Answer:
I = 15,645. kg*m/s or 15,645 N*s
Explanation:
I = m(^v)
I = 1050kg((26.2m/s-11.3m/s)
I = 15,645. kg*m/s
a volcano that may erupt again at some time in the distant future is
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
For example, we can take Water
In (A) Water has same mass and great volume
In (B) Water has same mass and lower volume
Will there be any change in its density then?
Answer:
yes there will be change in its density
Electron A is fired horizontally with speed 1.00 Mm/s into a region where a vertical magnetic field exists. Electron B is fired along the same path with speed 2.00 Mm/s. (i) Which electron has a larger magnetic force exerted on it
B will have the greater force
Fc=MV2 /R=Fm
The A particle has less centipetal force and larger radius so larger curve
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 11.0 ft/s at point A and 18.0 ft/s at point C. The cart takes 5.00 s to go from point A to point C, and the cart takes 1.30 s to go from point B to point C. What is the cart's speed at point B
Answer:
The speed at B is 16.18 ft/s .
Explanation:
Speed at A, u = 11 ft/s
Speed at C, v' = 18 ft/s
Time from A to C = 5 s
Time from B to C = 1.3 s
Let the speed of car at B is v.
Let the acceleration is a.
From A to B
Use first equation of motion
v = u + a t
18 = 11 + a x 5
a = 1.4 ft/s^2
Let the time from A to B is t' .
t' = 5 - 1.3 = 3.7 s
Use first equation of motion from A to B
v = 11 + 1.4 x 3.7 = 16.18 ft/s
A painter sets up a uniform plank so that he can paint a high wall. The plank is 2 m long and weighs 400 N. The two supports holding up the plank are placed 0.2 m from either end. Show that the upwards force on each of the planks is 200 N. Draw a sketch.
The upward force on each supporting plank is 200 N
The given parameters include;
weight of the plank, W₁ = 400 Nlength of the plank, l = 2 mupward force of each supporting plank, = W₂ and W₃To show that the upward force of each supporting plank is 200 N, make the following sketch.
W₂ W₃
↑ ↑
-----------------------------------------------------------------------
0.2m ↓ 0.2m
400 N
The two supporting planks keeps the 2m plank in equilibrium position. If the plank is in equilibrium position the sum of the upward forces equals sum of the downward force.W₂ + W₃ = 400 N
But the distance of each supporting plank from the end is equal, (0.2m).
Then, W₂ = W₃
2W₂ = 400 N
W₂ = 400N/2
W₂ = 200 N
W₃ = 200 N
Therefore, the upward force on each supporting plank that keeps the plank in equilibrium position is 200 N.
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A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force of 9.14 × 10-17 N. Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.
Answer:
a) The proton's speed is 5.75x10⁵ m/s.
b) The kinetic energy of the proton is 1723 eV.
Explanation:
a) The proton's speed can be calculated with the Lorentz force equation:
[tex] F = qv \times B = qvBsin(\theta) [/tex] (1)
Where:
F: is the force = 9.14x10⁻¹⁷ N
q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C
v: is the proton's speed =?
B: is the magnetic field = 3.28 mT
θ: is the angle between the proton's speed and the magnetic field = 17.6°
By solving equation (1) for v we have:
[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]
Hence, the proton's speed is 5.75x10⁵ m/s.
b) Its kinetic energy (K) is given by:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
m: is the mass of the proton = 1.67x10⁻²⁷ kg
[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]
Therefore, the kinetic energy of the proton is 1723 eV.
I hope it helps you!
An ink-jet printer steers charged ink drops vertically. Each drop of ink has a mass of 10-11 kg, and a charge due to 500,000 extra electrons. It goes through two electrodes that gives a vertical acceleration of 104 m/s2. The deflecting electric field is _____ MV/m.
Answer:
E = 1.25 MV / m
Explanation:
For this exercise let's use Newton's second law
F = m a
where the force is electric
F = q E
we substitute
q E = m a
E = m a / q
indicate there are 500,000 excess electrons
q = 500000 e
q = 500000 1.6 10⁻¹⁹
q = 8 10⁻¹⁴ C
the mass is m = 10⁻¹¹ kg and the acceleration a = 10⁴ m / s²
let's calculate
E = 10⁻¹¹ 10⁴ / 8 10⁻¹⁴
E = 0.125 10⁷ V / m = 1.25 10⁶ V / m
E = 1.25 MV / m
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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Rays of light coming from the sun (a very distant object) are near and parallel to the principal axis of a concave mirror. After reflecting from the mirror, where will the rays cross each other at a single point?
The rays __________
a. will not cross each other after reflecting from a concave mirror.
b. will cross at the center of curvature.
c. will cross at the point where the principal axis intersects the mirror.
d. will cross at the focal point. will cross at a point beyond the center of curvature.
A concave mirror is an example of curved mirrors. So that the appropriate answer to the given question is option D. The rays will cross at the focal point.
A concave mirror is a type of mirror in which its inner part is the reflecting surface, while its outer part is the back of the mirror. This mirror reflects all parallel rays close to the principal axis to a point of convergence. It can also be referred to as the converging mirror.
In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.
Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.
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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
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.
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]
1 A thing ring has a mass of 6kg and a radius of 20cm. calculate the rotational inertia.
Answer:
2400kgm²
Explanation:
Rotational inertia=mass x radius²
One charge is fixed q1 = 5 µC at the origin in a coordinate system, a second charge q2 = -3.2 µC the other is at a distance of x = 90 m from the origin.
What is the potential energy of this pair of charges?
