A cylindrical 5.00-kg reel with a radius of 0.600 m and a frictionless axle, starts from rest and speeds up uniformly as a 3.00-kg bucket falls into a well, making a light rope unwind from the reel. The bucket starts from rest and falls for 4.00 s.
Required:
a. What is the linear acceleration of the falling bucket?
b. How far does it drop?
c. What is the angular acceleration of the reel?
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,
[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]
I assume the path itself is a line segment, which can be parameterized by
[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]
with 0 ≤ t ≤ 1. Then the work performed by F along C is
[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]
Cho dòng điện xoay chiều trong sản xuất và sinh hoạt ở nước ta có tần số f = 50Hz. Tính chu kỳ T và tần số góc ω?
Answer:
T = 1/f = 1/50(s)
ω = 2πf = 100π (rad/s)
(vote 5 sao nhó :3 )
9. From this lab, we learn that the electric field and electric potential depend on both, the magnitude of the source charge (q), and the distance from the source charge (r). If we were to increase the magnitude of our source charge from 1 nC to 5 nC, then the magnitudes of the electric field and electric potential would be ____.(you can test this on the animation by dragging five 1 nC charges on top of each other and measuring E and V at a distance of 1 m)
Answer:
the electric field and the electric potential increase 5 times
Explanation:
The electric field created by a point charge is
E = k q / r²
in this case the charge changes from q₁ = 1 10⁺⁰ C to q₂ = 5 10⁻⁹ C
with the electric field is proportional to the charge
E₅ = 5 E₁
the electrical power for a point charge is
V = k q / r
as the electric power is proportional to the charge
V₅ = 5 V₁
consequently both the electric field and the electric potential increase 5 times
An electric field is produced by a charged object:
[tex]\to E = \frac{k q}{r^2}\\\\[/tex]
In this situation, the charge shifts:
[tex]\to q_1 = 1 10^{\circ}\ C \\\\ \to q_2 = 5 \times 10^{-9}\ C[/tex]
so with electric field remaining proportional to the charge
[tex]\to E_5 = 5 E_1[/tex]
The electrical power consumed for a point charge:
[tex]\to V = \frac{k q}{ r}[/tex]
since the electric power is related to the charge
[tex]\to V_5 = 5 V_1[/tex]
As a result, both the electric field as well as the electric potential increase by 5 times.
Learn more:
brainly.com/question/2017486
If a proton and electron both move through the same displacement in an electric field, is the change in potential energy associated with the proton equal in magnitude and opposite in sign to the change in potential energy associated with the electron?
a. The magnitude of the change is smaller for the proton.
b. The magnitude of the change is larger for the proton.
c. The signs Of the two changes in potential energy are opposite.
d. They are equal in magnitude.
e. The signs of the two changes in potential energy are the same.
Answer: They are equal in magnitude.
- The signs of the two changes in potential energy are opposite
Explanation:
When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.
Also, it should be noted that the signs of the two changes in potential energy are opposite.
any four difference between speed and acceleration
Speed:
Measured in distance / timeIt tells the speed of a bodyIt is a scalar quantitySpeed is always positiveIt's unit is meter per secondAcceleration:
Measured in velocity / timeIt tells the change in the velocityImplied on change in a velocityIt is a vector quantityIt's unit is meter per second square
From the given picture What's the force? And where did it happen? (at least 2 forces)
Answer:
the force happens on the wall and couch
Explanation:
she is using her arm strength to lift and hold
In December of 2011 they announced that a planet has been discovered in a habitable zone around a
star! It has clouds! It has twice the radius of the earth, but with the same density as earth, about 5.515 × 10^3kg/m3
. Find the new acceleration of gravity on the surface of this planet.
