^2/5[/tex]
[tex]a_c = 12.337\ m/s^2[/tex]
D.
Their angular speed is given by the equation:
[tex]w = \theta/t = 2\pi/4 = \pi/2 = 1.57 \ rad/s[/tex]
A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.61 rad/sec. The moment of inertia of the student plus the stool is 6 kg m^2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.39 m from the rotation axis.
Required:
a. Calculate the final angular speed of the student. Answer in units of rad/s.
b. Calculate the change in kinetic energy of the system. Answer in units of J.
Answer:
a) the final angular speed is 0.738 rad/s
b) the change in kinetic energy = 0.3 J
Explanation:
the two 1 kg objects have a total mass of 2 x 1 = 2 kg
radius of rotation of the objects = 0.9 m
moment of inertial of the student and the chair = 6 kg-m^2
initial angular speed of rotation of the sitting student and object system ω1 = 0.61 rad/s
final angular speed of rotation of the sitting student and object system ω2 = ?
moment of inertia of the rotating object is
[tex]I = mr^{2}[/tex] = 2 x [tex]0.9^{2}[/tex] = 1.62 kg-m^2
total moment of inertia of sitting student and object system will be
==> 6 + 1.62 = 7.62 kg-m^2
The initial angular momentum of the sitting student and object system will be calculated from
==> Iω1 = 7.62 x 0.61 = 4.65 kg-rad/s-m^2
if the radius of rotation of the object is reduced to 0.39 m,
new moment of inertia of the rotating object will be
[tex]I = mr^{2}[/tex] = 2 x [tex]0.39^{2}[/tex] = 0.304 kg-m^2
new total moment of inertia of the sitting student and object system will be
==> 6 + 0.304 = 6.304 kg-m^2
The final momentum of the sitting student and object system will be calculated from
==> Iω2 = 6.304 x ω2 = 6.304ω2
According to conservation of angular momentum, initial momentum of the system must be equal to the final momentum of the system. Therefore,
4.65 = 6.304ω2
ω2 = 4.65/6.30 = 0.738 rad/s
b) Rotational kinetic energy of the system = [tex]\frac{1}{2} Iw^{2}[/tex]
for the initial conditions, kinetic energy is
==> [tex]\frac{1}{2} Iw1^{2}[/tex] = [tex]\frac{1}{2}* 7.62*0.61^{2}[/tex] = 1.417 J
for the final conditions, kinetic energy is
==> [tex]\frac{1}{2} Iw1^{2}[/tex] = [tex]\frac{1}{2}*6.304*0.738^{2}[/tex] = 1.717 J
change in kinetic energy = final KE - initial KE
==> 1.717 - 1.417 = 0.3 J
what is the largest star in our night sky
Which pieces of information does the National Weather Service produce?
Answer:
1. It collects weather data as part of a network around the country.
2. its territories, adjacent waters and ocean areas for the protection of life and property and the enhancement of the national economy.
Answer:
Maps with isotherms
Explanation:
A physician orders Humulin R 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?
The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration
Complete question:
A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?
Answer:
the total units of insulin admistered each morning
= 22 units of qam and humulin
Explanation:
given
44 units and Humnlin N
with concentration 50/100 = 1/2 = 0.5
∴ 44 × 0.5 ≈ 22 units in the morning
regular insulin administered each day
(22 + 35)units of qam and humulin
= 57units
On Apollo missions to the Moon, the command module orbited at an altitude of 160 km above the lunar surface. How long did it take for the command module to complete one orbit?
Answer:
T = 2.06h
Explanation:
In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:
[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex] (1)
T: time for a complete orbit = ?
r: radius of the orbit
G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2
Mm: mass of the moon = 7.34*10^22 kg
The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:
[tex]r=R_m+160km\\\\[/tex]
Rm: radius of the moon = 1737.1 km
[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]
Then, you replace all values of the parameters in the equation (1):
[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]
In hours you obtain:
[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]
The time that the Apollo takes to complete an orbit around the moon is 2.06h
Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)
Answer:
*If the particles are deflected in opposite directions, it implies that their charges must be opposite
*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Explanation:
When a charged particle enters a magnetic field, it is subjected to a force given by
F = q v x B
where bold letters indicate vectors
this expression can be written in the form of a module
F = qv B sin θ
and the direction of the force is given by the right-hand rule.
