Answer:
A. The fan does work on the air in the room leading to an increase in its thermal energy and temperature.
Explanation:
Fans move fluid, typically a gas, such as air, through a room or an enclosure. A fan consists of a rotating arrangement of vanes or blades (impeller), which acts on the air. The action of the impeller does work on the fan by compressing and moving the air forward, doing work on it in the process. The work done on the fan leads to an increase in the thermal energy of the air.
A plate having an area of 0.6 m2 is sliding down the inclined plane at 300 to the horizontal with a velocity of 0.36 m/s. There is a cushion of fluid 1.8 mm thick between the plane and the plate. Calculate the viscosity of the fluid if the weight of the plate is 280 N.
Answer:
The viscosity of the fluid is 1.16 N-s/m²
Explanation:
Given that,
Area = 0.6 m²
Angle = 30°
Velocity = 0.36 m/s
Thickness = 1.8 mm
Weight = 280 N
We need to calculate the viscosity of the fluid
Using balance equation
[tex]w\sin\theta=\dfrac{\mu\times v}{t}\times A[/tex]
Put the value in the equation
[tex]280\sin30=\dfrac{\mu\times0.36}{1.8\times10^{-3}}\times(0.6)[/tex]
[tex]140=\mu\times120[/tex]
[tex]\mu=1.16\ N-s/m^2[/tex]
Hence. The viscosity of the fluid is 1.16 N-s/m².
A cyclist is standing still at the top of a hill and then begins to coast down the hill. The mass of the cyclist and bicycle is 64 kg total. The cyclist’s gravitational potential energy is converted into kinetic energy with an efficiency of 52%. What is her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started? Show all your work.
Her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started would be 10.1 m/s
What is mechanical energy?Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total potential energy stored energy in the system which is represented by total potential energy.
As given in the problem A cyclist is standing still at the top of a hill and then begins to coast down the hill. The mass of the cyclist and bicycle is 64 kg in total. The cyclist’s gravitational potential energy is converted into kinetic energy with an efficiency of 52%
The potential energy is getting converted into kinetic energy with an efficiency of 52 %
1/2mv² = 0.52 (mgh)
v²= 1.04gh
v = √(1.04gh)
v= √(1.04×9.81×10)
v = 10.1 m/s
Thus, her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started would be 10.1 m/s
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A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stretched to a tension of 122 N. During what time interval will a transverse wave travel the entire length of the two wires
Answer:
The time taken is [tex]t = 0.356 \ s[/tex]
Explanation:
From the question we are told that
The length of steel the wire is [tex]l_1 = 31.0 \ m[/tex]
The length of the copper wire is [tex]l_2 = 17.0 \ m[/tex]
The diameter of the wire is [tex]d = 1.00 \ m = 1.0 *10^{-3} \ m[/tex]
The tension is [tex]T = 122 \ N[/tex]
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as
[tex]t = t_s + t_c[/tex]
Where [tex]t_s[/tex] is the time taken to transverse the steel wire which is mathematically represented as
[tex]t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ][/tex]
here [tex]\rho_s[/tex] is the density of steel with a value [tex]\rho_s = 8920 \ kg/m^3[/tex]
So
[tex]t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]
[tex]t_s = 0.235 \ s[/tex]
And
[tex]t_c[/tex] is the time taken to transverse the copper wire which is mathematically represented as
[tex]t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ][/tex]
here [tex]\rho_c[/tex] is the density of steel with a value [tex]\rho_s = 7860 \ kg/m^3[/tex]
So
[tex]t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]
[tex]t_c =0.121[/tex]
So
[tex]t = t_c + t_s[/tex]
[tex]t = 0.121 + 0.235[/tex]
[tex]t = 0.356 \ s[/tex]
Determine the inductance L of a 0.65-m-long air-filled solenoid 3.2 cm in diameter containing 8400 loops.
