Answer:
Qsinθ/4πε₀R²θ
Explanation:
Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.
Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.
So dq = λRdθ.
Substituting dq into dE', we have
dE' = dqcosθ/4πε₀R²
= λRdθcosθ/4πε₀R²
= λdθcosθ/4πε₀R
E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ
E' = (λ/4πε₀R)[sinθ] from -θ to θ
E' = (λ/4πε₀R)[sinθ]
= (λ/4πε₀R)[sinθ - sin(-θ)]
= (λ/4πε₀R)[sinθ + sinθ]
= 2(λ/4πε₀R)sinθ
= (λ/2πε₀R)sinθ
Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ
Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ
Q = 2λRθ
λ = Q/2Rθ
Substituting λ into E', we have
E' = (Q/2Rθ/2πε₀R)sinθ
E' = (Q/θ4πε₀R²)sinθ
E' = Qsinθ/4πε₀R²θ where θ is in radians
Air bags greatly reduces the chance og injury in a car accident.explain how they do si in terms of energy transfer
Answer:
in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.
3. What are the first steps that you should take if you are unable to get onto the Internet? (1 point)
O Check your router connections then restart your router.
O Plug the CPU to a power source and reboot the computer.
O Adjust the display properties and check the resolution.
Use the Control Panel to adjust the router settings.
Answer:
Check your router connections then restart your router.
Explanation:
Answer:
Check your router connections then restart your router.
Explanation:
Most internet access comes from routers so the problem is most likely the router.
An LR circuit consists of a 35-mH inductor, ac resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed
Answer:
I = 1.23 A
Explanation:
In an RL circuit current passing is described by
I = E / R (1 - [tex]e^{-Rt/L}[/tex])
Let's reduce the magnitudes to the SI system
L = 35 mH = 35 10⁻³ H
t = 5.0 ms = 5.0 10⁻³ s
let's calculate
I = 18/12 (1 - [tex]e^{-12 .. 5 {10}^{-3}/35 .. {10}^{-3} }[/tex]e (- 5 10-3 12/35 10-3))
I = 1.5 (1- [tex]e^{-1.715}[/tex])
I = 1.23 A
What would happen in a State if its citizens lack relevant knowledge, skills
and positive attitude?
Answer:
If the older generation is lacking, the younger generation would likely have knowledge, skill, or a positive attitude in some combination, but it is relative to the culture.
The simple reason is the desirability for genetic variation using recessive genes.
In other words, if the older generation lacks something, it tends to be something they don’t need, but something that will look good on young people. But mostly relative to the culture and education system.
Hope this helps
A block of mass M rests on a block of mass M1 which is on a tabletop. A light string passes over a frictionless peg and connects the blocks. The coefficient of kinetic friction between the blocks and between M1 and the tabletop is the same. A force F pulls the upper block to the left and the lower block to the right. The blocks are moving at a constant speed.
Determine the mass of the upper block. (Express your answer to three significant figures.)
Answer:
M = F/3μ g - M₁/3
Explanation:
To solve this exercise we must use the equilibrium conditions translations
∑ F = 0
In the attachment we can see a free body diagram of each block
Block M (upper)
X axis
fr₁ + F₂ -F = 0
F = fr₁ + F₂ (1)
axis
N₁-W = 0
N₁ = Mg
the friction force has the formula
fr₁ = μ N₁
F = μ Mg + F₂
bottom block
X axis
F₂ - fr₁ - fr₂ = 0
F₂ = fr₁ + fr₂
Y axis
N - W₁ -W = 0
N = g (M + M₁)
we substitute
F₂ = μ Mg + μ (M + M1) g
F₂ = μ g (2M + M₁)
we substitute in 1
F = μ M g + μ g (2M + M₁)
F = μ g (3M + M₁)
we look for mass M
M = (F - μ g M₁)/ 3μ g
M = F/3μ g - M₁/3
the exercise does not have numerical data
Two instruments produce a beat frequency of 5 Hz. If one has a frequency of 264 Hz, what could be the frequency of the other instrument
Answer:
259 Hz or 269 Hz
Explanation:
Beat: This is the phenomenon obtained when two notes of nearly equal frequency are sounded together. The S.I unit of beat is Hertz (Hz).
