Answer:
Explanation:
Coefficient of static friction μs = .830216
Coefficient of kinetic friction μk = .326245
a ) The angle at which the box starts sliding depends upon coefficient of static friction . If θ be the required angle
tanθ = μs
tanθ = .830216
θ = 39.7°
b )
When the box starts sliding , kinetic friction will be acting on it .
frictional force on the box = μk mg cos 39.7
net force on the box
= mg sin39.7 - μk mg cos 39.7
Applying Newton's law of motion
mg sin39.7 - μk mg cos 39.7 = m a
a = g sin39.7 - μk g cos 39.7
= 9.8 x sin 39.7 - .326245 x 9.8 x cos 39.7
= 6.26 - 2.46
= 3.8 m /s² .
A 36.5 mA current is carried by a uniformly wound air-core solenoid with 430 turns, a 18.5 mm diameter, and 11.5 cm length. (a) Compute the magnetic field inside the solenoid. (b) Compute the magnetic flux through each turn. Tm2 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current?i) magnetic field inside the solenoid ii) magnetic flux through each turn inductance of the solenoid
Answer:
Explanation:
a )
magnetic field inside the solenoid B = μ₀ n I where n is no of turns per unit length , I is current and μ₀ = 4 π x 10⁻⁷ .
Putting the values in the equation
B = 4 π x 10⁻⁷ x (430 / .115 ) x 36.5 x 10⁻³
= 1.7 x 10⁻⁴ T .
b ) magnetic flux through each turn
= B x A where A is cross sectional area of solenoid .
= 1.7 x 10⁻⁴ x π x 9.25² x 10⁻⁶
= 456.73 x 10⁻¹⁰ Tm² .
c ) Inductance of solenoid
L = flux associated with all turns / current
= 456.73 x 10⁻¹⁰ x 430 / (36.5 x 10⁻³)
= 5381 x 10⁻⁷
= 538 x 10⁻⁶ H
= 538 μH .
d )
magnetic field inside the solenoid depends upon current
magnetic flux through each turn depends upon current
inductance of the solenoid does not depend upon current because current is divided from total flux with solenoid.
Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+βF(x)=αx3+β.
α=6.1×10^−9 N/m^3
β=−4.1×10^6 N
Answer:
K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x
Explanation:
To find the variation of kinetic energy, let's use the work energy theorem
W = ΔK
∫ F .dx = K -K₀
If the body starts from rest K₀ = 0
∫ F dx cos θ = K
Since the force and displacement are in the same direction, the angle is zero, so the cosine is 1
we substitute and integrate
α ∫ x³ dx + β ∫ dx = K
α x⁴ / 4 + β x / 1 = K
we evaluate from the lower limit F = 0 to the upper limit F
α (x⁴ / 4 -0) + β (x -0) = K
K = αX⁴ / 4 + β x
K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x
in order to finish the calculation we must know the displacement
Answer:
1.1 x 10^10J
Explanation:
∫x2,x1F(x)dx = ∫7.5 x 10^4 m ,0 (αx3+β)dx.
(αx4/4+βx) 7.5 x 10^4 m, 0
((6.1×10−9N/m3)( 7.5×104m)^4)/4 - (4.1×106N)( 7.5×104m) -0)
= 4.825 x 10^10 - 30.75 x 10^10
= 25.925 x 10^10J
= 2.5925 x 10^11J
The kinetic energy KE2 is,
KE2 = KE1 + ∫x2,x1F(x)dx
= 2.7×1011J - .5925 x 10^11J
= 0.1065 x 10^11J
= 1.1 x 10^10J
Monochromatic light with wavelength 588 nm is incident on a slit with width 0.0351 mm. The distance from the slit to a screen is 2.7 m. Consider a point on the screen 1.3 cm from the central maximum. Calculate (a) θ for that point, (b) α, and (c) the ratio of the intensity at that point to the intensity at the central maximum.