Answer:
The inductance is [tex]L = 0.1097 \ H[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 0.65 \ m[/tex]
The diameter is [tex]d = 3.2 cm = 0.032 \ m[/tex]
The number of loops is [tex]N = 8400[/tex]
Generally the radius is evaluated as
[tex]r = \frac{ 0.032 }{2} = 0.016 \ m[/tex]
The inductance is mathematically represented as
[tex]L = \frac{ \mu_o * N^2 * A }{ l }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
A is the cross-sectional area which is mathematically evaluated as
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * (0.016)^2[/tex]
=> [tex]A = 0.000804 \ m^2[/tex]
=> [tex]L = \frac{ 4\pi * 10^{-7} * 8400^2 * 0.000804 }{ 0.65 }[/tex]
=> [tex]L = 0.1097 \ H[/tex]
A metal sphere of radius 19 cm has a net charge of 2.4 × 10–8 C. (a) What is the electric field at the sphere's surface? (b) If V = 0 at infinity, what is the electric potential at the sphere's surface? (c) At what distance from the sphere's surface has the electric potential decreased by 370 V?
Answer:
1.29*10^6 N/C
1135.6 V
9.18 cm
Explanation:
Given that
radius of the metal, r = 19 cm
charge of the metal, q = 2.4*10^-8 C
coulomb's constant, k = 8.99*10^9
to find the electric field, we use the formula E = kq/r², where
E = electric field
k = coulomb constant
q = charge on the metal and
r = radius of the metal
E = (8.99*10^9 * 2.4*10^-8) / 0.19²
E = 215.76 / 0.0361
E = 1.29*10^6 N/C
to find the electric potential, we use this relation
V = kq/r
V = (8.99*10^9 * 2.4*10^-8) / 0.19
V = 215.76 / 0.19
V = 1135.6 V
V = kq/r,
r = kq/V
r = (8.99*10^9 * 2.4*10^-8)/ (1135.6 - 370
r = 215.76 / 765.6
r = 0.281 = 28.1 cm
distance from the sphere
28.18 - 19 = 9.18 cm
a. What quantum number of the hydrogen atom comes closest to giving a 61-nm-diameter electron orbit?
b. What are the electron's speed and energy in this state?
Answer:
a
[tex]n = 23[/tex]
b
[tex]v = 87377.95 \ m/s[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 61\ nm = 61 *10^{-9} \ m[/tex]
Generally the radius electron orbit is mathematically represented as
[tex]r = \frac{61 *10^{-9}}{2}[/tex]
=> [tex]r = 3.05*10^{-8} \ m[/tex]
This radius can also be represented mathematically as
[tex]r = n^2 * a_o[/tex]
Here n is the quantum number and [tex]a_o[/tex] is the Bohr radius with a value
[tex]a_o = 0.0529 *10^{-9} \ m[/tex]
So
[tex]n = \sqrt{\frac{3.05*10^{-8}}{ 0.059*10^{-9}} }[/tex]
=> [tex]n = 23[/tex]
Generally the angular momentum of the electron is mathematically represented as
[tex]L = m * v * r = \frac{n * h }{2 \pi}[/tex]
Here h is the Planck constant and the value is [tex]h = 6.626*10^{-34} J \cdot s[/tex]
m is the mass of the electron with values [tex]m = 9.1*10^{-31} \ kg[/tex]
So
[tex]v = \frac{23 * 6.626*10^{-34} }{2\pi * 9.1 *10^{-31} * 3.05*10^{-8} }[/tex]
[tex]v = 87377.95 \ m/s[/tex]
An airplane is in level flight over Antarctica, where the magnetic field of the earth is mostly directed upward away from the ground. As viewed by a passenger facing toward the front of the plane, is the left or the right wingtip at higher potential? Does your answer depend on the direction the plane is flying?
Answer:
It does not depend on direction of plane and the left windtips more potential
Explanation:
Because if the Fleming right hand rule is applied the the right han is pointed in the direction of flight, and the fingers are curled in the direction of the magnetic lines. Thus , the lines are vertical and so are pointing down, thus by the right hand rule, the electrons move to the left hand side of the plane, although the potentials are equal and opposite in direction
Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.
What was his average speed in mph over the last 400 m?
Answer: His average speed in mph over the last 400 m is 7.7 m/s.
Explanation:
Given: Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.
We know that , speed = [tex]\dfrac{distance}{time}[/tex]
Here , distance = 400m and time = 51.9 s
Then, speed = [tex]\dfrac{400}{51.9}\approx7.7\ m/s[/tex]
Hence, his average speed in mph over the last 400 m is 7.7 m/s.
