Answer:
Explanation:
Since energy is conserved:
2
mu
2
=
2
mv
2
+mgh
⇒u
2
=v
2
+2gh
⇒(3)
2
=v
2
+2(9.8)(0.5−0.5cos60)
⇒v=2m/s
Acceleration of the simple pendulum is 2.62 m/s².
What is meant by a simple pendulum ?When a point mass is suspended from a fixed support by a light, non-extensible string, the instrument is said to be a simple pendulum.
Here,
Let the mass of the bob be m. The simple pendulum is attached to the fixed support with a string having length l. The pendulum makes an angle of 15° with the vertical from the equilibrium point.
Let T be the tension acting on the string.
As, the bob passes through the angle,
The weight of the bob becomes equal to the vertical component of the tension.
mg = T cos15°
Also, the horizontal component of the tension,
T sin15° = ma
By solving these two equations, we get that,
Acceleration of the simple pendulum,
a = g tan15°
a = 9.8 x 0.267
a = 2.62 m/s²
Hence,
Acceleration of the simple pendulum is 2.62 m/s².
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A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance . At a time , the solenoid is connected to a battery that supplies a potential . At a time , what current flows through the outer loop
This question is incomplete, the complete question is;
A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?
Answer:
the current flows through the outer loop is 1.3 × 10⁻⁵ A
Explanation:
Given the data in the question;
Length [tex]l[/tex] = 11.34 cm = 0.1134 m
radius a = 1.85 cm = 0.0185 m
turns N = 1627
Net resistance [tex]R_{sol[/tex] = 144.9 Ω
radius b = 3.77 cm = 0.0377 m
[tex]R_o[/tex] = 1651.6 Ω
ε = 34.95 V
t = 2.58 μs = 2.58 × 10⁻⁶ s
Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]
so
L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134
L = 0.003576665 / 0.1134
L = 0.03154
Now,
ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) = [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]
so
ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )
ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]
ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
so we substitute in our values;
I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]
I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]
I = 1.3 × 10⁻⁵ A
Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A
Answer:
1.28 *10^-5 A
Explanation:
Same work as above answer. Needs to be more precise
Mass A, 2.0 kg, is moving with an initial velocity of 15 m/s in the x-direction, and it collides with mass M, 4.0 kg, initially moving at 7.0 m/s in the x-direction. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision, in joules
Answer:
the change in the kinetic energy of the system is -42.47 J
Explanation:
Given;
mass A, Ma = 2 kg
initial velocity of mass A, Ua = 15 m/s
Mass M, Mm = 4 kg
initial velocity of mass M, Um = 7 m/s
Let the common velocity of the two masses after collision = V
Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;
[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]
The initial kinetic of the two masses;
[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]
The final kinetic energy of the two masses;
[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]
The change in kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]
Therefore, the change in the kinetic energy of the system is -42.47 J
state the laws of reflection
Answer:
Explanation:
The law of reflection says that the reflected angle (measured from a vertical line to the surface called the normal) is equal to the reflected angle measured from the same normal line.
All other properties of reflection flow from this one statement.
Air is compressed polytropically from 150 kPa, 5 meter cube to 800 kPa. The polytropic exponent for the process is 1.28. Determine the work per unit mass of air required for the process in kilojoules
a) 1184
b) -1184
c) 678
d) -678
Answer:
wegkwe fhkrbhefdb
Explanation:B
HELP ME PLZ FAST
There is more than 1 answer,
The picture is down
Answer:
test her prototype and collect data about its flight
Describe sound and record
Answer:
record is information created, received and maintained as evidence and information by an organization or person.in simpler terms it's a collection of of fields probably of different data types.
sound is however something loud or soft.which can be defined as vibrations that travel through the air or another medium.
I hope this helps
What is utilization of energy
Explanation:
Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector
How do you know that a liquid exerts pressure?
Answer:
The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.
trong cùng một nhiệt độ, lượng năng lượng trên mỗi mol của chất khí nào lớn nhất
a) Khí đơn nguyên tử
b) Khí có từ ba nguyên tử
c) Khí lưỡng nguyên tử
A diffraction grating has 6000 lines per centimeter ruled on it. What is the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe when the grating is illuminated with a beam of light of wavelength 500 nm
Explanation:
Hope it Will help he hsuejwoamxgehanwpalasmbwfwfqoqlmdbehendalmZbgevzuxwllw. yeh we pabdvddxhspapalw. X
The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
What is diffraction?Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.
