Answer:
The minimum angular velocity necessary to assure that the riders will not slide down the wall is 1.58 rad/second.
Explanation:
The riders will experience a centripetal force from the cylinder
[tex]F_{C}[/tex] = mrω^2 .... equ 1
where
m is the mass of the rider
r is the inner radius of the cylinder = 3.4 m
ω is the angular speed of of the rider
For the riders not to slide downwards, this centripetal force is balanced by the friction between the riders and the cylinder. The frictional force is given as
[tex]F_{f}[/tex] = μR ....equ 2
where
μ = coefficient of friction = 0.87
R is the normal force from the rider = mg
where
m is the rider's mass
g is the acceleration due to gravity = 9.81 m/s
substitute mg for R in equ 2, we'll have
[tex]F_{f}[/tex] = μmg ....equ 3
Equating centripetal force of equ 1 and frictional force of equ 3, we'll get
mrω^2 = μmg
the mass of the rider cancels out, and we are left with
rω^2 = μg
ω^2 = μg/r
ω = [tex]\sqrt{\frac{ug}{r} }[/tex]
ω = [tex]\sqrt{\frac{0.87*9.81}{3.4} }[/tex]
ω = 1.58 rad/second
The minimum angular velocity necessary so that the riders will not slide down the wall is 1.58 rad/s
The riders will experience a centripetal force from the cylinder
[tex]F = mrw^2[/tex]
where m is the mass of the rider
r is the inner radius of the cylinder = 3.4 m
ω is the angular speed of the rider
For the riders not to slide downwards, this centripetal force must be balanced by friction. The frictional force is given as
f = μN
where
μ = coefficient of friction = 0.87
N is the normal force = mg
f = μmg
Equating centripetal force of and frictional force of we'll get
[tex]mrw^2 = umg[/tex]
[tex]rw^2 = ug[/tex]
[tex]w^2 = ug/r[/tex]
[tex]w= \sqrt{ug/r}[/tex]
[tex]w= \sqrt{0.87*9.8/3.4}[/tex]
ω = 1.58 rad/s is the minimum angular velocity needed to prevent the rider from sliding.
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mention two similarities of citizen and aliens
Answer:
The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily
Explanation:
Monochromatic coherent light shines through a pair of slits. If the wavelength of the light is decreased, which of the following statements are true of the resulting interference pattern? (There could be more than one correct choice.)
a. The distance between the maxima decreases.
b. The distance between the minima decreases.
c. The distance between the maxima stays the same.
d. The distance between the minima increases.
e. The distance between the minima stays the same.
Answer:
he correct answers are a, b
Explanation:
In the two-slit interference phenomenon, the expression for interference is
d sin θ= m λ constructive interference
d sin θ = (m + ½) λ destructive interference
in general this phenomenon occurs for small angles, for which we can write
tanθ = y / L
tan te = sin tea / cos tea = sin tea
sin θ = y / La
un
derestimate the first two equations.
Let's do the calculation for constructive interference
d y / L = m λ
the distance between maximum clos is and
y = (me / d) λ
this is the position of each maximum, the distance between two consecutive maximums
y₂-y₁ = (L 2/d) λ - (L 1 / d) λ₁ y₂ -y₁ = L / d λ
examining this equation if the wavelength decreases the value of y also decreases
the same calculation for destructive interference
d y / L = (m + ½) κ
y = [(m + ½) L / d] λ
again when it decreases the decrease the distance
the correct answers are a, b
distributed uniformly over the surface of a metal sphere with a radius 24.0 cm. If the potential is zero at a point at infinity, find the value of the pote my jobntA total electric charge of 3.50 nC is distributed uniformly over the surface of a metal sphere with a radius 24.0 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) 48.0 cm (b) 2ial at the following distances from the center of the sphere: (a) 48.0 cm (b) 24.0 cm (c) 12.0 cm
Answer:
(a) V = 65.625 Volts
(b) V = 131.25 Volts
(c) V = 131.25 Volts
Explanation:
Recall that:
1) in a metal sphere the charges distribute uniformly around the surface, and the electric field inside the sphere is zero, and the potential is constant equal to:
[tex]V=k\frac{Q}{R}[/tex]
2) the electric potential outside of a charged metal sphere is the same as that of a charge of the same value located at the sphere's center:
[tex]V=k\frac{Q}{r}[/tex]
where k is the Coulomb constant ( [tex]9\,\,10^9\,\,\frac{N\,m^2}{C^2}[/tex] ), Q is the total charge of the sphere, R is the sphere's radius (0.24 m), and r is the distance at which the potential is calculated measured from the sphere's center.