From the question,
Beat = f₂-f₁................ Equation 1
Note: The frequency of the other instrument is either f₁ or f₂.
If the unknown instrument's frequency is f₁,
Then,
f₁ = f₂-beat............ equation 2
Given: f₂ = 264 Hz, Beat = 5 Hz
Substitute into equation 2
f₁ = 264-5
f₁ = 259 Hz.
But if the unknown frequency is f₂,
Then,
f₂ = f₁+Beat................. Equation 3
f₂ = 264+5
f₂ = 269 Hz.
Hence the beat could be 259 Hz or 269 Hz
In the lab , you have an electric field with a strength of 1,860 N/C. If the force on a particle with an unknown charge is 0.02796 N, what is the value of the charge on this particle.
Answer:
The charge is [tex]q = 1.50 *10^{-5} \ C[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 1860 \ N/C[/tex]
The force is [tex]F = 0.02796 \ N[/tex]
Generally the charge on this particle is mathematically represented as
[tex]q = \frac{F}{E}[/tex]
=> [tex]q = \frac{0.02796}{ 1860}[/tex]
=> [tex]q = 1.50 *10^{-5} \ C[/tex]
The equivalent resistance of two resistors connected in series is always greater than the equivalent resistance of the same two resistors connected in parallel. True False
Answer:
True
Explanation:
Because the resistors in series is the sum of the two resistors given as
R= R1+R2
While that of resistors in parallel is the sum of the reciprocal of the resistance given as
1/R = 1/ R1+ 1/R2
So that of series connection will be greater
wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength
Complete question:
Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:
a) the midpoint between the two rings?
b) the center of the left ring?
Answer:
a) the electric field strength at the midpoint between the two rings is 0
b) the electric field strength at the center of the left ring is 2712.44 N/C
Explanation:
Given;
distance between the two rings, d = 25 cm = 0.25 m
diameter of each ring, d = 10 cm = 0.1 m
radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]
the charge on each ring, q = 20 nC
Electric field strength for a ring with radius r and distance x from the center of the ring is given as;
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]
The electric field strength at the midpoint;
the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]
[tex]E_{left} = 9210.5 \ N/C[/tex]
The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;
[tex]E_{right} = -9210.5 \ N/C[/tex]
The electric field strength at the midpoint;
[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]
(b)
The distance from the right ring to center of the left ring, x = 0.25 m.
[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.
Answer:
e. It is neither attracted nor repelled.
Explanation:
Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.
Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R
Answer:
The value of resistance of each resistor, R is 2.25 Ω
Explanation:
Given;
voltage across the three resistor, V = 1.5 V
power dissipated by the resistors, P = 3.00 W
the resistance of each resistor, = R
The effective resistance of the three resistors is given by;
R(effective) = R/3
Apply ohms law to determine the current delivered by the source;
V = IR
I = V/R
I = 3V/R
Also, power is calculated as;
P = IV
P = (3V/R) x V
P = 3V²/R
R = 3V² / P
R = (3 x 1.5²) / 3
R = 2.25 Ω
Therefore, the value of resistance of each resistor, R is 2.25 Ω
A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds. What is its rotation speed
Answer:
v = 6.28 m/s
Explanation:
It is given that,
A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0.5 seconds,
Number of revolution is half. It means angular velocity is 3.14 radians.
Let v is the angular speed. So,
[tex]v=\dfrac{\omega}{t}\\\\v=\dfrac{3.14}{0.5}\\\\v=6.28\ m/s[/tex]
So, the rotation speed is 6.28 m/s.
The angular velocity is the rotation speed, which is the angle of rotation
of the windmill per second, which is 2·π radians.