In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1
If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.
Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 10.0 rad/s. The bug crawls to the center of the disk.
(a) What is the new angular velocity of the disk?
(b) What is the change in the kinetic energy of the system?
(c) If the bug crawls back to the outer edge of the disk, what is the angular velocity of the disk then?
(d) What is the new kinetic energy of the system? (e) What is the cause of the increase and decrease of kinetic energy?
Answer:
Id = 1/2 Md * R^2 = 1/2 * .1 * .1^2 = .0005 kg m^2 inertia of disk
Ib = Mb * R^2 = .02 * .1^2 = .0002 kg m^2 inertia of bug at edge
(Id + Ib) w1 = Id w2 conservation of angular momentum
w2 = .0007 / .0005 * 10 = 14 /sec angular speed with bug at center
KE1 = 1/2 I1 * w1^2 = 1/2 * .0007 * 10^2 = .035 kg m^2 / s^2
KE2 = 1/2 * I2 w2*2 = (.0005 / 2 ) ^ 14^2 = .049 kg m^2 / s^2
The bug has to exert radial force on the disk to maintain its
centripetal acceleration. As the bug crawls to the center of the disk it
does work against this centripetal force which appears as an increase
of rotational energy of the disk. As the the bug crawls back to the edge
of the disk, the disk does work on the bug and loses KE.
"Consider the Earth and the Moon as a two-particle system. (a) How far from the center of the Earth is the gravitational field of this two-particle system zero?"
a is the correct
Explanation:
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An electromagnetic wave is propagating towards the west. At a certain moment the direction of the magnetic field vector associated with this wave points vertically up. What is the direction of the electric field vector?
Answer:
the electric field is pointing horizontal direction and in south direction
Explanation:
In an electromagnetic wave, the magnetic field and electrical field are perpendicular to each other and these are perpendicular to the direction of the waves.
Unpolarized light is incident upon two polarization filters that do not have their transmission axes aligned. If 38 % of the light passes through this combination of filters, what is the angle between the transmission axes of the filters
Answer:
64°
Explanation:
See attached file
Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 40.0 mm, and the potential difference between them is 370 V
A. What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
B. What is the magnitude of the force this field exerts on a particle with a charge of 2.40 nC ?
C. Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
D. Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.
Answer:
Explanation:
A )
electric field E = V / d where V is potential difference between plates separated by distance d .
putting the given values
E = 370 / .040 V / m
= 9250 V / m
B )
Force on charged particle of charge q in electric field E
F = q E
F = 2.4 x 10⁻⁹ x 9250
= 22200 x 10⁻⁹
= 222 x 10⁻⁷ N .
C ) since field is uniform , force will be constant
work done by electric field putting up this force
= force x displacement
= 222 x 10⁻⁷ x 40 x 10⁻³
= 888 x 10⁻⁹ J
D )
change in potential energy
= q ( V₁ - V₂ )
= 2.40 X 10⁻⁹ x 370
= 888 x 10⁻⁹ J .
(a) The magnitude of electric field in the region between the plates is 9,250 V/m.
(b) The magnitude of the force the field exerts on a particle with the given charge is 2.22 x 10⁻⁵ N.
(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is [tex]8.88 \times 10^{-7} \ J[/tex].
(d) the change of the potential energy is [tex]8.88 \times 10^{-7} \ J[/tex].
The given parameters;
distance between the two metal plates, d = 40 mmpotential difference between the plates, V = 370 V(a) The magnitude of electric field in the region between the plates is calculated as;
[tex]E = \frac{V}{d} \\\\E = \frac{370 }{40 \times 10^{-3} } \\\\E = 9,250 \ V/m[/tex]
(b) The magnitude of the force the field exerts on a particle with the given charge is calculated as follows;
F = Eq
F = 9,250 x 2.4 x 10⁻⁹
F = 2.22 x 10⁻⁵ N
(c) The work done by the field on the particle as it moves from the higher potential plate to the lower is calculated as follows;
[tex]W = Fd\\\\W = 2.22 \times 10^{-5} \times 40\times 10^{-3} \\\\W =8.88 \times 10^{-7} \ J[/tex]
(d) the change of the potential energy is calculated as;
[tex]\Delta U = q \Delta V\\\\\Delta U = q(V_1 - V_2)\\\\\\Delta U = 2.4 \times 10^{-9}(370)\\\\\Delta U = 8.88 \times 10^{-7} \ J[/tex]
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Determining the Mass of a Mystery Mystery object Object distance (m) Brick mass (kg) Brick distance (m) Brick torque (Nm) Object mass (kg) Fobject (N) Fbricks (N) Fpivot point (N) A. 1.00 20 B. 1.00 20 C. 1.00 20 D. 1.00 20
Answer:
Explanation:
according to resultant of two parallel forces,
Fpivot = Fobject + Fbricks
so that, the net force is zero
Three flat layers of transparent material are stacked upon one another. The top layer has index of refraction n1, the middle has n2 and the bottom one has n3. If n1 > n2 > n3, and if a ray of light strikes the top layer at an angle of incidence, in which layer is the angle of refraction the greatest? Why?