Answer:
0.276
0.9
0.756
Explanation:
Given that
Wavelength of the light, λ = 588 nm
Distance from the slit to the screen, L = 2.7 m
Width of the slit, a = 0.0351 mm
a point on the screen, y = 1.3 cm = 0.013 m
Sinθ = y/L
Sinθ = 0.013/2.7
sinθ = 0.0081
θ = sin^-1 0.00481
θ = 0.276°
α = (π.a.sinθ)/λ
α = (3.142 * 3.51*10^-5 * sin 0.276) / 588*10^-9
α = 5.3*10^-7 / 588*10^-9
α = 0.9 rad
I/i(m) = ((sinα)/α)²
I/I(m) = ((sin 0.9) / 0.9)²
I/I(m) = (0.783/0.9)²
I/I(m) = 0.87²
I/I(m) = 0.756
Note, our calculator has to be set in Rad instead of degree for part C, to get the answer
A truck filled with gravel parks on a roadside scale that measures 8.00m by 6.80m. If the gravel and truck has a mass of 7,400kg. What pressure does the scale put on the spring below?
Explanation:
Pressure = force / area
P = (7400 kg × 10 m/s²) / (8.00 m × 6.80 m)
P = 1360 Pa
A 2-kW electric heater takes 15 min to boil a quantity of water. If this is done once a day and power costs 10 cents per kWh, what is the cost of its operation for 30 days?
Answer:
$1.50Explanation:
Given data
power P= 2 kW
time t= 15 min to hours = 15/60= 1/4 h
cost of power consumption per kWh= 10 cent = $0.1
We are expected to compute the cost of operating the heater for 30 days
but let us computer the energy consumption for one day
Energy of heater for one day= 2* 1/4 = 0.5 kWh
the cost of operating the heater for 30 days= 0.5*0.1*30= $1.50
Hence it will cost $1.50 for 30 days operation
A 0.20-kg baseball is thrown with a speed of 20 m / s. If the speed of the ball at the start of the throw is zero, calculate the net work during the throw.
Explanation:
Work = change in energy
W = ½ mv²
W = ½ (0.20 kg) (20 m/s)²
W = 40 J
What is the wavelength (in 10-15 m) of a proton traveling at 13.2% of the speed of light?
Answer:
The wavelength is [tex]\lambda = 10.01 *10^{-15} \ m[/tex]
Explanation:
From the question we are told that
The speed is [tex]v = 0.132 c[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
Generally the wavelength is mathematically represented as
[tex]\lambda = \frac{h}{m* v }[/tex]
where m is the mass of the proton with the value [tex]m = 1.6726 ^{-27} \ kg[/tex]
h is the Planck's constant with value [tex]h = 6.626 *10^{-34} \ J\cdot s[/tex]
=> [tex]\lambda = \frac{6.626 *10^{-34}}{1.6726 *10^{-34}* 0.132*3.0*10^8 }[/tex]
=> [tex]\lambda = 10.01 *10^{-15} \ m[/tex]
A snail can move approximately 0.30 meters per minute.how many meters can the snail cover in 15 minutes?
Answer:
4.5 meters is your Answer goood luck please give 5 star
Explanation:
A thin rod of mass M and length l hangs from a pivot at its upper end. A ball of clay of mass m and of horizontal velocity v strikes the lower end at right angles andremains stuck (total inelastic collision).
Required:
How high will the rod swing after this collision?