Given:
The number of lines = 6000 per cm,
The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m
Calculate the diffraction grating,
[tex]d = 1 / no\ of\ lines[/tex]
d = 10⁻² / 6000 m,
Calculate the second-order maxima angle and third-order maxima angle by the formula given below,
[tex]dsin\theta_1 = n_1 \lambda[/tex]
[tex]sin\theta_1 = n_1\lambda / d[/tex]
[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]
θ₁ = sin⁻¹(0.6)
θ₁ = 36.87°
Similarly, for θ₂,
θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)
θ₂ = sin⁻¹(0.9)
θ₂ = 64.16°
Calculate the separation as follows,
θ₂ - θ₁ = 64.16° - 36.87°
θ₂ - θ₁ = 27.29°
Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
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which characteristic of nuclear fission makes it hazardous?
Answer:The radioactive waste
Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88
is the correct answer
a bullet is dropped from the same height when another bullet is fired horizontally they will hit the ground
Answer:
simultaneously
Time taken to reach the ground depends on the vertical component of velocity, not horizontal component of velocity.
As the speed of a particle approaches the speed of light, the momentum of the particle Group of answer choices approaches zero. decreases. approaches infinity. remains the same. increases.
Answer:
approaches infinity
Explanation:
There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.
The relativistic momentum can be written as:
[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]
where
u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.
m = mass of the object
c = speed of light.
So, as u tends to c, we will have:
[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]
Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.
Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of 1/50 times the weight of the object is pushing the object up (weight=mg). If we assume that water resistance exerts a force on the abject that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 40 m/s? Assume that the acceleration due to gravity is 9.81 m/sec^2.
Answer:
a) Fnet = mg - Fb - Fr
b) 8.67 secs
Explanation:
mass of object = 80 kg
Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696
Proportionality constant = 10 N-sec/m
a) Calculate equation of motion of the object
Force of resistance on object due to water = Fr ∝ V
= Fr = Kv = 10 V
Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object while mg ( weight ) acts in the negative y - direction on the object.
Fnet = mg - Fb - Fr
∴ Equation of motion of the object ( Ma = mg - Fb - Fr )
b) Calculate how long before velocity of the object hits 40 m/s
Ma = mg - Fb - Fr
a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V
V = u + at ---- ( 1 )
u = 0
V = 40 m/s
a = 9.6138 - 0.125 V
back to equation 1
40 = 0 + ( 9.6138 - 0.125 (40) ) t
40 = 4.6138 t
∴ t = 40 / 4.6138 = 8.67 secs
A roller coaster uses 800 000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy and pauses (velocity = 0) for a fraction of a second at the very top before heading down the other side.
a) Draw a sankey diagram for a roller coaster's climb.
A roller coaster uses 800 000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy and pauses for a fraction of a second at the very top before heading down the other side. At the top of the hill total, the kinetic energy of the roller coaster would be zero as the velocity is zero at the top of the hill, therefore the total mechanical energy is only because of potential energy.
What is mechanical energy?Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total stored energy in the system which is represented by total potential energy.
The expression for total mechanical energy is as follows
ME= KE+PE
As total mechanical energy is the sum of all the kinetic as well as potential energy stored in the system.As given in the problem a roller coaster uses 800000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy which means 300000 J of energy is lost in the frictional energy while climbing the hill,
Thus at the top of the hill, the total energy of the roller coasters is only due to the potential energy.
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Vector a has a magnitude of 8 and makes an angle of 45 with positive x axis vector B has also the same magnitude of 8 units and direction along the
Answer:
prove that Sin^6 ϴ-cos^6ϴ=(2Sin^2ϴ-1)(cos^2ϴ+sin^4ϴ)
please sove step by step with language it is opt maths question
I need help with this please!!!!
Answer:
1.84 hours
I hope it's helps you
A car accelerates from 0 m/s to 25 m/s in 5 seconds. What is the average acceleration of the car.
Answer:
5 m/s I hope it will help you
Explanation:
mark me as a brainlist answer
Convert 385k to temperature of
Answer:
233.33°F
Explanation:
(385K - 273.15) * 9/5 + 32 = 233.33°F
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How far horizontally will it travel in 2 seconds?
A. 30 m
B. 90 m
C. 45 m
D. 60 m
Answer:
It will travel Vx * t = 30 m/s * 2 s = 60 m
herical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.010-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass tran
Answer: Below is the complete question
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)
answer:
mass transfer coefficient = 9.56 * 10^-5 m/s
Explanation:
Candy density = 1950 kg/m^3
Candy diameter = 1 cm
Velocity of water = 1 m/s
water density = 1000 kg/m^3
Viscosity of water = 1 * 10^-3 kg/m/s
diffusion coefficient of candy in water = 2 * 10^-9 m^2/s
solubility of candy = 2 kg/m^3
Determine the mass transfer coefficient ( m/s )
( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3
where : Re = vdp / μ , Sh = KLd / Deff
attached below is the remaining solution .
mass transfer coefficient = 9.56 * 10^-5 m/s
how will be electric lines of force where intensity of electric field is maximum ?
a. wider
b. +ve to -ve
c. narrow
d. -ve to +ve
i'm pretty sure the answer is A wider
Electric lines of force where intensity of electric field is maximum when its wider.