Response:
The rotation speed is 2·π rad/sHow can the rotational speed of the windmill be calculated?The given parameter are;
The angle of rotation the windmill rotates in 0.5 seconds = One-half a
rotation.
Required:
The rotational speed (angular velocity)
Solution:
The angle of one rotation = 2·π radians
Angle of one-half ration = [tex]\frac{1}{2}[/tex] × 2·π radians = π radians
[tex]Rotational \ speed = \mathbf{\dfrac{Angle \ of \ rotation}{Time}}[/tex]
Which gives;
[tex]Rotational \ speed, \omega = \dfrac{\pi}{0.5 \ s} = \mathbf{2 \cdot \pi \ rad/s}[/tex]
The rotation speed is 2·π rad/sLearn more about rotational speed here:
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A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume that the loop is perpendicular to the magnetic field. What is the average induced emf?
Answer:
9.88 milivolt
Explanation:
Given: diameter d = 5.2 cm
magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T
t = 0.29 sec.
we know emf = - dΦ/dt
and flux Φ = BA
A= area
therefore emf ε = -A(B_2-B_1)/Δt
[tex]=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV[/tex]
Water is pumped with a 120 kPa compressor entering the lower pipe (1) and flows upward at a speed of 1 m/s. Acceleration due to gravity is 10 m/s and water density is1000 kg/m-3. What is the water pressure on the upper pipe (II).
Answer:
The water pressure on the upper pipe is 92.5 kPa.
Explanation:
Given that,
Pressure in lower pipe= 120 kPa
Speed of water in lower pipe= 1 m/s
Acceleration due to gravity = 10 m/s²
Density of water = 1000 kg/m³
Radius of lower pipe = 12 m
Radius of uppes pipe = 6 m
Height of upper pipe = 2 m
We need to calculate the velocity in upper pipe
Using continuity equation
[tex]A_{1}v_{1}=A_{2}v_{1}[/tex]
[tex]\pi r_{1}^2\times v_{1}=\pi r_{2}^2\times v_{2}[/tex]
[tex]v_{2}=\dfrac{r_{1}^2\times v_{1}}{r_{2}^2}[/tex]
Put the value into the formula
[tex]v_{2}=\dfrac{12^2\times1}{6^2}[/tex]
[tex]v_{2}=4\ m/s[/tex]
We need to calculate the water pressure on the upper pipe
Using bernoulli equation
[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]
Put the value into the formula
[tex]120\times10^{3}+\dfrac{1}{2}\times1000\times1^2+1000\times10\times0=P_{2}+\dfrac{1}{2}\times1000\times(4)^2+1000\times10\times2[/tex]
[tex]120500=P_{2}+28000[/tex]
[tex]P_{2}=120500-28000[/tex]
[tex]P_{2}=92500\ Pa[/tex]
[tex]P_{2}=92.5\ kPa[/tex]
Hence, The water pressure on the upper pipe is 92.5 kPa.
If we compare the force of gravity to strong nuclear force, we could conclude that
O gravity is the weaker force; it is related to mass
O gravity is the stronger force; it is related to distance
strong nuclear is the stronger force; it is related to mass
O strong nuclear is the weaker force; it is related to distance
Answer:
strong nuclear is the stronger force; it is related to mass
Explanation:
If we compare the force of gravity to strong nuclear force, we could conclude that strong nuclear is the stronger force; it is related to mass, therefore the correct answer is option C
What are nuclear forces?The nuclear force is the interaction between the subatomic particles that make up a nucleus. There are two types of nuclear forces: the strong nuclear force and the weak nuclear force. Depending on the separation between the proton neutron and proton pairs, these nuclear forces can be both attracting and positive.
Both types of nuclear forces come under the four fundamental forces of nature. There are mainly four fundamental forces of nature electromagnetic force, gravitational force, strong nuclear force, and weak nuclear force.
Thus, Option C is the appropriate response since, when compared to the force of gravity, the strong nuclear force is the greater force because it is tied to mass.