Then, at a distance of:
(a) 48 cm = 0.48 m, the electric potential is:
[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.48} =65.625\,\,V[/tex]
(b) 24 cm = 0.24 m, - notice we are exactly at the sphere's surface - the electric potential is:
[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]
(c) 12 cm (notice we are inside the sphere, and therefore the potential is constant and the same as we calculated for the sphere's surface:
[tex]V=k\frac{Q}{R}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]
Answer:
c) a difference in electric potential
Explanation:
my insta: priscillamarquezz
A skater on ice with arms extended and one leg out spins at 3 rev/s. After he draws his arms and the leg in, his moment of inertia is reduced to 1/2. What is his new angular speed
Answer:
The new angular speed is [tex]w = 6 \ rev/s[/tex]
Explanation:
From the question we are told that
The angular velocity of the spin is [tex]w_o = 3 \ rev/s[/tex]
The original moment of inertia is [tex]I_o[/tex]
The new moment of inertia is [tex]I =\frac{I_o}{2}[/tex]
Generally angular momentum is mathematically represented as
[tex]L = I * w[/tex]
Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so
[tex]I * w = constant[/tex]
=> [tex]I_o * w _o = I * w[/tex]
where w is the new angular speed
So
[tex]I_o * 3 = \frac{I_o}{2} * w[/tex]
=> [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]
=> [tex]w = 6 \ rev/s[/tex]
An aging coyote cannot run fast enough to catch a roadrunner. He purchases on eBay a set of jet-powered roller skates, which provide a constant horizontal acceleration of 15.0 m/s2. The coyote starts at rest 70.0 m from the edge of a cliff at the instant the roadrunner zips past in the direction of the cliff.
Required:
a. Determine the minimum constant speed the roadrunner must have to reach the cliff before the coyote. At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. The coyote’s skates remain horizontal and continue to operate while he is in flight.
b. The cliff is 100 m above the flat floor of the desert. Determine how far from the base of the cliff the coyote lands.
c. Determine the components of the coyote’s impact velocity
Answer:
a) v_correcaminos = 22.95 m / s , b) x = 512.4 m ,
c) v = (45.83 i ^ -109.56 j ^) m / s
Explanation:
We can solve this exercise using the kinematics equations
a) Let's find the time or the coyote takes to reach the cliff, let's start by finding the speed on the cliff
v² = v₀² + 2 a x
they tell us that the coyote starts from rest v₀ = 0 and its acceleration is a=15 m / s²
v = √ (2 15 70)
v = 45.83 m / s
with this value calculate the time it takes to arrive
v = v₀ + a t
t = v / a
t = 45.83 / 15
t = 3.05 s
having the distance to the cliff and the time, we can find the constant speed of the roadrunner
v_ roadrunner = x / t
v_correcaminos = 70 / 3,05
v_correcaminos = 22.95 m / s
b) if the coyote leaves the cliff with the horizontal velocity v₀ₓ = 45.83 m / s, they ask how far it reaches.
Let's start by looking for the time to reach the cliff floor
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
in this case y = 0 and the height of the cliff is y₀ = 100 m
0 = 100 + 45.83 t - ½ 9.8 t²
t² - 9,353 t - 20,408 = 0
we solve the quadratic equation
t = [9,353 ±√ (9,353² + 4 20,408)] / 2
t = [9,353 ± 13] / 2
t₁ = 11.18 s
t₂ = -1.8 s
Since time must be a positive quantity, the answer is t = 11.18 s
we calculate the horizontal distance traveled
x = v₀ₓ t
x = 45.83 11.18
x = 512.4 m
c) speed when it hits the ground
vₓ = v₀ₓ = 45.83 m / s
we look for vertical speed
v_{y} = [tex]v_{oy}[/tex] - gt
v_{y} = 0 - 9.8 11.18
v_{y} = - 109.56 m / s
v = (45.83 i ^ -109.56 j ^) m / s
Use Coulomb’s law to derive the dimension for the permittivity of free space.