Explanation:
The density of earth [tex]\rho_E[/tex] is given by
[tex]\rho_E = \dfrac{M_E}{\left(\frac{4\pi}{3}R_E^3\right)}[/tex]
and in terms of this density, we can write the acceleration due to gravity on earth as
[tex]g_E =G\dfrac{M_E}{R_E^2} = \dfrac{4\pi G}{3}\rho_ER_E[/tex]
Similarly, the acceleration due to gravity [tex]g_P[/tex] on this new planet is given by
[tex]g_P = G\dfrac{M_P}{R_P^2} = G\dfrac{\frac{4\pi}{3}R_p^3\rho_P}{R_P^2}[/tex]
[tex]\:\:\:\:\:= \dfrac{4\pi G}{3}\rho_PR_P[/tex]
We know that this planet has the same density as earth and has a radius 2 times as large. We can then rewrite [tex]g_P[/tex] as
[tex]g_P = \dfrac{4\pi G}{3}\rho_E(2R_E)[/tex]
[tex]\:\:\:\:\:= 2\left(\dfrac{4\pi G}{3}\rho_ER_E\right) = 2g_E[/tex]
[tex]\:\:\:\:\:= 2(9.8\:\text{m/s}^2) = 19.6\:\text{m/s}^2[/tex]
A capacitor is connected to an ac generator that has a frequency of 3.2 kHz and produces a rms voltage of 2.0 V. The rms current in the capacitor is 28 mA. When the same capacitor is connected to a second ac generator that has a frequency of 4.7 kHz, the rms current in the capacitor is 70 mA. What rms voltage does the second generator produce
Answer:
The rms voltage of new generator is 3.4 V.
Explanation:
f = 3200 Hz
rms voltage, V = 2 V
rms current, i = 28 mA
Now
f' = 4700 Hz
rms current, i' = 70 mA
let the new rms voltage is V'.
[tex]i = \frac{V}{Xc} = V \times 2\pi fC....(1)\\\\i' = V' \times 2 \pi f' C..... (2)\\\\\frac{i}{i'} =\frac{V f}{V' f'}\\\\\frac{28}{70}=\frac{2\times 3200}{V'\times 4700}\\\\V' = 3.4 V[/tex]
Using the given temperature and pressures, determine: a) the diameter of the water scale model balloon (m), b) the weight of the scale model, c) the specific gravity of the buoyant material such that the model conditions will be similar to the full-scale balloon.
Complete Question
Complete Question is attached below
Answer:
a) [tex]D=0.7[/tex]
b) [tex]W=1787.5N[/tex]
c) [tex]\rho'=998.19kg/m^3[/tex]
Explanation:
From the question we are told that:
Hot air:
Temperature [tex]T_a=360K[/tex]
Pressure [tex]P_a=100kPa[/tex]
Distance [tex]d=12m[/tex]
Weight [tex]W=1400N[/tex]
Water:
Temperature [tex]T_w=300K[/tex]
Pressure [tex]P_w=100kPa[/tex]
Since we have The same Reynolds number
a)
Generally the equation for equal Reynolds number is mathematically given by
Re_{air}=Re_{water}
Therefore
[tex]\frac{\rho V D}{\mu_{air}}=\frac{p_{water}*V*D}{\mu_{water]}}[/tex]
[tex]\frac{100*12}{300*0.28*1.81*10^{-3}}}=\frac{998*D}{0.000890}[/tex]
[tex]D=0.7[/tex]
b)
Generally the equation for Weight of scale is mathematically given by
[tex]W=\rho*V*g[/tex]
[tex]W=998*\frac{4}{3}*\pi*0.35^3*9.81[/tex]
[tex]W=1787.5N[/tex]
c)
Generally the equation for Density of buoyant material is mathematically given by
[tex]\rho'=\frac{w}{g*V}[/tex]
[tex]\rho'=\frac{1781.5}{\frac{4}{3}*\pi*0.35^3*9.81}[/tex]
[tex]\rho'=998.19kg/m^3[/tex]
Identify the factors that affect the intensity of radiation detected from a radioactive source. Select one or more: The color of the source Type of emission from the source Distance of the detector from the source Type of materials between the source and the detector
The intensity of radiation is the defined as amount of energy per surface angle which can be used to determine the amount of energy emitting from a source that will hit another surface.