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Two hoops, staring from rest, roll down identical incline planes. The work done by nonconservative forces is zero. The hoops have the same mass, but the larger hoop has twice the radius. Which hoop will have the greater total kinetic energy at the bottom
Answer:
They both have the same total K.E at the bottom
Explanation:
This Is because If assuming no work is done by non conservative forces, total mechanical energy must be conserved
So
K1 + U1 = K2 + U2
But If both hoops start from rest, and and at the bottom of the incline the level for gravitational potential energy is zero for reference
thus
K1 = 0 , U2 = 0
ΔK = ΔU = m g. h
But if the two inclines have the same height, and both hoops have the same mass m,
So difference in kinetic energy, must be the same for both hoops.
Which is a “big idea” for space and time? Energy can be transferred but not destroyed. Forces describe the motion of the universe. The universe is very big and very old. The universe consists of matter.
Answer:
Explanation:
That Universe Consists of Matter
What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV.
Answer:
The maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.
Explanation:
Given;
work function of silver, Φ = 2.93 eV = 2.93 x 1.602 x 10⁻¹⁹ J = 4.6939 x 10⁻¹⁹ J
Apply Einstein Photo electric effect;
E = K.E + Ф
Where;
E is the energy of the incident light
K.E is the kinetic of electron
Ф is the work function of silver surface
For the incident light to have maximum wavelength, the kinetic energy of the electron will be zero.
E = Ф
hf = Ф
[tex]h\frac{c}{\lambda} = \phi[/tex]
where;
c is speed of light = 3 x 10⁸ m/s
h is Planck's constant, = 6.626 x 10⁻³⁴ J/s
λ is the wavelength of the incident light
[tex]\lambda = \frac{hc}{\phi}\\\\\lambda =\frac{6.626*10^{-34} *3*10^8}{4.6939*10^{-19}} \\\\\lambda = 4.235 *10^{-7} \ m\\\\\lambda = 423.5 *10^{-9} \ m\\\\\lambda = 423.5 \ nm[/tex]
Therefore, the maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.
At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to:_____
(a) 3 times that of the Celsius and
(b) 1/5 times that of the Celsius
Answer:
C = 26.67° and F = 80°C = -20° and F = -4°Explanation:
Find:
3 times that of the Celsius and 1/5 times that of the CelsiusComputation:
F = (9/5)C + 32
3 times that of the Celsius
If C = x
So F = 3x
So,
3x = (9/5)x + 32
15x = 9x +160
6x = 160
x = 26.67
So, C = 26.67° and F = 80°
1/5 times that of the Celsius
If C = x
So F = x/5
So,
x/5 = (9/5)x + 32
x = 9x + 160
x = -20
So, C = -20° and F = -4°
Object A, with heat capacity CA and initially at temperature TA, is placed in thermal contact with object B, with heat capacity CB and initially at temperature TB. The combination is thermally isolated. If the heat capacities are independent of the temperature and no phase changes occur, the final temperature of both objects is
Answer:
d) (CATA + CBTB) / (CA + CB)
Explanation:
According to the given situation, the final temperature of both objects is shown below:-
We assume T be the final temperature
while m be the mass
So it will be represent
m CA (TA - T) = m CB (T - TB)
or we can say that
CATA - CA T = CB T - CBTB
or
(CA + CB) T = CATA + CBTB
or
T = (CA TA + CBTB) ÷ (CA + CB)
Therefore the right answer is d
The final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
The given parameters;
heat capacity of object A = CAinitial temperature of object A = TAheat capacity of object B = CBinitial temperature of object B = TBThe final temperature of both objects is calculated as follows;
heat lost by object A is equal to heat gained by object B
[tex]mC_A (T_A - T) = mC_B(T- T_B)\\\\C_AT_A-C_AT = C_BT - C_BT_B\\\\C_BT+C_AT = C_AT_A+ C_BT_B\\\\T(C_B + C_A) = C_AT_A+ C_BT_B \\\\T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex]
Thus, the final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
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"A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.33. Wavelengths of 479 nm and 798 nm and no wavelengths between are intensified in the reflected beam. The thickness of the film is"
Answer:
t = 8.98 10⁻⁷ m
Explanation:
This is an exercise in interference by reflection, let's analyze what happens on each surface of the film.