Answer:
P = 2923.89 W
Explanation:
Power is
P = F v
so This exercise will solve them in parts using the conservation of momentum and then using the conservation of energy
To use the conservation of the momentum we must define a system, formed by the bodies, so that the forces during the collision have internal forces and the moment is conserved
initial instant, before the crash
p₀ = m v
final instant. Right after the crash
[tex]p_{f}[/tex] = (M + m) v₂
p₀ =p_{f}
m v = (M + m) v₂
v₂ = m / (m + M) v
this is the speed with which two come out, now we can apply the conservation of energy to the system formed by the two bodies together
Starting point. Lower
Em = K = ½ (M + m) v²
Final point. Highest point
Em = U = (M + m) g h
Eo₀ = [tex]Em_{f}[/tex]
½ (M + m) v2 = (M + m) g h
h = 1/2 v2 / g
h = ½ [m / (m + M) v] 2 / g
h = 1/2 (m / m + M) 2 / g we must calculate the force, let's use Newton's second law, let's set a coordinate system with a parallel axis flat and the other axis (y) perpendicular to the plane
X Axis
Fe - Wₓ = 0
F = Wₓ
Y Axis
N - [tex]W_{y}[/tex] = 0
let's use trigonometry for the components of the weight
sin 6 = Wₓ / W
cos 6 = [tex]W_{y}[/tex] / W
Wx = W sin 6
W_{y}= W cos 6
F = mg cos 6
F = 75 9.8 cos 6
F = 730.97 N
let's calculate the power
P = F v
P = 730.97 4.0
P = 2923.89 W
In certain metal, the stopping potential is found to be 3.70 V. When 235 nm light is incident on the metal, electrons are emitted. What is the maximum kinetic energy given to the electrons in eV and J?
Answer:
3.7 eV
5.92*10^-19 J
Explanation:
Given that.
Potential difference of the metal, V = 3.7 V
Wavelength of the light, n = 235 nm
maximum kinetic energy given to the electrons is giving them the formula
K(max) = e.V(s), where
KE(max) is the maximum kinetic energy needed
V = potential difference of the metal
KE(max) = e * 3.7
KE(max) = 3.7eV
converting our answer to Joules, we have
3.7eV = 3.7eV * 1.6*10^-19 J/eV
3.7eV = 5.92*10^-19 J
Therefore, the maximum kinetic energy in both eV and Joules is 3.7eV and 5.92*10^-19 Joules respectively
Answer:
Explanation:
d dnnd
Water flows through a valve with inlet and outlet velocities of 3 m/s. If the loss coefficient of the valve is 2.0, and the specific weight of water is 9800 N/m3, the pressure drop across the valve is most nearly:
Answer:
9,000 kg/ms^2
Explanation:
The computation of the pressure fall across the valve is shown below:
It is to be computed by using the following formula
[tex]\Delta P = \frac{1}{2}\times K\times P\times v^2[/tex]
where,
[tex]\Delta P[/tex] = Fall in pressure
k = Coefficent loss
P = Loss of density
V = velocity of water
But before reach to the final solution first we have to determine the loss of density which is
[tex]P = \frac{r}{g}\\\\ = \frac{9,800 N/m^{3}}{9.81 m/s^{2}}\\\\ = 999kg/m^{3}\\\\ = 1000kg/m^{3}[/tex]
Now put all other values to the given formula
So,
[tex]= 2 \times \frac{1}{2} \times 1000 \times 3^2 \\\\ = 9,000 kg/ms^2[/tex]
A large, cylindrical water tank with diameter 3.00 m is on a platform 2.00 m above the ground. The vertical tank is open to the air and the depth of the water in the tank is 2.00 m. There is a hole with diameter 0.420 cm in the side of the tank just above the bottom of the tank. The hole is plugged with a cork. You remove the cork and collect in a bucket the water that flows out the hole.