The factors that affect the intensity of radiation are
Type of emission from the source :This can be alpha, gamma, beta or electromagnetic rays etc
Distance of the detector from the source: The shorter the distance between the source and the detector, the more the effect and vice versa for the longer the distance.
Type of materials between the source and the detector: The type of material between the source and the detector will tell how absorbing and penetrating the radiation is.
Read more on Radiation Intensity here: https://brainly.com/question/10148635
A motorcyclist start from rest to reaches 6m/s with uniform acceleration for 3s what his acceleration?
Answer:
[tex]\boxed {\boxed {\sf 2 \ m/s^2}}[/tex]
Explanation:
Acceleration is the rate of change in velocity with respect to time. It is calculated by dividing the change in velocity by the change in time. The formula is:
[tex]a= \frac{ \Delta v}{\Delta t}[/tex] or [tex]a= \frac{v_f-v_i}{\Delta t}[/tex]
The change in velocity is the difference between the initial velocity and the final velocity. The motorcycle starts at rest, or 0 meters per second and reaches 6 meters per second. The change in time is 3 seconds.
[tex]\bullet \ v_f= 6 \ m/s\\\bullet \ v_i= 0 \ m/s \\\bullet \ \Delta t = 3 \ s[/tex]
Substitute the values into the formula
[tex]a= \frac { 6 \ m/s - 0 m/s}{3 \ s}[/tex]
Solve the numerator.
[tex]a= \frac{6 \ m/s}{3 \ s}[/tex]
[tex]a= 2 \ m/s^2[/tex]
The motorcyclist's acceleration is 2 meters per second squared.
What is the incorrect statement regarding the isotopes of the same element?
1) Electronic configuration is equal
2) Mass number is equal
3) Number of protons are equal
4) Number of electrons are equal
Answer:
1231
Explanation:
Two charged objects attract each other with a force 1.0 N. What happens to the force between them if one charge is increased by a factor of 2, the other charge is increased by a factor of 4, and the separation distance between their centers is reduced to 1/4 its original value
Answer:
F' = 128 N
Explanation:
The electrostatic force of attraction between two charges is given by Colomb's Law, as follows:
[tex]F = \frac{kq_1q_2}{r^2}\\\\[/tex]
where,
F = Force of attraction = 1 N
G = universal gravitational constant
q₁ = magnitude of the first charge
q₂ = magnitude of the second charge
r = distance between charges
Therefore,
[tex]1\ N = \frac{kq_1q_2}{r^2}[/tex] --------------------- eq(1)
Now, we apply the changes given in the question:
[tex]F' = \frac{k(2q_1)(4q_2)}{(\frac{1}{4}r)^2}\\\\F' = 128(\frac{kq_1q_2}{r^2})[/tex]
using eq (1):
F' = 128(1 N)
F' = 128 N
At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magnetic B=B k and electric E=−E k fields with the initial velocity v=v0 i. The magnitudes of the fields: B=0.18 T, E=278 V/m, and the initial speed v0=2.1 m/s are given. Find at what time t, the particle's speed would become equal to v(t)=3.78·v0:
Answer:
10.78 s
Explanation:
The force on the charge is computed by using the equation:
[tex]F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j[/tex]
F = ma
∴
[tex]a ^{\to}= \dfrac{F^{\to}}{m}[/tex]
[tex]a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)[/tex]
[tex]a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j[/tex]
At time t(sec; the partiCle velocity becomes [tex]v(t) = 3.78 v_o[/tex]
The velocity of the charge after the time t(sec) is expressed by using the formula:
[tex]v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}[/tex]
CHEGG Over the course of a multi-stage 4820-km bicycle race, the front wheel of an athlete's bicycle makes 2.40x106 revolutions. How many revolutions would the wheel have made during the race if its radius had been 1.4 cm larger?