* When the light ray shifts from a medium with a lower refractive index to a medium with a higher refractive index, the reflected ray has a reflection of 180
* The beam when passing to the middle its wavelength changes
λ = λ₀ / n
if we take this into account, the constructive interference equation for normal incidence is
2t = (m + ½) λ₀ / n
let's apply this equation to our case
for λ₀ = 479 nm = 479 10⁻⁹ m
t = (m + ½) 479 10⁻⁹ / 1.33
(m + ½) = 1.33 t / 479 10⁻⁹
for λ₀ = 798 nm = 798 10⁻⁹ m
t = (m' + ½) 798 10⁻⁹ /1.33
(m' + ½) = 1.33 t / 798 10⁻⁹
as they tell us that no other constructive interference occurs between the two wavelengths, the order of interference must be consecutive, let's write the two equat⁻ions
(m + ½) = 1.33 t / 479 10⁻⁹
((m-1) + ½) = 1.33 t / 798 10⁻⁹
(m + ½) = 1.33 t / 798 10⁻⁹ +1
resolve
1.33 t / 479 10⁻⁹ = 1.33 t / 798 10⁻⁹ +1
1.33 t / 479 10⁻⁹ = (1.33t + 798 10⁻⁹) / 798 10⁻⁹
1.33t = (1 .33t + 798 10⁻⁹) 479/798
1.33t = (1 .33t + 798 10⁻⁹) 0.6
1.33 t = 0.7983 t + 477.6 10⁻⁹
t (1.33 - 0.7983) = 477.6 10⁻⁹
t = 477.6 10⁻⁹ /0.5315
t = 8.98 10⁻⁷ m
A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.
Answer: Ф = 17.2657 ≈ 17°
Explanation:
we simply apply ET =0 about the ending of the rod
so In.g.L/2sinФ - In.a.L/2cosФ = 0
g.sinФ - a.cosФ = 0
g.sinФ = a.cosФ
∴ tanФ = a/g
Ф = tan⁻¹ a / g
Ф = tan⁻¹ ( 10 / 32.17405)
Ф = tan⁻¹ 0.31080948777
Ф = 17.2657 ≈ 17°
Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°
3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000 rpm, completando un total de 12 revoluciones.
d. Determina la aceleración tangencial, centrípeta y la total en m.s-2 cuando el tambor a alcanzado los 60000 rpm
e. Explica lo que ocurre con la magnitud y dirección de los vectores aceleración tangencial, aceleración centrípeta, aceleración total, aceleración angular, velocidad angular cuando la lavadora ha girado desde 3000 rpm hasta 6000 rpm.
Answer:
d) α = 1693.5 rad / s² , a = 392.7 m / s² , a_total = α √(R² +1) ,
e) tan θ = a / α
Explanation:
This is an exercise in linear and angular kinematics.
We initialize reduction of all the magnitudes to the SI system
w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s
w = 6000 rev / mi = 628.32 rad / s
θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad
d) ask for centripetal, tangential and total acceleration.
Let's start by looking for centripetal acceleration, let's use the formula
w² = w₀² + 2 α θ
α = (w²- w₀²) / 2θ
we calculate
α = (628.32²2 - 314.16²) / 2 75.398
α = 1693.5 rad / s²
the quantity is linear and angular are related
the linear or tangential acceleration is
a = α R
where R is the radius of the drum
a = 1693.5 R
Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m
a = 1693.5 0.20
a = 392.7 m / s²
the total acceleration is
a_total = √(a² + α²)
a_total = √ (α² R² + α²)
a_total = α √(R² +1)
e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant
Tangential acceleration is tangency to radius and its value varies proportionally radius
the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry
tan θ = a / α
the angular velocity increases linearly when with centripetal acceleration
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
B. Which skater, if either, has the greater speed after the push-off? Explain.
Answer:
the two ice skater have the same momentum but the are in different directions.