Answer: 17.2 MPH
Explanation:
Average speed = distance/time
400m÷ 51.9s= 7.7 m/s
Now convert m/s →MPH
m→km→mi and s→min→hr
[tex]\frac{7.7m}{s} x[/tex] [tex]\frac{1 km}{1000 m} x\frac{0.6214 mi}{1 km} x\frac{60 s}{1 min} x\frac{60 mins}{1 hr} = 17.2 mph[/tex]
A 2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
a. Determine the pressure Pb.
b. Determine the work done on the gas during the process ab.
c. Determine the change in internal energy of the gas during the process ab.
d. Determine the heat transferred to the gas during the process ab.
Answer:
a) Pb= 200 PA
b).work done= -3600 joules
c).3600joules
D).the system works under isothermal condition so no heat was transferred
Explanation:
2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
a). PbVb= PaVa
Pb= (PaVa)/VB
Pb= (600*3)/9
Pb= 1800/9
Pb= 200 PA
b). work done= n(Pb-Pa)(Vb-Va)
Work done= 2*(200-600)(9-3)
Work done= -600(6)
Work done=- 3600 Pam³
work done= -3600 joules
C). Change in internal energy I the work done on the system
= 3600joules
D).the system works under isothermal condition so no heat was transferred
A man starts from rest and accelerates at 4.00 m/s2. If he covers a distance of 525 m, how long does he accelerate?
Answer:
16.2 s
Explanation:
Given:
Δx = 525 m
v₀ = 0 m/s
a = 4.00 m/s²
Find: t
Δx = v₀ t + ½ at²
525 m = (0 m/s) t + ½ (4.00 m/s²) t²
t = 16.2 s
A 1500-kg car accelerates from 0 to 25 m/s in 7.0s with negligible friction and air resistance. What is the average power delivered by the engine? (1 hp 746 W)
Answer:
90 hp
Explanation:
Power = work / time
P = ½ (1500 kg) (25 m/s)² / 7.0 s
P = 67,000 W
P = 90 hp
Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the direction in which such pressure fluctuations travel
Answer:
the pressure fluctuation is LONGITUDINAL
Explanation:
Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.
The expression for the wave is
ΔP = Δo sin (kx - wt)
Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL
what single property was the most important in jesseca's material
Answer:
Jesseca wanted to create a material that reflected most of the light that fell on it.
Explanation: The Graphite was the material in the passage that had reflected most of the light.
For the microscope to be in focus, how far should the objective lens be from the specimen?
Answer:
p ≈ f_ objective Therefore for the object to be in focus it must be close to the focal length
Explanation:
A microscope is an optical instrument that uses two lenses, or a long focal length objective lens that forms a real image of the object and an eyepiece that forms a virtual image of the object. Therefore for the object to be in focus it must be close to the focal length
p ≈ f_ objective
p distance objetive
A 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds. During the 10 seconds, the truck has traveled a distance of
Answer:
250 m
Explanation:
1. First, we have to calculate the acceleration:
[tex]\boxed{\mathsf{v=v_o+a\cdot t}}[/tex]
where:
v = present velocity
vo = initial velocity
a = acceleration
t = time
2. Let us use given information and substitute in the expression above. Hence:
[tex]\mathsf{50=0+a\cdot10}\\\\\mathsf{50=10\cdot a}\\\\\mathsf{a=\dfrac{50}{10}}\\\\\therefore \boxed{\mathsf{a=5\,m/s^2}}}[/tex]
3. Now we can calculate traveled distance with Torricelli's equation:
[tex]\boxed{\mathsf{v^2=v_o^2+2\cdot a \cdot d}}[/tex]
where:
v = present velocity
vo = initial velocity
a = accelerarion
d = distance
4. So, we get:
[tex]\mathsf{50^2=0^2+2\cdot 5 \cdot d}\\\\\mathsf{2500=10\cdot d}\\\\\mathsf{d=\dfrac{2500}{10}}\\\\\therefore \boxed{\mathsf{d=250\,m}}[/tex]
Conclusion: during 10 seconds the truck has traveled a distance of 250 m.