Answer:
θ' = 14.44 × [tex]10^{6}[/tex]
Explanation:
given data
total distance is d = 4820
radius = 1.4 cm
solution
we get here total angle by which the wheel rotates traveling is express as
⇒ [tex]\theta=2.40\times10^6\ \rm{rev}=2.40\times 2\pi\times10^6\ \rm{rad}[/tex] ................1
and
total angle (θ) and the total distance (d) express as
⇒ d = r × θ ...............2
here r is radius
and here rotated through some other angle θ' so put value in given equation and find revolutions
⇒ d = (r+r)θ' ........3
here r = d/θ
so
⇒ [tex]d = ( \frac{d}{\theta}+r) \theta'[/tex]
so put value and get θ'
⇒ θ' = 2.40 × 2π × [tex]10^{6}[/tex] × [tex]\frac{4820 \times 10^3}{4820 \times 10^3 +0.014 \times 2.40 \times 2 \times \pi \times 10^6}[/tex]
⇒ θ' = 14.44 × [tex]10^{6}[/tex] revSusan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to view distant objects with her eye relaxed. With her glasses on, what is the closest object on which she can focus?
Answer:
[tex]u=34cm[/tex]
Explanation:
From the question we are told that:
Far point is [tex]V=34 cm[/tex]
Near point is [tex]u=17 cm[/tex]
Therefore
Focal Length
[tex]f=-34cm[/tex]
Generally the equation for the Lens is mathematically given by
[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]
[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]
[tex]u=34cm[/tex]
g A small object of mass 2.5 g and charge 18 uC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What is the magnitude and direction of the electric field
Answer: [tex]1361.11\ N/C,\text{upward}[/tex]
Explanation:
Given
Mass of particle is [tex]m=2.5\ gm[/tex]
Charge of particle is [tex]q=18\ \mu C[/tex]
Electrostatic force must balance the weight of the particle
[tex]\lim_{n \to \infty} a_n \Rightarrow mg=qE\\\\\Rightarrow E=\dfrac{2.5\times 9.8\times 10^{-3}}{18\times 10^{-6}}\\\\\Rightarrow E=1361.1\ N/C[/tex]
Direction of the electric field is in upward direction such that it opposes the gravity force.
Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m. What is the work done on the ball by the 34.4 N tension force in the string during one half-revolution of the ball
Answer:
the work done on the ball is 0
Explanation:
Given the data in the question;
Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m.
circle circumference = 2πr = 2.1 m
radius r will be; r = 2.1 m / 2π = 0.33 m
Tension force = 34.4 N
one half revolution means, displacement of the ball is;
d = 2r = 2 × 0.33 = 0.66 m
Now, Work done = force × displacement × cosθ
we know that, the angle between the tension force on string and displacement of object is always 90.
so we substitute
Work done = 34.4 N × 0.66 m × cos(90)
Work done = 34.4 N × 0.66 m × 0
Work done = 0 J
Therefore, the work done on the ball is 0
The kinetic theory of gases states that the kinetic energy of a gas is directly proportional to the temperature of the gas.
a. True
b. False
Answer:
true
Explanation:
The kinetic energy of a gas is directly proportional to the temperature of the gas.because temperature is the average kinetic energy of a substance
I hope this helps
Astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c relative to an observer on Earth. When Jill left Earth, the spaceship was equipped with all kinds of scientific instruments, including a meter stick. Now that Jill is underway, how long does she measure the meter stick to be?
A) 0.280 m
B) 1.00 m
C) 0.960 m
D) 1.28 m
E) 1.04 m
(B) 1.00 m
Explanation:
Since the meter stick is traveling with Jill, it will have the same speed as she does so relative to Jill, the meter stick is stationary so its length remains 1.00 m as measured by her.
When astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c and measure the meter stick to be 1 meter. Hence, option B is correct.
What is length contraction?Length contraction is defined as the phenomenon of the moving object being shorter than its appropriate length, measured in the object's rest frame.
When the object travels with the speed of light, the length of the object gets more contracted than its original length, relative to the observer. It is also known as the Lorentz-Fitgerald contraction.
Length contraction, L = L₀√(1-v²/c²), where L is the original length, L₀ is the contracted length. c² is known as the velocity of light. v² is the velocity of the speed of the object.
