Answer:
The angular speed of the blade is of 166.67 rad/min
If a radial saw has a blade with a 9-in. radius. Suppose that the blade spins at 1000 rpm then the angular speed of the blade would be 104.72 rad/seconds.
What is speed?
The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.
The mathematical expression for speed is given by
speed = total distance /Total time
As given in the problem If a radial saw has a blade with a 9-in. radius. Suppose that the blade spins at 1000 rpm then we have to find the angular speed of the blade,
the angular speed of the blade (ω) = 2πN/60
=2×π×1000/60
= 104.72 rad/seconds
Thus, the angular speed of the radial saw would be 104.72 rad/seconds
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How would doubling the mass of Earth affect the gravity we experience?
Explanation:
if the mass or both of the objects is doubled, then the force of gravity between them is quadrupled and so on. Since gravayional force is inversely proportional to the square of separation distance between two interacting objects, more separation distance will result in weaker gravitational forces.
name the three major types of clouds
Answer:
Cumulus, Stratus, and Cirrus. There are three main cloud types.
Explanation:
hopes it help^_^
An air-filled pipe is found to have successive harmonics at 980 Hz , 1260 Hz , and 1540 Hz . It is unknown whether harmonics below 980 Hz and above 1540 Hz exist in the pipe. What is the length of the pipe
Answer:
L = 0.7 m
Explanation:
This is a resonance exercise, in this case the air-filled pipe is open at both ends, therefore we have bellies at these points.
λ / 2 = L 1st harmonic
λ = L 2nd harmonic
λ = 2L / 3 3rd harmonic
λ = 2L / n n -th harmonic
the speed of sound is related to wavelength and frequencies
v =λ f
f = v /λ
we substitute
f = v n / 2L
the speed of sound in air is v = 343 m / s
suppose that the frequency of f = 980Hz occurs in harmonic n
f₁ = v n / 2L
f₂ = v (n + 1) / 2L
f₃ = v (n + 2) / 2L
we substitute the values
2 980/343 = n / L
2 1260/343 = (n + 1) / L
2 1540/343 = (n + 2) / L
we have three equations, let's use the first two
5.714 = n / L
7.347 = (n + 1) / L
we solve for L and match the expressions
n / 5,714 = (n + 1) / 7,347
7,347 n = 5,714 (n + 1)
n (7,347 -5,714) = 5,714
n = 5,714 / 1,633
n = 3.5
as the number n must be integers n = 4 we substitute in the first equation
L = n / 5,714
L = 0.7 m
An electron of kinetic energy 1.59 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 35.4 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.
Answer:
a) [tex] v = 2.36 \cdot 10^{7} m/s [/tex]
b) [tex] B = 3.80 \cdot 10^{-4} T [/tex]
c) [tex] f = 1.06 \cdot 10^{7} Hz [/tex]
d) [tex] T = 9.43 \cdot 10^{-8} s [/tex]
Explanation:
a) We can find the electron's speed by knowing the kinetic energy:
[tex] K = \frac{1}{2}mv^{2} [/tex]
Where:
K: is the kinetic energy = 1.59 keV
m: is the electron's mass = 9.11x10⁻³¹ kg
v: is the speed =?
[tex] v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2*1.59 \cdot 10^{3} eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}}{9.11 \cdot 10^{-31} kg}} = 2.36 \cdot 10^{7} m/s [/tex]
b) The electron's speed can be found by using Lorentz's equation:
[tex] F = q(v\times B) = qvBsin(\theta) [/tex] (1)
Where:
F: is the magnetic force
q: is the electron's charge = 1.6x10⁻¹⁹ C
θ: is the angle between the speed of the electron and the magnetic field = 90°
The magnetic force is also equal to:
[tex] F = ma_{c} = m\frac{v^{2}}{r} [/tex] (2)
By equating equation (2) with (1) and by solving for B, we have:
[tex] B = \frac{mv}{rq} = \frac{9.11 \cdot 10^{-31} kg*2.36 \cdot 10^{7} m/s}{0.354 m*1.6 \cdot 10^{-19} C} = 3.80 \cdot 10^{-4} T [/tex]
c) The circling frequency is:
[tex] f = \frac{1}{T} = \frac{\omega}{2\pi} = \frac{v}{2\pi r} [/tex]
Where:
T: is the period = 2π/ω
ω: is the angular speed = v/r
[tex] f = \frac{v}{2\pi r} = \frac{2.36 \cdot 10^{7} m/s}{2\pi*0.354 m} = 1.06 \cdot 10^{7} Hz [/tex]
d) The period of the motion is:
[tex] T = \frac{1}{f} = \frac{1}{1.06 \cdot 10^{7} Hz} = 9.43 \cdot 10^{-8} s [/tex]
I hope it helps you!