Have a nice day! : )
If a 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds. During the 10 seconds, the truck traveled a distance of 250 meters.
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
By using the first equation of motion,
v = u + at
50 = 0 + 10 a
a = 50/10
a= 5 meters/second²
As given in the problem a 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds.
The total distance traveled by truck,
S = ut + 0.5at²
S = 0 + 0.5 ×5 ×10²
S = 250 meters
Thus, the total distance traveled by truck would be 250 meters.
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A man walks 7 km, east in 2 hours and 2 km in 1 hour in the same direction. a) what is
the man's average speed for the whole journey? b) what is the man's average velocity
for the whole journey?
Explanation:
Average speed = distance / time
|v| = (7 km + 2 km) / (2 hr + 1 hr)
|v| = 3 km/hr
Average velocity = displacement / time
v = (7 km east + 2 km east) / (2 hr + 1 hr)
v = 3 km/hr east
In this picture the bike rider starts at point A rides to his friend's house at point B(4 miles away),rides to point C (3 miles away)
and then returns to point A (5 miles away).
Explain-- What is his displacement and why? What is his total distance and how did you calculate it? In general, what is the
difference between distance and displacement?
BONUS QUESTION-Why can displacement never be greater than distance?
Answer:
bonus questions:becz displacement work only with the help of any force or object and distance is the totl length of two points
The half-life of element X is 20 years. If there are 48 g initially a) How much is there after 80 years
Answer:
After 80 years there will be 3 g of element X remaining
Explanation:
Given;
the half life of element X = 20 year
initial mass of element X = 48 g
a) How much is there after 80 years
0 year --------------------------> 48 g
20 years -----------------------> (48g / 2) = 24 g
40 years ------------------------> 12 g
60 years ------------------------> 6 g
80 years --------------------------> 3 g
Therefore, after 80 years there will be 3 g of element X remaining.
Rank these significant figures numbers from the least to the most
a. 357
b. 0.006
c. 9520.00
d. 9256.0
e. 700.003
f. 6010
Answer:
0.006<357<700.003<6010<9256.0<9520.00
a 16.0 kg cart is being pulled by a 95.4 N force to the right and a 36.0 N force to the left. What is the acceleration of the cart?
Answer:
The cart's acceleration is [tex]\approx 3.71\,\,\frac{m}{s^2}[/tex]
Explanation:
Let's start by finding the net force acting on the cart, and then find its acceleration using Newton's 2nd Law.
Net force = 95.4 N -36.0 N = 59.4 N
Now, since we know the cart's mass, we can use Newton's 2nd Law to find the cart's acceleration:
[tex]F=m\,*\,a\\a=\frac{F}{m} \\a=\frac{59.4}{16} \,\,\frac{m}{s^2} \\a\approx 3.71\,\,\frac{m}{s^2}[/tex]
Answer:
3.71 m/s²
Explanation:
A solenoid with a certain number of turns N and carrying a current of 2.000 A has a length of 34.00 cm. If the magnitude of the magnetic field generated at the center of the solenoid is 9.000 mT, what is the value of N?
Answer:
The number of turns of the solenoid is 1217 turns
Explanation:
Given;
current in the solenoid, I = 2 A
length of the solenoid, L = 34 cm = 0.34 m
magnitude of the magnetic field, B = 9 mT = 0.009 T
Number of turns of the solenoid = N
The magnitude of magnetic field at the center of the solenoid is given by;
B = μnI
Where;
μ is permeability of free space = 4π x 10⁻⁷ m/A
I is the current in the solenoid
n is the number of turns per length
n = B/μI
n = (0.009) / (4π x 10⁻⁷)(2)
n = 3580.52 turns/m
N = nL
N =(3580.52 turns/m) x (0.34 m)
N = 1217 turns
Therefore, the number of turns of the solenoid is 1217 turns
An electron moves along the z-axis with vz=4.5×10^7m/s. As it passes the origin, what are the strength and direction of the magnetic field at the following (x, y, z) positions?
a. (1 cm , 0 cm, 0 cm)
b. (0 cm, 0 cm, 1 cm )
c. (0 cm, 2 cm , 1 cm )
IN A FORCE COMPRESSION GRAPH, WHAT IS THE STORED POTENTIAL ENERGY OF THE SPRING WHEN IT'S COMPRESS 0.60M ?