Answer:
Coulomb's law is:
[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]
First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:
N = (1/{e0})*C^2/m^2
then we have:
{e0} = C^2/(m^2*N)
And we know that N = kg*m/s^2
then the dimensions of e0 are:
{e0} = C^2*s^2/(m^3)
(current square per time square over cubed distance)
And knowing that a Faraday is:
F = C^2*S^2/m^2
The units of e0 are:
{e0} = F/m.
An 88.0 kg spacewalking astronaut pushes off a 645 kg satellite, exerting a 110 N force for the 0.450 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 1.40 min?
Answer:
The astronaut and the satellite are 53.718 m apart.
Explanation:
Given;
mass of spacewalking astronaut, = 88 kg
mass of satellite, = 645 kg
force exerts by the satellite, F = 110N
time for this action, t = 0.45 s
Determine the acceleration of the satellite after the push
F = ma
a = F / m
a = 110 / 645
a = 0.171 m/s²
Determine the final velocity of the satellite;
v = u + at
where;
u is the initial velocity of the satellite = 0
v = 0 + 0.171 x 0.45
v = 0.077 m/s
Determine the displacement of the satellite after 1.4 m
d₁ = vt
d₁ = 0.077 x (1.4 x 60)
d₁ = 6.468 m
According to Newton's third law of motion, action and reaction are equal and opposite;
Determine the backward acceleration of the astronaut after the push;
F = ma
a = F / m
a = 110 / 88
a = 1.25 m/s²
Determine the final velocity of the astronaut
v = u + at
The initial velocity of the astronaut = 0
v = 1.25 x 0.45
v = 0.5625 m/s
Determine the displacement of the astronaut after 1.4 min
d₂ = vt
d₂ = 0.5625 x (1.4 x 60)
d₂ = 47.25 m
Finally, determine the total separation between the astronaut and the satellite;
total separation = d₁ + d₂
total separation = 6.468 m + 47.25 m
total separation = 53.718 m
Therefore, the astronaut and the satellite are 53.718 m apart.
An asteroid that has an orbit with a semi-major axis of 4 AU will have an orbital period of about ______ years.
Answer:
16 years.
Explanation:
Using Kepler's third Law.
P2=D^3
P=√d^3
Where P is the orbital period and d is the distance from the sun.
From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.
P=?
P= √4^3
P= √256
P= 16 years.
Orbital period is 16 years.
An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
The new voltage between the plates of the capacitor is 18 V
Explanation:
The charge on parallel plate capacitor is calculated as;
q = CV
Where;
V is the battery voltage
C is the capacitance of the capacitor, calculated as;
[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]
[tex]q = \frac{\epsilon _0A V}{d}[/tex]
where;
ε₀ is permittivity of free space
A is the area of the capacitor
d is the space between the parallel plate capacitors
If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;
[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]
Therefore, the new voltage between the plates of the capacitor is 18 V
A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ????=0.929 kg·m2 ?
Answer:
The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]
Explanation:
The Diagram for this question is gotten from the first uploaded image
From the question we are told that
The mass of the rod is [tex]M = 3.41 \ kg[/tex]
The mass of each small bodies is [tex]m = 0.249 \ kg[/tex]
The moment of inertia of the three-body system with respect to the described axis is [tex]I = 0.929 \ kg \cdot m^2[/tex]
The length of the rod is L
Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as
[tex]I = I_r + 2 I_m[/tex]
Where [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as
[tex]I_r = \frac{ML^2 }{12}[/tex]
And [tex]I_m[/tex] the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as
[tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]
Thus [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]
Hence
[tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]
=> [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]
substituting vales we have
[tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]
[tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]
[tex]L = 1.5077 \ m[/tex]
A "laser cannon" of a spacecraft has a beam of cross-sectional area A. The maximum electric field in the beam is 2E. The beam is aimed at an asteroid that is initially moving in the direction of the spacecraft. What is the acceleration of the asteroid relative to the spacecraft if the laser beam strikes the asteroid perpendicularly to its surface, and the surface is not reflecting
Answer:
Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m
Explanation:
The maximum electric field in the beam = 2E
cross-sectional area of beam = A
The intensity of an electromagnetic wave with electric field is
I = cε[tex]E_{0} ^{2}[/tex]/2
for [tex]E_{0}[/tex] = 2E
I = 2cε[tex]E^{2}[/tex] ....equ 1
where
I is the intensity
c is the speed of light
ε is the permeability of free space
[tex]E_{0}[/tex] is electric field
Radiation pressure of an electromagnetic wave on an absorbing surface is given as
P = I/c
substituting for I from above equ 1. we have
P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex] ....equ 2
Also, pressure P = F/A
therefore,
F = PA ....equ 3
where
F is the force
P is pressure
A is cross-sectional area
substitute equ 2 into equ 3, we have
F = 2ε[tex]E^{2}[/tex]A
force on a body = mass x acceleration.