In a Young's double-slit experiment, two parallel slits with a slit separation of 0.165 mm are illuminated by light of wavelength 560 nm, and the interference pattern is observed on a screen located 4.05 m from the slits. (a) What is the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen
Answer:
the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m
Explanation:
Given the data in the question;
slit separation d = 0.165 mm = 0.165 × 10⁻³ m
wavelength λ = 560 nm = 560 × 10⁻⁹ m
distance between the screen and slits D = 4.05 m
now,
for fifth-order bright fringe path difference = mλ
where m is 5
so, the difference in path lengths from each of the slits will be;
Δr = mλ
we substitute
Δr = 5( 560 × 10⁻⁹ m )
Δr = 28 × 10⁻⁷ m
Therefore, the difference in path lengths from each of the slits to the location of the center of a fifth-order bright fringe on the screen is 28 × 10⁻⁷ m
Question 3
By what volume would 25 L of alcohol increase if its temperature was
increased from 20°C to 30°C? (3 marks)
Answer:
V2 = 37.5 L
Explanation:
Given the following data;
Initial volume, V1 = 25 L
Initial temperature, T1 = 20°C
Final temperature,T2 = 30°C
To find the final volume V2, we would use Charles' law;
Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Charles is given by;
[tex] \frac {V}{T} = K[/tex]
[tex] \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}} [/tex]
Making V2 as the subject formula, we have;
[tex] V_{2}= \frac{V_{1}}{T_{1}} * T_{2}[/tex]
[tex] V_{2}= \frac{25}{20} * 30 [/tex]
[tex] V_{2}= 1.25 * 30 [/tex]
V2 = 37.5 L
describe the concept of force represent it quantiatively and derive unit of force
important please answer it as the question says and only if you know the answer
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Force is defined as the rate of change of momentum. For an unchanging mass, this is equivalent to mass x acceleration. So, 1 N = 1 kg m s-2, or 1 kg m/s2.
Does an infrared wave or an x-ray travel faster in the vacuum of space?
Answer:
All electromagnetic radiation, of which radio waves and X-rays are examples, travels at the speed c in a vacuum. The only difference between the two is that the frequency of X-rays is very much higher than radio waves
What is a centripetal acceleration of a greyhound running on a circular track with a radius of 50 m at 12.5 m/s
Answer:
Centripetal acceleration = 3.125 m/s²
Explanation:
Given the following data;
Radius, r = 50 m
Velocity, V = 12.5 m/s
To find the centripetal acceleration;
Centripetal acceleration = Velocity²/radius
Centripetal acceleration = 12.5²/50
Centripetal acceleration = 156.25/50
Centripetal acceleration = 3.125 m/s²
Therefore, the centripetal acceleration of the greyhound running on a circular track is 3.125 meters per seconds square.
Two children are balanced on a seesaw that has a mass of 18.0 kg. The first child has a mass of 26.0 kg and sits 1.60 m from the pivot. The center of mass of the seesaw is 0.133 m from the pivot (on the side of the first child). (a) If the second child has a mass of 34.4 kg, how far (in m) is she from the pivot
Answer:
1.28 m
Explanation:
As shown in the diagram attached,
According to the principle of moment,
For a body at equilibrium,
Sum of clockwise moment = sum of anticlockwise moment.
Taking moment about the pivot,
W₁(1.6)+W(0.133) = W₂(x)............... Equation 1
Where W₁ = Weight of the first child, Wₓ = Weight of the seesaw, W₂ = weight of the second child, x = distance of the second child from the pivot.