Paula will have a greater speed than Ricardo after the push-off.
Explanation:
Given that:
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off
The law of conservation of momentum states that the total momentum of two or more objects acting upon one another will not change, provided there are no external forces acting on them.
So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.
Momentum is the product of mass and velocity.
SO, from the information given:
Let represent the mass of Paula with [tex]m_{Pa}[/tex] and its initial velocity with [tex]u_{Pa}[/tex]
Let represent the mass of Ricardo with [tex]m_{Ri}[/tex] and its initial velocity with [tex]u_{Ri}[/tex]
At rest ;
their velocities will be zero, i.e
[tex]u_{Pa}[/tex] = [tex]u_{Ri}[/tex] = 0
The initial momentum for this process can be represented as :
[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = 0
after push off from each other then their final velocity will be [tex]v_{Pa}[/tex] and [tex]v_{Ri}[/tex]
The we can say their final momentum is:
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0
Using the law of conservation of momentum as states earlier.
Initial momentum = final momentum = 0
[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
Since the initial velocities are stating at rest then ; u = 0
[tex]m_{Pa}[/tex](0) + [tex]m_{Pa}[/tex](0) = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = - [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.
B. Which skater, if either, has the greater speed after the push-off? Explain.
Given that Ricardo weighs more than Paula
So [tex]m_{Ri} > m_{Pa}[/tex] ;
Then [tex]\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}[/tex]
The magnitude of their momentum which is a product of mass and velocity can now be expressed as:
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
The ratio is
[tex]\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1[/tex]
[tex]v_{Pa} >v_{Ri}[/tex]
Therefore, Paula will have a greater speed than Ricardo after the push-off.
(A) Both the skaters have the same magnitude of momentum.
(B) Paula has greater speed after push-off.
Conservation of momentum:Given that two skaters Paula and Ricardo are initially at rest.
Ricardo weighs more than Paula.
Let us assume that the mass of Ricardo is M, and the mass of Paula is m.
Let their final velocities be V and v respectively.
(A) Initially, both are at rest.
So the initial momentum of Paula and Ricardo is zero.
According to the law of conservation of momentum, the final momentum of the system must be equal to the initial momentum of the system.
Initial momentum = final momentum
0 = MV + mv
MV = -mv
So, both of them have the same magnitude of momentum, but in opposite directions.
(B) If we compare the magnitude of the momentum of Paula and Ricardo, then:
MV = mv
M/m = v/V
Now, we know that M>m
so, M/m > 1
therefore:
v/V > 1
v > V
So, Paula has greater speed.
Learn more about conservation of momentum:
https://brainly.com/question/2141713?referrer=searchResults
on which principle does water pump work ?
Answer:
The working principle of a water pump mainly depends upon the positive displacement principle as well as kinetic energy to push the water.
Explanation:
it mainly depends upon the positive displacement principle and also kinetic energy to push water. hope this hepls!
A rigid container holds 4.00 mol of a monatomic ideal gas that has temperature 300 K. The initial pressure of the gas is 6.00 * 104 Pa. What is the pressure after 6000 J of heat energy is added to the gas?
Answer:
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
Explanation:
When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.
In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:
[tex]\bar c_{v} = \frac{3}{2}\cdot R_{u}[/tex]
Where:
[tex]R_{u}[/tex] - Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.
If [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], then:
[tex]\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)[/tex]
[tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]
And change in heat energy ([tex]Q[/tex]), measured by joules, by:
[tex]Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})[/tex]
Where:
[tex]n[/tex] - Molar quantity, measured in moles.