that is
F = ma
therefore,
a = F/m
acceleration of the asteroid will then be
a = 2ε[tex]E^{2}[/tex]A/m
where m is the mass of the asteroid.
can I get help please?
When a particular wire is vibrating with a frequency of 6.3 Hz, a transverse wave of wavelength 53.3 cm is produced. Determine the speed of wave pulses along the wire.
Answer:
335.79cm/s
Explanation:
When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;
v = fλ ------------------(ii)
From the question;
f = 6.3Hz
λ = 53.3cm
Substitute these values into equation (i) as follows;
v = 6.3 x 53.3
v = 335.79cm/s
Therefore, the speed of the wave pulses along the wire is 335.79cm/s
Which characteristic gives the most information about what kind of element an atom is ?
Answer:
The atomic number
Explanation:
a ring with a clockwise current is situated with its center directly above another ring. The current in the top ring is decreasing. What is the directiong of the induced current in the bottom ring
Answer:
clockwise
Explanation:
when current flows through a ring in a clockwise direction, it produces the equivalent magnetic effect of a southern pole of a magnet on the coil.
Since the current is decreasing, there is a flux change on the lower ring; generating an induced current on the lower ring. According to Lenz law of electromagnetic induction, "the induced current will act in such a way as to oppose the motion or the action producing it". In this case, the induced current will have to be the same polarity to the polarity of the current change producing it so as to repel the two rings far enough to stop the electromagnetic induction. The induced current will then be in the clockwise direction on the lower ring.
The direction of the induced current in the bottom ring is in the clockwise direction.
The given problem is based on the concept and fundamentals of the induced current and the direction of flow of the induced current.
When current flows through a ring in a clockwise direction, it produces the equivalent magnetic effect of a southern pole of a magnet on the coil. Since the current is decreasing, there is a flux change on the lower ring; generating an induced current on the lower ring. According to Lenz law of electromagnetic induction, "the induced current will act in such a way as to oppose the motion or the action producing it". In this case, the induced current will have to be the same polarity to the polarity of the current change producing it so as to repel the two rings far enough to stop the electromagnetic induction. The induced current will then be in the clockwise direction on the lower ring.Thus, we can conclude that the direction of the induced current in the bottom ring is in the clockwise direction.
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A 5.0-Ω resistor and a 9.0-Ω resistor are connected in parallel. A 4.0-Ω resistor is then connected in series with this parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors. What is the current through (a) the 4.0-Ω resistor? (b) the 5.0-Ω resistor? (c) the 9.0-Ω resistor?
Answer:
Explanation:
The current through the resistor is 0.83 A
.
Part b
The current through resistor is 0.53 A
.
Part c
The current through resistor is 0.30 A
The voltage between the cathode and the screen of a television set is 30 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed (m/s) just before it hits the screen
Answer:
The speed is [tex]v =10.27 *10^{7} \ m/s[/tex]
Explanation:
From the question we are told that
The voltage is [tex]V = 30 kV = 30*10^{3} V[/tex]
The initial velocity of the electron is [tex]u = 0 \ m/s[/tex]
Generally according to the law of energy conservation
Electric potential Energy = Kinetic energy of the electron
So
[tex]PE = KE[/tex]
Where
[tex]KE = \frac{1}{2} * m* v^2[/tex]
Here m is the mass of the electron with a value of [tex]m = 9.11 *10^{-31} \ kg[/tex]
and
[tex]PE = e * V[/tex]
Here e is the charge on the electron with a value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]e * V = \frac{1}{2} * m * v^2[/tex]
=> [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]
substituting values
[tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]
[tex]v =10.27 *10^{7} \ m/s[/tex]
An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)
Answer:
K_{f} / K₀ =1.12
Explanation:
This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.