Answer:
5.4uC
Explanation:
A parallel-plate capacitor consists of two plates, each with an area of 29 cm2cm2 separated by 3.0 mmmm. The charge on the capacitor is 7.8 nCnC . A proton is released from rest next to the positive plate. Part A How long does it take for the proton to reach the negative plate
Answer:
t = 2.09 10⁻³ s
Explanation:
We must solve this problem in parts, first we look for the acceleration of the electron and then the time to travel the distance
let's start with Newton's second law
∑ F = m a
the force is electric
F = q E
we substitute
q E = m a
a = [tex]\frac{q}{m} \ E[/tex]
a = [tex]\frac{1.6 \ 10^{-19}}{ 9.1 \ 10^{-31} } \ 7.8 \ 10^{-9}[/tex]
a = 1.37 10³ m / s²
now we can use kinematics
x = v₀ t + ½ a t²
indicate that rest starts v₀ = 0
x = 0 + ½ a t²
t = [tex]\sqrt{\frac{2x}{a} }[/tex]
t = [tex]\sqrt{\frac {2 \ 3 \ 10^{-3}}{ 1.37 \ 10^3} }[/tex]
t = 2.09 10⁻³ s
A large metal sphere has three times the diameter of a smaller sphere and carries three times the charge. Both spheres are isolated, so their surface charge densities are uniform. Compare (a) the potentials (relative to infinity) and (b) the electric field strengths at their surfaces.
Answer:
A. Equals to that of the smaller sphere
B. 3 times less than that of the smaller sphere
Explanation:
(a) Equals to that of the smaller sphere
The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential
b) 3 times less than that of the smaller sphere
However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3
The voltage in an EBW operation is 45 kV. The beam current is 50 milliamp. The electron beam is focused on a circular area that is 0.50 mm in diameter. The heat transfer factor is 0.87. Calculate the average power density in the area in watt/mm2.
Answer:
[tex]P_d=6203.223062W/mm^2[/tex]
Explanation:
From the question we are told that:
Voltage [tex]V=45kV[/tex]
Current [tex]I=50mAmp[/tex]
Diameter [tex]d=0.50mm[/tex]
Heat transfer factor [tex]\mu= 0.87.[/tex]
Generally the equation for Power developed is mathematically given by
[tex]P=VI\\\\P=45*10^3*50*10^{-3}[/tex]
[tex]P=2.250[/tex]
Therefore
Power in area
[tex]P_a=1400*0.87[/tex]
[tex]P_a=1218watt[/tex]
Power Density
[tex]P_d=\frac{P_a}{Area}[/tex]
[tex]P_d=\frac{1218}{\pi(0.5^2/4)}[/tex]
[tex]P_d=6203.223062W/mm^2[/tex]
elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extension of 2.0cm
Answer:
40g
Explanation:
20g range > 1.0cm
Therefore,
40g range > 2.0cm
g Light that is incident upon the eye is refracted several times before it reaches the retina. As light passes through the eye, at which boundary does most of the overall refraction occur?
Answer
Explanation
:giác mạc
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
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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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
A mass is tired to spring and begins vibration periodically the distance between it's lowest position is 48cm what is the Amplitude of the vibration
Answer:
The amplitude of vibration of the spring is "24 cm"
The periodic vibrating body's motion follows a sinusoidal path. This sinusoidal path is illustrated in the attached picture.
From the picture, it can be clearly seen that the amplitude of the periodic vibration motion is the distance from its mean position to the highest point.
Since the distance of both the highest and the lowest points from the mean position is the same. Therefore, the distance between the lowest and the highest point must be equal to two times the amplitude of the wave.
Amplitude = 24 cm
A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.
A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted. What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.30 cm?
A.
52.3 kN
B.
62.3 kN
C.
72.3 kN
D.
42.3 N
Answer:
cobina
Explanation:
me 2
Derive the dimension of coefficient of linear expansivity
Answer:
The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.
In the diagram, the crest of the wave is show by:
A
B
C
D
Answer:
D.
Explanation:
The crest of a wave refers to the highest point of a wave. This is illustrated by D.
Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be
Answer:
a) [tex]F_g=1.5*10^9Ibf[/tex]
b) [tex]F_t=12490Ibf/ft^2[/tex]
[tex]F_b=0[/tex]
Explanation:
From the question we are told that:
Height [tex]h=200ft[/tex]
Width [tex]w=1200ft[/tex]
a)
Generally the equation for Dam's Hydro static force is mathematically given by
[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]
Where
[tex]\rho=Density\ of\ water[/tex]
[tex]\rho=62.4Ibm/ft^3[/tex]
Therefore
[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]
[tex]F_g=1.5*10^9Ibf[/tex]
b)
Generally the equation for Dam's Force per unit area is mathematically given by
[tex]F=\rho*g*h[/tex]
For Top
[tex]F_t=\rho*g*h[/tex]
[tex]F_t=62.4*32.2*200[/tex]
[tex]F_t=12490Ibf/ft^2[/tex]
For bottom
[tex]Here \\H=0 zero[/tex]
Therefore
[tex]F_b=0[/tex]
The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].
The force per unit area near the top is 86.74 psi.
The force per unit area near the bottom is zero.
Hydrostatic force
The hydrostatic force on the dam is the force exerted on the dam by the column of the water.
[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]
Force per unit area near the topThe force per unit area is the pressure exerted near the top of the dam.
[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]
where;
P is pressure in PSI
ρ is density of water in lb/gal
h is the vertical height in ft
[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]
The pressure near the bottom is zero, become the vertical height is zero.
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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