A. When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank? Express your answer in millimeters.
B. How long does it take you to collect 1.00 gal of water in the bucket? Express your answer in seconds.
Answer:
1999.46 mm
45.59 s
Explanation:
given that
cylindrical water tank with diameter, D = 3 m
Height of the tank above the ground, h = 2 m
Depth of the water in the tank, d = 2 m
Diameter of hole, d = 0.420 cm
We start by calculating the volume of water in the tank, which is given as
Volume = πr²h
V = (πD²)/4 * h
V = (3.142 * 3²)/4 * 2
V = 28.278/4 * 2
V = 7.07 * 2
V = 14.14 m³
If 1.0 gal of water is equal to 0.0038m³, then
1 gal is 0.0038 = A * h
the area of the tank is 7.07 m²
therefore, 0.0038 = 7.07 * h
h₁ =0.00054 m = 0.54 mm is the height of water that flow out
the change in height of water in the tank = h - h₁ = 2 - 0.00054 = 1.99946 m
b)
Like we stated earlier, 1.0 gal of water is 0.0038m³
to solve this we use the formula
Q = Cd * A * √2gH
where Cd is a discharge coefficient, and is given by 0.9 for water
A is the area of the small hole
A = (πD²)/4
A = (π * 0.0042²)/4
A = 5.54*10^-5 / 4
A = 1.39*10^-5 m²
H= height of the hole from the tank water level = 2m - 0.0042 = 1.9958 m
g = 9.8 m/s²
Q = 0.9 * 1.39*10^-5 m² * √2 * 9.8 * 1.9958
Q = 1.251*10^-5 * 6.25
Q = 7.82*10^-5 m³/s
Q = V/t
t = V/Q = 0.0038m³ / 7.82*10^-5 m³/s
t = 45.59 s
A ball bearing of radius of 1.5 mm made of iron of density
7.85 g cm is allowed to fall through a long column of
glycerine of density 1.25 g cm. It is found to attain a
terminal velocity of 2.25 cm s-'. The viscosity of glycerine is
Answer:
[tex] \boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise} [/tex]
Given:
Radius of ball bearing (r) = 1.5 mm = 0.15 cm
Density of iron (ρ) = 7.85 g/cm³
Density of glycerine (σ) = 1.25 g/cm³
Terminal velocity (v) = 2.25 cm/s
Acceleration due to gravity (g) = 980.6 cm/s²
To Find:
Viscosity of glycerine ([tex] \sf \eta [/tex])
Explanation:
[tex] \boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}[/tex]
[tex] \sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v} [/tex]
Substituting values of r, ρ, σ, v & g in the equation:
[tex] \sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25} [/tex]
[tex]\sf \implies \eta = \frac{2}{9} \times 64.7196[/tex]
[tex]\sf \implies \eta = 2 \times 7.191[/tex]
[tex]\sf \implies \eta = 14.382 \: poise[/tex]
In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separation between adjacent bright fringes on a screen 5 m from the slits is: CONVERT FIRST
Answer:
Δx = 2.5 x 10⁻³ m = 2.5 mm
Explanation:
The distance between two consecutive fringes, also known as fringe spacing, in Young's Double Slit Experiment, is given as follows:
Δx = λL/d
where,
Δx = distance between consecutive fringes = ?
λ = wavelength of light = 500 nm = 5 x 10⁻⁷ m
L = Distance between slits and screen = 5 m
d = slit separation = 1 mm = 1 x 10⁻³ m
Therefore,
Δx = (5 x 10⁻⁷ m)(5 m)/(1 x 10⁻³ m)
Δx = 2.5 x 10⁻³ m = 2.5 mm
The separation between adjacent bright fringes on a screen 5 m from the slits is: 2.5 mm
We are given;
Wavelength of light; λ = 500 nm = 500 × 10⁻⁹ m
Distance of slit separation; d = 1mm = 0.001 m
Distance between slit and the screen; D = 5 m
Now, formula for fringe width is;
β = λD/d
Plugging in the relevant values gives;
β = (500 × 10⁻⁹ × 5)/0.001
β = 2.5 × 10⁻³ m
Converting to mm gives;
β = 2.5 mm
Read more about fringe widths at; https://brainly.com/question/16749356
A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is released from rest near the stationary charge and is free to move. Which statement best describes the motion of q after it is released
The complete question is;
A positive point charge Q is fixed on a very large horizontal frictionless tabletop. A second positive point charge q is released from rest near the stationary charge and is free to move. Which statement best describes the motion of q after it is released?