Initial moment. With arms outstretched
L₀ = I₀ w₀
the wo value is 5.0 rad / s
final moment. After he shrugs his arms
[tex]L_{f}[/tex] = I_{f} w_{f}
indicate that the moment of inertia decreases by 11%
I_{f} = I₀ - 0.11 I₀ = 0.89 I₀
L_{f} = L₀
I_{f} w_{f} = I₀ w₀
w_{f} = I₀ /I_{f} w₀
let's calculate
w_{f} = I₀ / 0.89 I₀ 5.0
w_{f} = 5.62 rad / s
Having these values we can calculate the change in kinetic energy
[tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)
K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²
K_{f} / K₀ =1.12
A 100 kg lead block is submerged in 2 meters of salt water, the density of which is 1096 kg / m3. Estimate the value of the hydrostatic pressure.
Answer:
21,920 Pascals
Explanation:
P = ρgh
P = (1096 kg/m³) (10 m/s²) (2 m)
P = 21,920 Pa
A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is
The lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.
What is diffraction?The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.
Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So, The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.
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The Law of Biot-Savart shows that the magnetic field of an infinitesimal current element decreases as 1/r2. Is there anyway you could put together a complete circuit (any closed path of current-carrying wire) whose field exhibits this same 1/r^2 decrease in magnetic field strength? Explain your reasoning.
Answer and Explanation:
There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e. [tex]B \alpha \frac{1}{r^2}[/tex]
The magnetic field is a volume of vectors
And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit
Therefore for a closed path, we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.
When looking at the chemical symbol, the charge of the ion is displayed as the
-superscript
-subscript
-coefficient
-product
Answer:
superscript
Explanation:
When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.
An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.
Answer:
superscript
Explanation:
What is the one single most important reason that human impact on the planet has been so great?
Answer:
Increasing population
Explanation:
As we can see that the death rate is decreasing while at the same time the birth rate is increasing due to which it increased the population that directly impact the planet so great
Day by day the population of the villages, cities, states, the country is increasing which would create a direct human impact on the planet
Therefore the increasing population is the one and single most important reason
Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.
a) Is the momentum of Cart 1 conserved?
Yes
No
It depends on M
b) Is the momentum of Cart 2 conserved?
Yes
No
It depends on M
c) Is the total momentum of Carts 1 and 2 conserved?
Yes
No
It depends on M
d) Which cart ends up moving faster?
Cart 1
Cart 2
They move at the same speed
e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?
0 m/s
4.0 m/s
5.3 m/s
16 m/s
48 m/s
64 m/s
Answer:
a) yes
b) no
c) yes
d)Cart 2 with mass [tex]\frac{M}{3}[/tex] is expected to be more faster
e) u₂ = 48 m/s
Explanation:
a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.
b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved
c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction
note: m₁u₁ + m₂u₂ = (m₁ + m₂)v
d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.
e) Given
m₁= M
u₁ = 16m/s
m₂ =[tex]\frac{M}{3}[/tex]
u₂ = ?
from law of conservation of momentum
m₁u₁= m₂u₂
M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)
therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)
∴u₂ = 3×16 = 48 m/s
When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F
Answer:
F ’= F 0.25
Explanation:
This problem refers to the electric force, which is described by Coulomb's law
F = k q₁ q₂ / r²
where k is the Coulomb constant, q the charges and r the separation between them.
The initial conditions are
F = k q_A q_B / d²
they indicate that the loads are reduced to ¼ q and the distance is reduced to ½ d
F ’= k (q / 4 q / 4) / (0.5 d)²
F ’= k q / 16 / 0.25 d²
F ’= k q² / d² 0.0625 / 0.25
F ’= F 0.25
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. This can be explained through Coulomb's law.