a. the bottom layer
b. the top layer
c. Once the ray enters the touching layers, the angle of refraction remains constant.
d. the middle layer
Answer:
a. the bottom medium
Explanation:
it has the least index of refraction and hence most rarer.
The electric field strength is 1.70 × 104 N/C inside a parallel-plate capacitor with a 0.800 m spacing. An electron is released from rest at the negative plate. What is the electron's speed when it reaches the positive plate?
Answer:
Here, "v" is the velocity of electron and "V" is the potential.
What is the length of a contention slot in CSMA/CD for (a) a 2-km twin-lead cable (signal propagation speed is 82% of the signal propagation speed in vacuum)
Answer:
1.99*10-4sec
Explanation:
Signal propagation speed=0.82∗2.46∗108m/s
d=2000 m
Tp=20000/0.82∗2.46∗108 sec
ContentionPeriod=2Tp=2∗20000/0.82∗2.46∗10^8
= 1.99* 10^-4seconds
A particle accelerator fires a proton into a region with a magnetic field that points in the x-direction. (a) If the proton is moving in the y-direction, what is the direction of the magnetic force on the proton
Answer:
The magnitude of the magnetic field will act in a direction towards me.
Explanation:
When a charged particle enters a magnetic field, it is deflected. The direction of travel of the particle is deflected, but the kinetic energy of the particle is not affected. The force experienced by a charged particle as it enters a magnetic field that acts perpendicular to the path of the velocity of the particle, will produce a force that is perpendicular to both the direction of travel of the particle and the direction of the magnetic field. In this case, the proton moves in the y-direction, the magnetic field is in the x-direction, therefore the force experienced by the particle will be towards me.
A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion, after which a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find
(a) the coefficient of static friction.
(b) the coefficient of kinetic friction between the block and the surface.
Answer:
(a) 0.31
(b) 0.245
Explanation:
(a)
F' = μ'mg.................... Equation 1
Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.
make μ' the subject of the equation above
μ' = F'/mg............. Equation 2
Given: F' = 75 N, m = 25 kg
constant: g = 9.8 m/s²
Substitute these values into equation 2
μ' = 75/(25×9.8)
μ' = 75/245
μ' = 0.31.
(b) Similarly,
F = μmg.................. Equation 3
Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.
make μ the subject of the equation
μ = F/mg.............. Equation 4
Given: F = 60 N, m = 25 kg, g = 9.8 m/s²
Substitute these values into equation 4
μ = 60/(25×9.8)
μ = 60/245
μ = 0.245
A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that time?
Answer:
the car have travelled 0.31 mile during that time
Explanation:
Applying the Equation of motion;
s = 0.5(u+v)t
Where;
s = distance travelled
u = initial speed = 0 mph
v = Final speed = 50 mph
t = time taken = 3/4 min = 3/4 ÷ 60 hours = 1/80 hour
Substituting the given values into the equation;
s = 0.5(0+50)×(1/80)
s = 0.3125 miles
s ~= 0.31 mile
the car have travelled 0.31 mile during that time
Unpolarized light enters a polarizer with vertical polarization axis. The light that passes through passes another polarizer with transmission axis at 40 degrees to the horizontal. What is the intensity of the light after the second polarizer expressed as a fraction of the original intensity
Answer:
I = 0.2934 I₀
Explanation:
The expression that governs the transmission of polarization is
I = I₀ cos² θ
Let's apply this to our case, when the unpolarized light enters the first polarized, the polarized light that comes out has the intensity of
I₁ = I₀ / 2
this is the light that enters the second polarizer
I = I₁ cos² θ
we substitute
I = I₀ / 2 cos² 40
I = I₀ 0.2934
I = 0.2934 I₀
A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance h, as the drawing shows. Using 1.013 105 Pa for the atmospheric pressure and 1200 kg/m3 for the density of the sauce, find the absolute pressure in the bulb when the distance h is (a) 0.15 m and (b) 0.10 m.