But,
W = mg
Where g = 9.8 m/s², m = mass of the body
Therefore,
W₁ = 26×9.8 = 254.8 N,
Wₓ = 18×9.8 = 176.4 N
W₂ = 34.4×9.8 = 337.12 N
Substitute these values into equation 1
(254.8×1.6)+(176.4×0.133) = 337.12(x)
407.68+23.4612 = 337.12x
337.12x = 431.1412
x = 431.1412/337.12
x = 1.2789
x ≈ 1.28 m
Which wave has the smallest wave
period? What is its period?
Answer:
C
Explanation:
The said wave takes the shortest time to move/get transmitted
An object that weighs 20 N encounters 20 N of air resistance. What is the acceleration in the y direction?
Answer:
Explanation:
F=ma.
a=m÷fcos 0
A compressed spring launches a block up an incline. Which objects should be included within the system in order to make an energy analysis as easy as possible
Answer:
Block, incline, spring and gravity.
Explanation:
For us to have an energy analysis involving gravitational and spring potential energy, we will need to have a block with specific mass that is held at rest on a frictionless incline plane. Now, the block will compress the spring by a specific length from its equilibrium position and then the block is released to travel a distance right up the slope.
So basically, we will need Block, incline, spring and then gravity for it to move.
1. How fast is a radio wave traveling if it is 4.0 meters long and its frequency is
82 Hz?
Answer:
Speed = 328 m/s
Explanation:
Given the following data:
Wavelength = 4 m
Frequency = 82 Hz
To find the speed of the radio wave;
Speed = wavelength * frequency
Speed = 4 * 82
Speed = 328 m/s
Therefore, the radio wave is travelling at 328 meters per seconds.
Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m. Therefore, radio waves are a series of repetitive valleys and peaks that are typically characterized of having the longest wavelength in the electromagnetic spectrum.
An Olympic skier moving at 20.0 m/s down a 30.0o slope encounters a region of wet snow, of
coefficient of friction μk = 0.740. How far down the slope does she go before stopping?
a.119 m
b.145 m
c.170 m
d.199 m
Answer:
Option B
Explanation:
Forces acting on the skier-
F1 [tex]= -mg sin(30)[/tex] down the slope
F2 [tex]= -mg cos(30)[/tex]
F3 = friction force [tex]= 0.74 mg cos(30)[/tex]
Net force, down the slope
[tex]= -mg sin(30)+ 0.74 mg cos(30)\\= mg(.74 cos(30)-sin(30))\\= 0.14mg\\= 1.38m[/tex]
Acceleration [tex]= F/m= 1.38[/tex] m/s2
Acceleration remains constant, initial speed is 20 m/s
Speed at time t is [tex]1.38t- 20[/tex] m/s
Distance down the slope at time t is [tex]0.69t^2- 20t[/tex]
When the skier stops, her speed is 0. Thus,
, [tex]1.38t- 20= 0\\t= 20/1.38\\= 14.5[/tex] seconds
Distance travelled in 14.5 seconds [tex]= (0.69)(14.52- 20(14.5)= -145 m[/tex](negative because it is down the slope).
Option B is correct
A rigid body consists of four bodies joined together, as drawn below to scale. The point is at one corner of the rectangle and the component bodies are: A uniform disk of radius and mass . A uniform rod of length and mass . A uniform rectangle with side lengths and , and mass . A point mass at with mass . What is the moment of inertia about the axis through the point
Answer:
I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) ^ {3/2}
Explanation:
The moment of inertia is a scalar quantity, therefore additive, therefore we can find the moments of inertia of each body with respect to the point and add them.
Let's use the parallel axis theorem for the moment of inertia.