The final temperature of the monoatomic ideal gas is now cleared:
[tex]T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}[/tex]
Given that [tex]T_{1} = 300\,K[/tex], [tex]Q = 6000\,J[/tex], [tex]n = 4\,mol[/tex] and [tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex], the final temperature is:
[tex]T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}[/tex]
[tex]T_{2} = 420.279\,K[/tex]
The final pressure of the system is calculated by the following relationship:
[tex]P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}[/tex]
If [tex]T_{1} = 300\,K[/tex], [tex]T_{2} = 420.279\,K[/tex] and [tex]P_{1} = 6.00\times 10^{4}\,Pa[/tex], the final pressure is:
[tex]P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)[/tex]
[tex]P_{2} = 8.406\times 10^{4}\,Pa[/tex]
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the resistor is
Answer:
At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage
Explanation;
Because at series connection the battery and resistor have equal voltage
Physical properties of a mineral are a result of the arrangement of the atoms in the mineral. Use this fact to explain the following:_________
A. One mineral has a density of 2.7 g/ml while another has a density of 5.1 g/ml
B. The mineral mica cleaves into thin flat sheets while olivine does not show cleavage
Explanations:
a) The physical properties of a mineral is as a result of the arrangement of the atoms in the minerals. The reason behind one mineral having a density of 2.1 g/ml which is lower than that of another mineral with density of 5.1 g/ml is the packing density of the minerals. Minerals with high density have their atoms more closely packed together, leaving less space between the atoms. This characteristics means that they have more atomic mass per unit volume for a given molecular space, when compared to another mineral with its atoms less closely packed.
b) The property of cleavage is due to the crystalline structure of a mineral species. Cleavage is used to describe the ease with which minerals cleaves. Cleavage is due to a weak bonding strength between the molecules of the mineral, or a due to a greater lattice spacing across the the cleavage plane than in other planes within the crystal. The greater the lattice spacing, the weaker the strength of the bond across a plane.
From these, we can clearly see that the property of cleavage is due to the physical properties of the crystalline structure of the mineral species.
You have two capacitors and want to connect them across a voltage source (battery) to store the maximum amount of energy. Should they be connected in series or in parallel?
Answer:
In parallel
Explanation:
Ctotal = C1 + C2 + ... + Cn
The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresponds to a 1 % change in length.
A. Young's modulus for bone is about Y = 1.4 x 10 N/m². The tibia (shin bone) of a human is 0.35 m long and has an average cross-sectional area of 2.9 cm. What is the effective spring constant of the tibia?
B. If a man weighs 750 N, how much is the tibia compressed if it supports half his weight?
C. What is the maximum force that can be applied to a tibia with a cross-sectional area, A = 2.90 cm?
Answer:
a
[tex]k = 11600000 N/m[/tex]
b
[tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]
c
[tex]F = 3750.28 \ N[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]E = 1.4 *10^{10} \ N/m^2[/tex]
The length is [tex]L = 0.35 \ m[/tex]
The area is [tex]2.9 \ cm^2 = 2.9 *10^{-4} \ m ^2[/tex]
Generally the force acting on the tibia is mathematically represented as
[tex]F = \frac{E * A * \Delta L }{L}[/tex] derived from young modulus equation
Now this force can also be mathematically represented as
[tex]F = k * \Delta L[/tex]
So
[tex]k = \frac{E * A }{L}[/tex]
substituting values
[tex]k = \frac{1.4 *10^{10} * 2.9 *10^{-4} }{ 0.35}[/tex]
[tex]k = 11600000 N/m[/tex]
Since the tibia support half the weight then the force experienced by the tibia is
[tex]F_k = \frac{750 }{2} = 375 \ N[/tex]
From the above equation the extension (compression) is mathematically represented as
[tex]\Delta L = \frac{ F_k * L }{ A * E }[/tex]
substituting values
[tex]\Delta L = \frac{ 375 * 0.35 }{ (2.9 *10^{-4}) * 1.4*10^{10} }[/tex]
[tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]
From the above equation the maximum force is
[tex]F = \frac{1.4*10^{10} * (2.9*10^{-4}) * 3.233*10^{-5} }{ 0.35}[/tex]
[tex]F = 3750.28 \ N[/tex]