What is Coulomb's law?Coulomb's law is a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
[tex]F = k \frac{q_Aq_B}{d^{2} } = k \frac{q^{2} }{d^{2} } [/tex]
where,
[tex]q_A [/tex] and [tex]q_B[/tex] are the charges of A and B (and equal to q).k is the Coulomb's constant.If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, the new force will be:
[tex]F_2 = k \frac{(0.25q_A)(0.25q_B)}{(0.5d)^{2} } = 0.25k\frac{q^{2} }{d^{2} } = 0.25 F[/tex]
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.
Learn more about Coulomb's law here: https://brainly.com/question/506926
Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!
Upon using Thomas Young’s double-slit experiment to obtain measurements, the following data were obtained. Use these data to determine the wavelength of light being used to create the interference pattern. Do this using three different methods.
The angle to the eighth maximum is 1.12°.
The distance from the slits to the screen is 302.0 cm.
The distance from the central maximum to the fifth minimum is 3.33 cm.
The distance between the slits is 0.000250 m.
The 3 equations I used were 1). d sin θ_m =(m)λ 2). delta x =λL/d and 3.) d(x_n)/L=(n-1/2)λ
but all my answers are different.
DID I DO SOMETHING WRONG!!!!!!!
Given info
d = 0.000250 meters = distance between slits
L = 302 cm = 0.302 meters = distance from slits to screen
[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])
[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )
---------------
Method 1
[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]
Make sure your calculator is in degree mode.
-----------------
Method 2
[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]
-----------------
Method 3
[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]
There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.
A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What will be the wavelength if the tension is decreased to 600 N and the frequency is kept constant
Answer:
The wavelength will be 33.9 cm
Explanation:
Given;
frequency of the wave, F = 1200 Hz
Tension on the wire, T = 800 N
wavelength, λ = 39.1 cm
[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]
Where;
F is the frequency of the wave
T is tension on the string
μ is mass per unit length of the string
λ is wavelength
[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]
when the tension is decreased to 600 N, that is T₂ = 600 N
[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]
Therefore, the wavelength will be 33.9 cm
A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \rhorho.
(a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rhorho.
(b) What is the electric field at a point outside the volume in terms of the charge per unit length \lambdaλ in the cylinder?
(c) Compare the answers to parts (a) and (b) for r = R.
(d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.
Answer:
the answers are provided in the attachments below
Explanation:
Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface
Applying Gauss law to the long solid cylinder
A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) E = 2K λ / r
C) Answers from parts a and b are the same
D) attached below
Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.
A ) The expression for the electric field inside the volume at a distance r
Gauss law : E. A = [tex]\frac{q}{e_{0} }[/tex] ----- ( 1 )
where : A = surface area = 2πrL , q = p(πr²L)
back to equation ( 1 )
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) Electric field at point Outside the volume in terms of charge per unit length λ
Given that: linear charge density = area * volume charge density
λ = πR²P
from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]
∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex] ----- ( 2 )
where : πR²P = λ
Back to equation ( 2 )
E = λ / 2e₀π*r where : k = 1 / 4πe₀
∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ
E = 2K λ / r
C) Comparing answers A and B
Answers to part A and B are similar
Hence we can conclude that Applying Gauss law to the long solid cylinder
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.
Learn more about Gauss's Law : https://brainly.com/question/15175106
The interference of two sound waves of similar amplitude but slightly different frequencies produces a loud-soft-loud oscillation we call __________.
a. the Doppler effect
b. vibrato
c. constructive and destructive interference
d. beats
Answer:
the correct answer is d Beats
Explanation:
when two sound waves interfere time has different frequencies, the result is the sum of the waves is
y = 2A cos 2π (f₁-f₂)/2 cos 2π (f₁ + f₂)/2
where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.
When examining the correct answer is d Beats
A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car
Answer:
9,375
Explanation:
Data provided
The mass of the car m = 1500 Kg.
The diameter of the circular track D = 200 m.
For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-
The radius of the circular path is
[tex]R = \frac{D}{2}[/tex]
[tex]= \frac{200}{2}[/tex]
= 100 m
after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula
[tex]Force F = \frac{mv^2}{R}[/tex]
[tex]= \frac{1500\times (25)^2}{100}[/tex]
= 9,375
Therefore for computing the magnitude of the centripetal force we simply applied the above formula.