I = [tex]I_{cm}[/tex] + m d²
the moments and inertia of the bodies are
disk Icm = ½ m R²
rod Icm = 1/12 m L²
rectangle Icm = 1/12 m (a² + b²)
where a and b are the sides of the rectangle
Let's fix a reference frame on the point body, the length of the rectangle is x and its height y, the total moment of inertia is
I_total = I_point + I_disco + I_rod + I_rectangular
the moment of inertia of the point is
I = m r² = m 0
I_point = 0
disk moment of inertia
suppose it is on the y-axis with x = 0
I_disco = ½ m R² + m y²
moment inertia rod
located in the opposite corner
The distance from the point to the center of the mass of the rod is
R = [tex]\sqrt{x^2 +y^2 }[/tex]
I_varrilla = 1/12 m L² + m ([tex]\sqrt{ x^2 + y^2 }[/tex])
rectangle moment inertia
located on the x axis
I_rectangle = ½ m (a² + b²) + m x²
we substitute
I_total = 0 + ½ m R² + m y² + 1/12 m L² + m (Ra x ^ 2 + y²) + ½ m (a² + b²) + m x²
I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) + m √(x^2 + y²)
I_total = ½ m R² + 1/12 m L² + ½ m (a² + b²) + m (x² + y²) ^ {3/2}
the titan mission scheduled for 2026 is a reconnaissance to explore the origin of life true or false
Answer:
its true
Explanation:
The titan mission scheduled for 2026 is a reconnaissance to explore the origin of life. Because, some scientific studies based on titan has showed that there is the presence of some biomolecules. Hence, the statement is true.
What is titan mission ?The distinctive, abundantly organic planet of Titan is where we will travel next in the solar system, according to NASA. The Dragonfly expedition will fly numerous sorties to sample and inspect locations around Saturn's icy moon, advancing our hunt for the components of life.
In 2026, Dragonfly will take off, and it will land in 2034. The rotorcraft will travel to a great number of Titan's promising sites in search of prebiotic chemical processes that are similar to those on Titan and Earth.
Dragonfly, which has eight rotors and flies like a big drone, is the first multi-rotor vehicle NASA will operate for science on another world. To become the first spacecraft to ever fly its full science payload, it will take advantage of Titan's dense atmosphere, which is four times denser than Earth's. Hence, the statement is true.
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13. The percent of Earth's surface covered by high clouds
in January 1987 was closest to which of the following?
A. 13.09
B. 13.5%
C. 14.0%
D. 14.5%
16. Which of the following figures best represents the
monthly average cover of high. middle, and low clouds
in January 19922
E.
H.
cloud cover
okud cover
middle cloud
high clouds
high cloud
low clouds
low clouds
midle clouds
G.
J.
14. Based on Table I. a cosmic ray flux of 440.000 parti-
cles/m he would correspond to a cover of low clouds
that is closest to which of the following?
F. 28.75
G. 29.09
H. 29.35
J. 29.6%
ckud cover
cloud cover
high clouds
low clouds
middle cloud
high clouds
middle clouds
low clouds
An 800 kg car finishes a 5000-meter road in 300 seconds. Considering the car was moving at constant speed, what was its momentum?
Answer:
Momentum(P) = 1336kgm/s
Explanation:
Momentum is the product of the mass and velocity, therefore Momentum(P) is = mass*velocity. since the mass is given 800kg, all we are left with is the displacement 500m and the time 300s which can be used to find the velocity. so velocity is the displacement divided by the time, ie 500/300 which will give us the velocity 1.67m\s. hence we multiply the mass and the velocity and we'll get 1336kgm/s.
9
A canoe with a mass of 120 kg is floating downriver at a speed of 2.5 m/s. What is the canoe's kinetic eergy?
Answer:
K.E. = ½ × mv²
= ½ × 120 × (2.5)²
= 60 × 6.25
= 375 J
After the water has boiled, the temperature of the water decreases by 22 °C.
The mass of water in the kettle is 0.50 kg.
The specific heat capacity of water is 4200 J/kg °C.
Calculate the energy transferred to the surroundings from the
water.
Energy=
Explanation:
Use the formula:
[tex] e = mc \delta \theta[/tex]
e is the energy released,
m is the mass of water,
c is the specific heat capacity,
δθ is the change in temperature ( 100 - 22)
A car of mass 800kg is travelling at 10m/s. How much work must be done to stop it?
Answer:
80
Explanation:
800/10
=80
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state and explain newton second law of motion also describe the concept of force, represent it quantiatively and derive the unit of force
this is a long question only answer if you know how to solve it you will be REWARDED with points
Explanation:
Newton's second law of motion states F=ma which means force is equal to mass multiplied by acceleration which in simple terms means If you give mass force it will accelerate the concept of force in physics is any interaction that when unopposed will change the motion of an object.
a boy cycles continuously through a distance of 1.0 km in 5 minutes calculate its average speed
Answer:
0.2 km/minute or 12 km/h
Explanation:
All you have to do is divide distance by time to find the speed. 1/5 km/min is the speed into which we simplify to 0.2 km/min. I then multiplied by 60(numerator and denominator) since there are 60 minutes in an hour to get 12 km/h.