Answer:
La energía potencial elástica es la energía asociada con los materiales elásticos. Por ejemplo, un resorte al ser comprimido o elongado almacena energía potencial elástica y, al ser soltado, puede realizar trabajo sobre un objeto.
Para mantener el resorte comprimido o alargado una cierta longitud x, a partir de su largo natural, es necesario que, en este caso, la mano aplique una fuerza F_{M} sobre el resorte; esta fuerza es directamente proporcional a x.
Explanation:
ón conocida como ley de Hooke.
Para encontrar una expresión que describa la energía potencial asociada con la fuerza del resorte, se determina el trabajo que se requiere para comprimir el resorte desde su posición de equilibrio hasta cierta posición final arbitraria x. Debido a que la fuerza varía desde O hasta kx, se utiliza la fuerza promedio \frac{(F_{0}+F_{X})}{2}.
\[ \bar{F}=\frac{0+K X}{2}=\frac{1}{2}kx \]
fuerza-sobre-un-resorte
Fuerza sobre un resorte. La fuerza para estirar un resorte aumenta linealmente con su elongación .
El trabajo realizado por la fuerza aplicada será: W=\bar{Fx}=\frac{1}{2}kx^{2}
El trabajo realizado se almacena en el resorte comprimido en forma de energía potencial elástica como:
\[ \boxed{ Ep_{elas}=\frac{1}{2}kx^{2}} \]
Una vez que se ha comprimido o estirado el resorte respecto a su posición de equilibrio, la energía potencial elástica se puede considerar como la energía almacenada en el resorte deformado. Esta energía siempre es positiva en un objeto deformado al depender de x^{2}.
Por ejemplo, en la figura se observa que un resorte realiza trabajo sobre un bloque. El resorte que se encuentra sin deformar (a) cuando es empujado por un bloque de masa m, se comprime una distancia x (b). Cuando el bloque se suelta (c), partiendo del reposo, la energía potencial plástica almacenada en el sistema se transforma en energía cinética del bloque.
energia-potencial
g Radiation of an unknown wavelength is used in a photoelectric effect experiment on a sodium surface. The maximum kinetic energy of the observed electrons is 0.7 eV. What is the wavelength of the light
Answer:
λ = 4.1638 10⁻⁷ m
Explanation:
The photoelectric effect was explained by Einstein assuming that the radiation acts like particles and the equation that describes the process is
K = h f -Ф
where K is the kinetic energy of the emitted electrons, hf the energy of the photons according to Planck's equation and Ф the work function of the material
In this case they give us the kinetic energy of the electrons
K = 0.7 eV
The sodium work function is tabulated Ф = 2.28 eV
Let's find the frequency of the photons
f = (K + Ф) / h
Planck's constant is
h = 6.626 10⁻³⁴ J s (1 eV / 1.6 10⁻¹⁹ J) = 4.136 10⁻¹⁵ eV s
f = (0.7 + 2.28) / 4.136 10⁻¹⁵
f = 7.2050 10¹⁴ Hz
let's find the wavelength using the relationship between speed and frequency and wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 7.205 10¹⁴
λ = 4.1638 10⁻⁷ m
A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t = 0.10 s is closest to:________.
Answer:
The value is [tex]v = -0.04 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m = 2.0 \ kg[/tex]
The force constant of the spring is [tex]k = 590 \ N/m[/tex]
The amplitude is [tex]A = + 0.080[/tex]
The time consider is [tex]t = 0.10 \ s[/tex]
Generally the angular velocity of this block is mathematically represented as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
=> [tex]w = \sqrt{\frac{590}{2} }[/tex]
=> [tex]w = 17.18 \ rad/s[/tex]
Given that the block undergoes simple harmonic motion the velocity is mathematically represented as
[tex]v = -A w sin (w* t )[/tex]
=> [tex]v = -0.080 * 17.18 sin (17.18* 0.10 )[/tex]
=> [tex]v = -0.04 \ m/s[/tex]
The velocity of the block at the given time is -0.04 m/s.