A) Its speed will be greatest just after it is released.
B) Its acceleration is zero just after it is released.
C) As it moves farther and farther from Q, its speed will keep increasing
D) As it moves farther and farther from Q, its acceleration will keep increasing.
Answer:
Option C - As it moves farther and farther from Q, its speed will keep increasing.
Explanation:
We are told that when a second positive point charge(q) is released from rest near the stationary charge and it's free to move. Thus, the stationary charge will now exert a repulsive force upon thus second positive point charge and it will go on decreasing because the mobile charge will move away from the stationary charge. Thus, it will have a decreasing but positive acceleration . So we can conclude that it's velocity will keep increasing but it will be at a declining rate.
Thus, the correct answer is;
Option C - As it moves farther and farther from Q, its speed will keep increasing.
an electromagnetic wave has an electric field with peak value 120. What is the averge energy delievered to a surface
Answer:
The average energy delivered to a surface is 19.116 W/m².
Explanation:
Given;
maximum electric field, E₀ = 120 v/m
The average energy delivered by the wave to a surface is given by
[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2}[/tex]
where;
c is the speed of light, = 3 x 10⁸ m/s
ε₀ is the permittivity of free space = 8.85 x 10⁻¹² c²/Nm²
[tex]I_{avg} = \frac{c\epsilon_ o E_o^2}{2} \\\\I_{avg} = \frac{(3*10^8)(8.85*10^{-12})( 120)^2}{2}\\\\ I_{avg} =19.116 \ W/m^2[/tex]
Therefore, the average energy delivered to a surface is 19.116 W/m².
If a car is travelling 120 miles southbound for 3 hours, what is the velocity?
Answer:
40 miles / hour south
Explanation:
120 miles/3 hours = 40 miles / hour
Answer:
v = 40 miles / hour
Explanation:
using velocity formula v = d / t
where d = distance and t = time
120 miles
v = --------------
3 hours
v = 40 miles / hour
If you ride quickly down a hill on a bicycle your eardrums are pushed in before they pop back. Why is this?
Answer:
The difference in pressure between the external air pressure, and the internal air pressure of the middle ear.
Explanation:
First of all, we should note that pressure decreases with height and increases with depth. The air within the middle ear (between the ear drum and the Eustachian tube) adjusts itself to respond to the atmospheric pressure, or when we yawn. At a high altitude like on the hill, the air pressure in the middle ear, is fairly low (this is to balance the low air pressure at this height). While riding down the hill quickly, there is little time for the air pressure in the ear to readjust itself to the increasing external air pressure, causing the external air to push into the ear drum. Along the way, the air within the middle ear is adjusted by the opening of the Eustachian tube, allowing more air into the space in the middle ear to balance the external air pressure. This readjustment causes the ear to pop.
How many significant figures are in 246.32
Answer:5
Explanation:
Decimal :2
Significant notation :2.4632× 10^2
Land, labor, and capital are examples of...
Answer:
The factors of production are resources that are the building blocks of the economy; they are what people use to produce goods and services. Economists divide the factors of production into four categories: land, labor, capital, and entrepreneurship
When stable air is forced to rise, any clouds that are produced are generally thin and flat lying.
a) true
b) false
Answer:
a) true
Explanation:
One of the important factors behind the formation of clouds is the stability of the atmosphere. Air gets condensed with the increase in the height, while it becomes warm with a decrease in its height. A stable air is the type of air that can sink. The air which has low temperature has more density than the air it is surrounded by. When clouds are formed with stable air, the clouds formed are thin and horizontal.
If a bicyclist, with initial speed of zero, steadily gained speed until reaching a final speed of 13m/s, how far would she travel during the race (in the same amount of time)?
Answer:
The distance travel during race is 13 m.