What kind of force allows a navigational compass to work?
Answer: Magnetic Fields
Explanation: A Compass works by detecting the earth's magnetic fields. The earth has a iron core that is part liquid and part solid crystal due to gravitational pressure.
Water is to be heated from 10°C to 80°C as it flows through a 2-cm-internal-diameter, 13-m-long tube. The tube is equipped with an electric resistance heater, which provides uniform heating throughout the surface of the tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube. If the system is to provide hot water at a rate of 5 L/min, determine the power rating of the resistance heater. Also, estimate the inner surface temperature of the pipe at the exit.
Answer:
- the power rating of the resistance heater is 24139.5 W
- the inner surface temperature of the pipe at the exit is 96.34°C
Explanation:
Given the data in the question;
Flow rate of water in the tube V" = 5L/min = 8.333 × 10⁻⁵ m³/s
The water is to be heated from 10°C to 80°C;
so Average or mean temperature [tex]T_{avg[/tex] will be;
[tex]T_{avg[/tex] = (T₁ + T₂) / 2 = (10 + 80) / 2 = 90/2 = 45°C
Now, from the Table " Properties of Water " at average temperature;
at [tex]T_{avg[/tex] = 45°C
density p = 990.1 kg/m³
specific heat [tex]C_p[/tex] = 4180 J/kg-k
thermal conductivity k = 0.637 W/m-°C
Now, we determine the mass flow;
m" = pV"
we substitute
m" = 990.1 × 8.333 × 10⁻⁵
m" = 0.08250 kg/s
we know that the power rating of the resistance heater is equal to the heat transfer rate to the water;
Q' = m"[tex]C_p[/tex]( T₂ - T₁ )
we substitute
Q' = (0.08250 × 4180 ) ( 80 - 10 )
Q' = 344.85 × 70
Q' = 24139.5 W
Hence, the power rating of the resistance heater is 24139.5 W
Next, we determine the average velocity of water in the tube;
[tex]V_{avg[/tex] = V" / [tex]A_c[/tex]
[tex]V_{avg[/tex] = V" / ( [tex]\frac{1}{4}[/tex]πD² )
given that; flows through a 2-cm-internal-diameter; D = 0.02 m
we substitute
[tex]V_{avg[/tex] = (8.333 × 10⁻⁵) / ( [tex]\frac{1}{4}[/tex]π × (0.02)² )
[tex]V_{avg[/tex] = (8.333 × 10⁻⁵) / ( 3.14159 × 10⁻⁴ )
[tex]V_{avg[/tex] = 0.265 m/s
Also, from table " saturated water property table "
At 45°C
viscosity μ = 0.596 × 10⁻³ kg/m-s
Prandtl number Pr = 3.91
Now, we determine the kinematic viscosity
v = μ / p
we substitute
v = ( 0.596 × 10⁻³ ) / 990.1
v = 6.01959 × 10⁻⁷ m²/s
so, Reynolds number in the flow region will be;
Re = ([tex]V_{avg[/tex] × D) / v
we substitute
Re = ( 0.265 × 0.02) / (6.01959 × 10⁻⁷)
Re = 8804.586
we can see that our Reynolds number ( 8804.586 ) more than 2300 and less than 10,000.
Hydraulic and thermal entry length are equal in this flow region,
such that;
[tex]L_h[/tex] = [tex]L_t[/tex]
⇒ 10 × D = 10 × 0.02 = 0.2 m
we can see that the entry length ( 0.2 m ) is smaller than the given length ( 13 m ) in the question; the flow is a turbulent flow.