Angular speed of the blockThe angular speed of the block is determined by using the following wave equation;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{590}{2} } \\\\\omega = 17.176 \ rad/s[/tex]
Velocity of the blockThe velocity of the block at the given time is calculated as follows;'
[tex]v = - A sin(\omega t)\\\\v =- 0.08 \times sin(17.176 \times 0.1)\\\\v = -0.04 \ m/s[/tex]
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The homogeneous beam of mass 5 kg indicated in the figure is in equilibrium and supported at points A and B. Calculate the reactions at the supports.
Explanation:
Sum of moments about point A:
∑τ = Iα
-mg (L/2) + Rb x = 0
-(5 kg) (10 m/s²) (0.75 m) + Rb (0.70 m) = 0
Rb = 53.6 N
Sum of forces in the y direction:
∑F = ma
Ra + Rb − mg = 0
Ra = mg − Rb
Ra = (5 kg) (10 m/s²) − 53.6 N
Ra = -3.6 N
A scientist claims that a certain chemical will make fabric waterproof. Which
option describes a controlled experiment that will produce evidence that will
support or refute her claim?
Answer: She adds different amounts of the chemical to the material and then puts them in the water
Explanation:
Answer: One group of fabric is treated with the chemical, and the other group is not. Then each group is exposed to water.
Explanation:
A discus thrower achieves a high throw of 100m with the discus released at an angle of 30° calculate the initial speed of the discuss.
Answer:
u = 88.54 m/s
Explanation:
Given that,
A discus thrower achieves a high throw of 100 m.
Angle of projection is 30°
We need to find the initial speed of the discuss. It is a cse of projectile motion. The maximum height reached by the discus is given by :
[tex]H=\dfrac{u^2\sin^2\theta}{2g}[/tex]
u is the initial speed of the discus
So,
[tex]u^2=\dfrac{2\times 9.8\times 100}{\sin^2(30)}\\\\u=\sqrt{7840}\\\\u=88.54\ m/s[/tex]
So, the initial speed of the discus is 88.54 m/s.
A cannonball is fired on flat ground at 420 m/s at a 53.0° angle. how far away does it land?
Answer:
17,300 m
Explanation:
Using kinematic equations, first find the time it takes to land.
Δy = v₀ t + ½ at²
0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²
t = 0 s or 68.5 s
The horizontal distance it moves in that time is:
Δx = v₀ t + ½ at²
Δx = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²
Δx = 17,300 m
Alternatively, you can use the range equation:
R = v₀² sin(2θ) / g
R = (420 m/s)² sin(2 × 53.0°) / (9.8 m/s²)
R = 17,300 m
The distance a cannonball will land if it is fired on flat ground at 420 m/s at a 53.0° angle is 17,300 meters.
What is the distance?The complete movement of an object, regardless of direction, is referred to as distance. The amount of ground a thing travels from its starting point to its destination is also referred to as distance.
Given:
A cannonball is fired on flat ground at 420 m/s at a 53.0° angle,
Calculate the time to land on the ground as shown below,
[tex]\Delta y = v_o t +1/2 at^2[/tex]
0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²
t = 0 s or 68.5 s
Calculate the distance as shown below,
[tex]\Delta x = v_o t +1/2 at^2[/tex]
Δ x = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²
Δ x = 17,300 m
Thus, the total distance covered by the cannonball fired with a speed of 420 m/s is 17300 meters.
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Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV. What is the maximum kinetic energy of the photoelectrons emitted from the metal
Answer:
1.41eV
Explanation:
Kinetic Energy of photoelectron(K. Emax)
E = Wo + K.Emax
E = hc/ λ
h = planck's constant = 6.63 * 10^-34
c = speed of light = 3×10^8 m/s
λ = 400nm
Work function (Wo) = 1.70eV
1 eV = 1×10^-19
E = [(6.63×10^-34) * (3×10^8)] / 400×10^-7
E = (19.89 × 10^(-26))/400×10^-7
E = 0.049725×10^-19
K.Emax = E - Wo
K.Emax = (0.049725×10^-19) - (1.7×10^-38)
0.049725×10^-19 interms of eV = (0.049725/1.6)×10^-19 =
K.Emax = 3.11eV - 1.70eV
K.Emax = 1.41eV