Explanation:
Given that,
Initial speed = 0
Final speed = 13 m/s
Unit time = 1 sec
We need to calculate the distance travel during race
Using formula of distance
[tex]d=vt[/tex]
Where, d = distance
v = velocity
t = time
Put the value into the formula
[tex]d=13\times1[/tex]
[tex]d=13\ m[/tex]
Hence, The distance travel during race is 13 m.
If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged?
Answer:
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Explanation:
The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.
The first minimum occurs for m = 1, so the diffraction equation of a slit remains
a sin θ = λ
in general, the diffraction patterns occur at very small angles, so
sin θ = θ
θ = λ / a
in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.
θ = 1.22 λ /a
In this exercise we are told that the opening changes
a’ = 2 a
we substitute
θ ‘= 1.22 λ / 2a
θ' = (1.22 λ / a) 1/2
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
800 g of water and 160 g of alcohol are placed in a container. Knowing that the density of alcohol is 0.8 g / mL then the density (in g / mL) of the mixture obtained is:
Answer:
0.96 g/mL
Explanation:
The volume of 800 g of water is:
(800 g) / (1 g/mL) = 800 mL
The volume of 160 g of alcohol is:
(160 g) / (0.8 g/mL) = 200 mL
Density = mass / volume
ρ = (800 g + 160 g) / (800 mL + 200 mL)
ρ = 0.96 g/mL
In the photoelectric effect, electrons are ejected from a metal surface when light strikes it. A certain minimum energy, Emin, is required to eject an electron. Any energy absorbed beyond that minimum gives kinetic energy to the electron. It is found that when light at a wavelength of 540 nm falls on a cesium surface, an electron is ejected with a kinetic energy of 260 x 10-20 1 When the wavelength is 400 nm, the kinetic energy is 1.54 x 10-19 J. (a) Calculate Emin for cesium in joules. (b) Calculate the longest wavelength, in nanometers, that will eject electrons from cesium.
Answer:
A) E_min = 36.21 × 10^(-20) J
B) 549 nm
Explanation:
A) The formula for energy of a photon is given as;
E = hc / λ
Where;
h is Planck's constant = 6.626 x 10^(-34) J.s
c is the speed of light = 3 × 10^(8) m/s
λ is wavelength
Wavelength is given as; 540 nm = 540 × 10^(-9) m
Thus;
E = (6.626 × 10^(-34) × 3 × 10^(8))/(540 × 10^(-9))
E = 36.81 × 10^(-20) J
We are given kinetic energy as;2.60 x 10^(-20) J
Now formula for E_min is;
E_min = E - K.E
E_min = (36.81 × 10^(-20)) - (2.60 x 10^(-20))
E_min = 36.21 × 10^(-20) J
B) the longest wavelength, in nanometers, that will eject electrons from cesium would have an energy that would be equal to E_min.
Thus,
36.21 × 10^(-20) = (6.626 × 10^(-34) × 3 × 10^(8))/λ
Making λ the subject gives;
λ = (6.626 × 10^(-34) × 3 × 10^(8))/36.21 × 10^(-20) = 549 x 10(-9) = 549 nm
The minimum energy of the electron is [tex]E_{min} = 34.2 \times 10^{(-20)} \;\rm J.[/tex]
The longest wavelength of the electron is 549 nm.
Given that, in the photoelectric effect, electrons are ejected from a metal surface when light strikes it. A certain minimum energy, Emin, is required to eject an electron. Also given wavelength of the light is 540 nm. The kinetic energy ejected by the electron is 260 x 10-20 J.
When the wavelength is 400 nm, the kinetic energy is 1.54 x 10-19 J.
The energy of the electron can be calculated as,
[tex]E = hc/\lambda[/tex]
Where, [tex]E[/tex] is the energy of the electron, [tex]h=6.626\times10^{(-34)} \;\rm Js[/tex] is plank's constant, [tex]c[/tex] is the speed of light that is [tex]3 \times 10^8 \;\rm m/s[/tex] and [tex]\lambda[/tex] is the wavelength.