Answer:
(a) P = 103064 Pa = 103.064 KPa
(b) P = 102476 Pa = 102.476 KPa
Explanation:
(a)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.15 m)
Pg = 1764 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1764 Pa
P = 103064 Pa = 103.064 KPa
(b)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.1 m)
Pg = 1176 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1176 Pa
P = 102476 Pa = 102.476 KPa
The absolute pressure in the bulb is approximately 1.031 x 10⁵ Pa when h = 0.15 m and 1.025 x 10⁵ Pa when h = 0.10 m.
Absolute pressure is the total pressure exerted by a fluid, including both the pressure from the fluid itself and the atmospheric pressure. It is the sum of the gauge pressure, which is the pressure above atmospheric pressure, and the atmospheric pressure. Absolute pressure is measured relative to a complete vacuum, where the pressure is zero.
In fluid mechanics, absolute pressure is important for determining the forces and behaviors of fluids in various systems. It is commonly expressed in units such as pascals (Pa), atmospheres (atm), pounds per square inch (psi), or torr.
The absolute pressure in the bulb can be calculated using the following formula:
P = P₀ + ρgh
where:
P is the absolute pressure in the bulb,
P₀ is the atmospheric pressure (1.013 x 10⁵ Pa),
ρ is the density of the sauce (1200 kg/m³),
g is the acceleration due to gravity (9.8 m/s²), and
h is the height of the sauce in the tube.
(a) When h = 0.15 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.15 m)
P ≈ 1.013 x 10⁵ Pa + 1764 Pa
P ≈ 1.031 x 10⁵ Pa
(b) When h = 0.10 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.10 m)
P ≈ 1.013 x 10⁵ Pa + 1176 Pa
P ≈ 1.025 x 10⁵ Pa
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A box experiencing a gravitational force of 600 N. is being pulled to the right with a force of 250 N. 825 N. frictional force acting on the box as it moves to the right what is the net force in the Y direction
Answer:A
Explanation:
Explanation:
Given that,
Gravitational force = 600 N
Frictional force = 25 N
Pulled by the Force = 250 N
We know that,
The gravitational force in downward and normal force act in upward. the frictional force in left side and the box pulled by the force to the right side.
The balance equation is along y-axis
The box will not move in y-axis therefore, the net force in the y-axis will be zero.
Hence, The net force in the y-direction will be zero.
A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva so you can better study its flow pattern in the laboratory. Your model must be able to move at 50 cm/s and you will place the model in honey instead of water. Honey has a density of 1400 kg/m3 and a viscosity of 600 Pa-s.
Required:
How long should your model be?
Answer:
Explanation:
For the problem, we should have same reynolds number
ρvd/mu = constant
1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600
d = 25.66 cm
A piece of tape is pulled from a spool and lowered toward a 100-mg scrap of paper. Only when the tape comes within 8.0 mm is the electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.
Requried:
Determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance.
Answer:
The magnitude of the electric force is [tex]F_e = 0.00098 \ N[/tex]
Explanation:
From the question we are told that
The mass of the paper is [tex]m= 100 mg = 100 *10^{-6} \ kg[/tex]
The position is [tex]d = 8.0\ mm = 0.008 \ m[/tex]
Generally the magnitude of the electric force at the point of equilibrium between the electric force and the gravitational force is mathematically represented as
[tex]F_e = F_g = mg[/tex]
Where [tex]F_g[/tex] is gravitational force
substituting values
[tex]F_e = 100 *10^{-6} * 9.8[/tex]
[tex]F_e = 0.00098 \ N[/tex]
Now generally the gravitational force acts downward (negative y axis ) hence the reason the electric force is same magnitude but opposite in direction (upward + y - axis )
Light of wavelength 575 nm falls on two double slits spaced 0.30 mm apart. What is the required distance from the slits to a screen if the spacing between the first and second dark fringe is to be 4.00 mm
Answer:
L = 2.1 m
Explanation:
From Young's Double Slit Experiment, the formula for the distance between two consecutive dark or bright fringes, called fringe spacing, is derived as:
Δx = λL/d
where,
Δx = distance between first and second dark fringe = 4 mm = 4 x 10⁻³ m
λ = wavelength of light = 575 nm = 5.75 x 10⁻⁷ m
d = distance between the slits = 0.3 mm = 3 x 10⁻⁴ m
L = Distance from slits to screen = ?