From the given,
speed of the spaceship = 0.280c (c is the speed of the light)
Length contraction, L = L₀ √(1-v²/c²)
The stick also travels in the spaceship. Hence, the length of the meter stick does not change. It remains at its original length of one meter. Thus, the ideal solution is option B.
To learn more about length contraction:
https://brainly.com/question/10272679
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Describe how the words Science and optics would appear when viewed in a plane mirror?
Answer:
Lateral inversion will occur in a plane mirror.
Explanation:
When words are displayed in a plane or flat mirror, the result is that if the words are displayed left, they change to right and if they were normally displayed right, they change to left. This phenomenon is known as lateral inversion. So, this will apply to the words, Science and optics. Only the sides will be interchanged.
A plane mirror reflects light, therefore, the image that is produced by it remains the same size. The image produced will not appear upside down. Only the sides will be interchanged.
Two harmonic sound waves reach an overseveer simulatenouslt. the obsever hears the sound intensity rise and fall with a time of 0.2 between the maximmu intensity and the successing minimum intensity. What is the difference in frequency of the two sound waves?
Answer:
[tex]dF=2.5Hz[/tex]
Explanation:
From the question we are told that:
Time [tex]T=0.2sec[/tex]
Generally the Period is given as
[tex]T= 2 * 0.2 = 0.4[/tex]
Therefore difference in frequency dF
[tex]dF=\frac{1}{T}[/tex]
[tex]dF=\frac{1}{0.4}[/tex]
[tex]dF=2.5Hz[/tex]
a cell phone is released from the top with the speed of 10ms what is the speed 3s after?
Answer:
30ms
Explanation:
you need to multiple the 10ms by 3s which gives you 30ms
What is the energy of a photon with a frequency of 3.6 × 1015 Hz? Planck’s constant is 6.63 × 10–34 J•s.
1.8 × 10–49 J
2.4 × 10–19 J
1.8 × 10–18 J
2.4 × 10–18 J
We know
[tex]\boxed{\sf E=hv}[/tex]
[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]
[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]
[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]
[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]
Answer:
D!!!!!
Explanation:
A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put
Answer:
875 Watts
Explanation:
P = W/t = mgh/t = 700(10)/8 = 875 Watts
Current is a measure of…
3) A lead bullet initially at 30 C just melts upon striking a target. Assuming that all of the initial kinetic energy of the bullet goes into the internal energy of the bullet to raise its temperature and melt it, calculate the speed of the bullet upon impact. (Specific heat of lead is 0.128 kJ/kg K and lead latent heat of fusion is 24.7 kJ/Kg and melting point of lead is 600 K).
Answer:
The speed of bullet is 354.2 m/s
Explanation:
initial temperature, T = 30 degree C
specific heat, c = 128 J/kg K
Latent heat, L = 24.7 x 1000 J/kg
melting point = 600 K = 327 degree C
Let the mass is m and the speed is v.
Kinetic energy = heat used to increase the temperature + Heat used to melt
[tex]\frac{1}{2} mv^2 = m c (T' - T) + m L\\\\0.5 v^2 = 128 \times (327 - 30) + 24.7\times 1000\\\\0.5 v^2 = 38016 + 24700 \\\\0.5 v^2 = 62716\\\\v = 354.2 m/s[/tex]
Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. Calculate the frequency of this light. Be sure to include units in your answer.
Answer:
5.71×10¹⁴ Hz
Explanation:
Applying,
v = λf................. Equation 1
Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency
make f the subject of the equation
f = v/λ............. Equation 2
From the question,
Given: λ = 525 nm = 5.25×10⁻⁷ m,
Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s
Substitute these values into equation 2
f = (3.0×10⁸)/(5.25×10⁻⁷)
f = 5.71×10¹⁴ Hz
Hence the frequency of light is 5.71×10¹⁴ Hz
Difference between uniform motion and non uniform motion
Answer:
When an object covers equal distance in an equal interval of time, it is uniform motion but when an object covers unequal distance in an equal interval of time, it is called non uniform motion.