So the energy of the electron is,
[tex]E = \dfrac{6.626\times 10^{(-34)} \times 3\times 10^8}{540\times 10^{(-9)}}[/tex]
[tex]E = 3.68 \times 10^{(-19)} \;\rm J[/tex]
The energy of the electron is [tex]E = 3.68 \times 10^{(-19)} \;\rm J[/tex].
The Emin can be calculated as given below.
[tex]E_{min} = E - KE[/tex]
Where [tex]KE[/tex] is the kinetic energy of the electron that is given as [tex]260 \times 10^{(-20)} \;\rm J.[/tex]
So [tex]E_{min} = 3.68\times 10^{(-19)} - 2.60\times 10^{(-20)}[/tex]
[tex]E_{min} = 36.8\times 10^{(-20)} - 2.60\times 10^{(-20)}[/tex]
[tex]E_{min} = 34.2 \times 10^{(-20)} \;\rm J.[/tex]
The minimum energy of the electron is [tex]E_{min} = 34.2 \times 10^{(-20)} \;\rm J.[/tex]
For the longest wavelength, the electron will have its minimum energy that is Emin.
Hence, the longest wavelength can be calculated as given below.
[tex]\lambda = \dfrac {h\times c} {E_{min}}[/tex]
[tex]\lambda=\dfrac{6.626\times 10^{(-34)} \times 3\times 10^8} {34.2 \times 10^{(-20)}}[/tex]
[tex]\lambda = 549 \times 10^{(-9)} \;\rm m\\\lambda = 549 \;\rm nm[/tex]
The longest wavelength of the electron is 549 nm.
For more details, follow the link given below.
https://brainly.com/question/19634968.
Two plastic projectiles have the same mass M and speed V when they hit a glass window pane at a right angle. The interaction between both projectiles and the glass pane takes time T. Projectile A bounces off of the window (in the opposite direction) with the same speed V it had at the beginning. Projectile B sticks to the window pane and stops. Describe the likely damage to the glass pane by selecting all the true statements below.
a. Projectile B has the greater likelihood of breaking the glass since its final momentum is smaller
b. Projectiles A & B have the same likelihood of breaking the glass since they have the same initial momentum
c. Projectile A has the greater likelihood of breaking the glass since its momentum change is larger
d. Projectile A has the greater likelihood of breaking the glass since its momentum change is smaller
Answer:
b. Projectiles A & B have the same likelihood of breaking the glass since they have the same initial momentum
.
c. Projectile A has the greater likelihood of breaking the glass since its momentum change is larger.
Explanation:
for option b, the two projectiles have the same initial mass and velocity, hence they posses the same amount of momentum that if sufficient enough could break the glass.
for option c, projectile A changes direction, maintaining the same speed v. Its momentum changes from from mv to -mv, since its speed changed direction.
the difference in momentum becomes
Δp = -mv - mv = -2mv
this is twice the initial momentum.
projectile B changes momentum from mv to 0
Δp = 0 - mv = -mv.
this is half of the final momentum of projectile A.
Also we know that force is proportional to to the rate of change of momentum, which is greater in projectile A, therefore projectile A impacts more force on the glass. Projectile A therefore has the greater likelihood of breaking the glass since its momentum change is larger.
Please answer all and will give Brainly and get points.
Answer:
16 is D, 17 is A, 18 is C, 19 is B ( would approve of brainliest)
Explanation:
16 is D, 17 is A, 18 is C, 19 is B
In the First option, distance is constant so D shows the correct graph,
In the second option, distance is increasing with time so, the velocity graph A is correct,
In the third option distance is constantly increasing with time, so C is the correct option.
In the last option distance is decreasing with time, so option B is the correct.
What is Distance time graph?