So we the Nuddelt number
Nu = [tex]0.023Re^{0.8} Pr^{0.4[/tex]
Nu = 0.023 × [tex]8804.586^{0.8[/tex] × [tex]3.91^{0.4[/tex]
Nu = 56.8
Hence, the heat transfer coefficient h will be;
h = [tex]\frac{k}{D}[/tex] × Nu
we substitute
h = [tex]\frac{0.637}{0.02}[/tex] × 56.8
h = 31.85 × 56.8
h = 1809.1 W/m²-°C
Now, area of the heat transfer will be
A[tex]_s[/tex] = πDL
we substitute
A[tex]_s[/tex] = π × 0.02 × 13
A[tex]_s[/tex] = 0.8168 m²
Finally we determine the inner temperature of the pipe at exit. using the relation;
Q' = hA[tex]_s[/tex]( T₃ - T₂ )
we substitute
24139.5 = 1809.1 × 0.8168( T₃ - 10 )
24139.5 = 1477.67288( T₃ - 80 )
24139.5 = 1477.67288T₃ - 118213.8304
24139.5 + 118213.8304 = 1477.67288T₃
1477.67288T₃ = 142353.3304
T₃ = 142353.3304 / 1477.67288T
T₃ = 96.34°C
Therefore, the inner surface temperature of the pipe at the exit is 96.34°C
The Earth Science students are making a human scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters (the length of a football field). How far from the school will the student representing Mars stand? A) 50 meters B) 105 meters 150 meters D) 1500 meters NEED HELP ASAP
In the model 1 AU = 100 meters (the length of a football field) , then mars would be 150 meters far from the school, therefore the correct answer is option C.
What is a unit of measurement?A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation.
As given in the problem the Earth Science students are making a human-scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters,
As given in the table marks is 1.5 AU far from Mars,
1 AU = 100 meters
1.5 AU = 1.5 × 100
= 150 meters
Thus, mars would be 150 meters far from the school, therefore the correct answer is option C.
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Exposure to a sufficient quantity of ultraviolet will redden the skin, producing erythema - a sunburn. The amount of exposure necessary to produce this reddening depends on the wavelength. For a 1.0 cm2 patch of skin, 3.7 mJ of ultraviolet light at a wavelength of 254 nm will produce reddening; at 300 nm wavelength, 13 mJ are required. Part A What is the photon energy corresponding to each of these wavelengths
Answer:
Energy = 7.83 x 10⁻¹⁹ J
Energy = 6.63 x 10⁻¹⁹ J
Explanation:
The energy of a photon in terms of wavelength can be calculated by the following formula:
[tex]Energy = \frac{hc}{\lambda}\\[/tex]
where,
h = Plank's Constant = 6.63 x 10⁻³⁴ Js
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light
Now, for λ = 254 nm = 2.54 x 10⁻⁷ m:
[tex]Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{2.54\ x\ 10^{-7}\ m}\\[/tex]
Energy = 7.83 x 10⁻¹⁹ J
Now, for λ = 300 nm = 3 x 10⁻⁷ m:
[tex]Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{3\ x\ 10^{-7}\ m}\\[/tex]
Energy = 6.63 x 10⁻¹⁹ J
Explain how to make aluminum magnetic
Answer:
Use some Glue.. . . Aluminium is not magnetic, so your magnet can't sticl to aluminium without using Glue. You cannot get a magnet to stick to aluminum, no matter how hard you try. Aluminum is diamagnetic, which means it repels a magnetic field.
Explanation:
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a horizontal 4-cm-diameter water jet with a velocity of 18 m/s. He impinges the jet normally upon a vertical plate of mass 750 kg. The plate rides on a nearly frictionless track and is initially stationary. When the jet strikes the plate, the plate begins to move in the direction of the jet. The water always splatters in the plane of the retreating plate. Determine (a)the acceleration of the plate when the jet first strikes it (time
Answer:
0.5429 m/s^2
Explanation:
velocity of waterjet = 18 m/s
diameter of water jet ( d ) = 4 cm = 0.04 m
mass of vertical plate(m) = 750 kg
Determine the acceleration of plate when the jet first strikes ( i.e. t = 0 )
first we will determine the impact force
F = β*A*V^2 ----- ( 1 )
where ; β = 1000 kg/m^3 , A = π/4 * d^2 , V = 18 m/s
input values into equation 1
F = 407.15 N
finally determine the acceleration at t = 0
F = m*a
a = F / m = 407.15 / 750 = 0.5429 m/s^2