What is Electric field?The physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field. It can also refer to a system of charged particles' physical field.
Electric charges and time-varying electric currents are the building blocks of electric fields. The electromagnetic field, one of the four fundamental interactions (also known as forces) of nature, manifests itself in both electric and magnetic fields.
Electrical technology makes use of electric fields, which are significant in many branches of physics. For instance, in atomic physics and chemistry, the electric field acts as an attracting force to hold atoms' atomic nuclei and electrons together.
Therefore, Electric lines of force where intensity of electric field is maximum when its wider.
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A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?
Answer:
The smallest and largest areas could be 6400 m and 7500 m, respectively.
Explanation:
The area of a rectangle is given by:
[tex] A = l*w [/tex]
Where:
l: is the length = 100 m
w: is the width
We can calculate the smallest area with the lower value of the width.
[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]
And the largest area is:
[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]
Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.
I hope it helps you!
Answer:
the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m
the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m
Explanation:
to the find the largest and the smallest area of the field measurement error is to be considered.
we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.
therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:
[tex]A_l[/tex]= (L+0.5)(W+0.5)
(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m
To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:
[tex]A_s[/tex]= (L-0.5)(W-0.5)
(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m
Express 6revolutions to radians
Answer:
About 37.70 radians.
Explanation:
1 revolution = 2[tex]\pi[/tex] radians
∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)
6 revolutions = 37.6991 or ≈ 37.70 radians
A circular parallel-plate capacitor whose plates have a radius of 25 cm is being charged with a current of 1.3 A. What is the magnetic field 11 cm from the center of the plates
The magnetic field at 11 cm from the center of the plates is 2.364 x 10⁻⁷ T.
Given;
radius of the circular plate, d = 25 cm = 0.25 m
current in the plate, I = 1.3 A
distance from the center of the circular plate, r = 11 cm = 0.11 m
To find:
magnetic field (B)The magnetic field from the given distance is calculated as from Biot Savart equation:
[tex]B = \frac{\mu_o I}{2\pi r} \\\\where;\\\\\mu_o \ is \ permeability \ of \ free \ space \ 4\pi \times 10^{-7} \ T.m/A\\\\B = \frac{(4\pi \times 10^{-7} ) \times (1.3)}{2\pi \times 0.11} \\\\B = 2.364 \ \times 10^{-6} \ T[/tex]
Therefore, the magnetic field 11 cm from the center of the plates is 2.364 x 10⁻⁷ T.
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What is cubical expansivity of liquid while freezing
Answer:
"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web
Explanation:
tbh up above ✅
Answer:
cubic meter
Explanation:
Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion
A refrigerator has a coefficient of performance equal to 4.00. The refrigerator takes in 110 J of energy from a cold reservoir in each cycle. (a) Find the work required in each cycle. J (b) Find the energy expelled to the hot reservoir. J
Answer:
The correct answer is:
(a) 27.5 Joules
(b) 141.5 Joules
Explanation:
Given:
Energy,
[tex]Q_c = 110 \ J[/tex]
Coefficient of performance refrigerator,
[tex]Cop(refrig)=4[/tex]
(a)
As we know,
⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]
or,
⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]
[tex]=\frac{110}{4}[/tex]
[tex]=27.5 \ Joules[/tex]
(b)
⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]
or,
⇒ [tex]Q_h = Q_c+Work[/tex]
[tex]=114+27.5[/tex]
[tex]=141.5 \ Joules[/tex]
A car is traveling at 118 km/h when the driver sees an accident 85 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup
Answer:
The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
Explanation:
From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h
The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,
Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s
First, convert 118 km/h to m/s
118 km/h = (118 × 1000) /3600 = 32.7778 m/s
∴ u = 32.7778 m/s
Now, to determine the deceleration, a, required to stop,
From one of the equations of motion for linear motion,
v² = u² + 2as
Then
0² = (32.7778)² + 2×a×85
0 = 1074.3841 + 170a
∴ 170a = - 1074.3841
a = - 1074.3841 / 170
a = - 6.3199
a ≅ - 6.32 m/s²
Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