A distance-time graph is defined as how far an object has traveled in a given amount of time which is a simple line graph that shows the plot of distance versus time on a graph. Distance is plotted on the Y-axis while time is plotted on the X-axis.
The graphs which is shown in the question is distance-time graphs for various types of body motion.
When the body is steady or stationary,When thebody is moving non-uniformly with increasing speed,When the body is moving at a uniform speed, andWhen the body is moving non-uniformly with decreasing speed.Thus, the correct options for 16, 17, 18 and 19 are D, A, C and B respectively.
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(II) A baseball pitcher throws a baseball with a speed of 43 m????s. Estimate the average acceleration of the ball during the throwing motion. In throwing the baseball, the pitcher accelerates it through a displacement of about 3.5 m, from behind the body to the point where it is released
Answer:
a = 264.14 m/s²
Explanation:
From the question;
Initial velocity; u will be 0 m/s since the ball will start from rest.
Final velocity; v = 43 m/s
distance covered by the motion; s = 3.5m
To get the acceleration, we will make use of Newton's third equation of motion which is;
v² = u² + 2as
Making a the subject, we have;
a = (v² - u²)/2s
Plugging in the relevant values to give;
a = (43² - 0)/(2 × 3.5)
a = 264.14 m/s²
The average acceleration of the ball during the throwing motion is 265.14m/s².
In order to get the acceleration, the Newton's third law of motion will be used. This will be:
v² = u² + 2as
We'll make a to be the subject of the formula and this will be:
a = (v² - u²) / 2s
We'll plug in the value into the equation and this will be:
a = (43² - 0) / (2 × 3.5)
a = 1849 / 7
= 264.14 m/s²
Therefore, the acceleration is 265.14m/s.
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A large, cylindrical water tank with diameter 2.40 mm is on a platform 2.00 mm above the ground. The vertical tank is open to the air and the depth of the water in the tank is 2.00 mm. There is a hole with diameter 0.600 cmcm in the side of the tank just above the bottom of the tank. The hole is plugged with a cork. You remove the cork and collect in a bucket the water that flows out the hole.
A) When 1.00 gal of water flows out of the tank, what is the change in the height of the water in the tank?
B) How long does it take you to collect 1.00 gal of water in the bucket?
Answer:
Explanation:
A ) radius of tank r = 1.2 m
depth of water in the tank = 2 m
1 gal of water = 1 / 264.17 m³
= 3.785 x 10⁻³ m³
Let h be the change in height of water in the tank .
volume of water flowing out
= π r² x h = 3.785 x 10⁻³
3.14 x 1.2² x h = 3.785 x 10⁻³
h = 83.71 x 10⁻⁵ m
= .84 mm .
B )
change in height is negligible .
velocity of efflux of water from the hole at the bottom
v = √ 2 gh
h is height of water level which is 2 m
v = √ (2 x 9.8 x 2 )
= 6.26 m / s
radius of hole = .3 x 10⁻² m
area of cross section
= π r²
= 3.14 x ( .3 x 10⁻² )²
= 28.26 x 10⁻⁶ m²
volume of water flowing through the hole per unit area
= area of cross section x velocity of efflux
= 28.26 x 10⁻⁶ x 6.26
If t be the time required ,
28.26 x 10⁻⁶ x 6.26 x t = 3.785 x 10⁻³
t = 21.4 s
A calculator draws a current of 0.0008 A for 8 min. How much charge flows through it?
Answer:
The charge that flow through the calculator is 0.384 C
Explanation:
Given;
current drawn by the calculator, I = 0.0008 A
time of current flow, t = 8 min = 8min x 60s = 480 s
The charge that flow through the calculator is given;
q = It
where;
q is the charge that flow through the calculator
I is the current drawn
t is the time
q = 0.0008 x 480
q = 0.384 Coulombs
Therefore, the charge that flow through the calculator is 0.384 C