Therefore,
4 x 10⁻³ m = (5.75 x 10⁻⁷ m)(L)/(3 x 10⁻⁴ m)
L = (4 x 10⁻³ m)/(1.92 x 10⁻³)
L = 2.1 m
A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block is 1.2 m to get to the bottom, calculate how fast it is moving at the bottom using Conservation of Energy.
Complete Question
The diagram for this question is showed on the first uploaded image (reference homework solutions )
Answer:
The velocity at the bottom is [tex]v = 11.76 \ m/ s[/tex]
Explanation:
From the question we are told that
The total distance traveled is [tex]d = 1.2 \ m[/tex]
The mass of the block is [tex]m_b = 0.3 \ kg[/tex]
The height of the block from the ground is h = 0.60 m
According the law of energy
[tex]PE = KE[/tex]
Where PE is the potential energy which is mathematically represented as
[tex]PE = m * g * h[/tex]
substituting values
[tex]PE = 3 * 9.8 * 0.60[/tex]
[tex]PE = 17.64 \ J[/tex]
So
KE is the kinetic energy at the bottom which is mathematically represented as
[tex]KE = \frac{1}{2} * m v^2[/tex]
So
[tex]\frac{1}{2} * m* v ^2 = PE[/tex]
substituting values
=> [tex]\frac{1}{2} * 3 * v ^2 = 17.64[/tex]
=> [tex]v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }[/tex]
=> [tex]v = 11.76 \ m/ s[/tex]
A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval, the angular velocity decreases to 10 rad/s. Assume that the angular acceleration is constant during the 5.0-s interval. How many radians does the wheel turn through during the 5.0-s interval
Answer:
The angular displacement of the wheel is 45 radians
Explanation:
Given;
initial angular velocity, ω₀ = 20 rad/s
final angular velocity, ωf = 10 rad/s
time interval, t = 5
Angular acceleration is calculated as;
[tex]\alpha = \frac{\omega _f - \omega_0}{t} \\\\\alpha = \frac{10 -20}{5} \\\\\alpha = -2 \ rad/s^2[/tex]
|α| = 2 rad/s²
Angular displacement is calculated as;
[tex]\theta = \omega_0 \ + \ \frac{1}{2} \alpha t^2\\\\\theta = 20 \ + \ \frac{1}{2} *(2)*5^2\\\\\theta = 20 \ + 25\\\\ \theta = 45 \ radians[/tex]
Therefore, the angular displacement of the wheel is 45 radians
g Doppler Radar gathers information about precipitation by sending out pulses of ______ energy that is reflected back by the precipitation towards the radar. Group of answer choices
Answer:
Doppler Radar gathers information about precipitation by sending out pulses of ___Radio wave___ energy
Please help! Calculate velocity. Show all work!
Answer:
v = 23.66 m/s
Explanation:
recall that one of the equations of motion may be expressed:
v² = u² + 2as,
Where
v = final velocity (we are asked to find this)
u = initial velocity = 0 m/s since we are told that it starts from rest
a = acceleration = 0.56m/s²
s = distance traveled = given as 500m
Simply substitute the known values into the equation:
v² = u² + 2as
v² = 0 + 2(0.56)(500)
v² = 560
v = √560
v = 23.66 m/s
When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated? When a force is applied directly to the pivot point of a balance, what is the torque due to that force?
Answer:
Explanation:
Net torque is calculated by multiplying the force with distance from the point of application of force to the point of pivot .
If more than 2 forces are present, then we either subtract the product of forces with their respective distances from pivot point or we add them . It depends on whether they both are present on opposite sides of pivot or on same side of pivot .
When a force is applied directly to the pivot point of balance, then the torque on due that force = 0 (zero) .
It is so because the torque is defined as the product of force and perpendicular distance from the pivot point but here the distance is 0 , therefore torque is zero.