The horizontal force applied to the lower block is approximately 1,420.85 Newtons
The known parameters are;
The mass of the block, m₁ = 400 kg, weight, W₁ = 3,924 N
The mass of the block resting on the first block, m₂ = 100 kg, weight, W₂ = 981 N
The length of the string attached to the block, W₂, l = 6 m
The horizontal distance from the point of attachment of the second block to the block W₂, x = 5 m
The coefficient of friction between the surfaces, μ = 0.25
Let T represent the tension in the string
The upward force on W₂ due to the string = T × sin(θ)
The normal force of W₁ on W₂, N₂ = W₂ - T × sin(θ)
The tension in the string, T = N₂ × μ × cos(θ)
∴ T = (W₂ - T × sin(θ)) × μ × cos(θ)
sin(θ) = √(6² - 5²)/6
cos(θ) = 5/6
∴ T = (981 - T × √(6² - 5²)/6) × 0.25 × 5/6
Solving, we get;
T ≈ 183.27 N
The normal reaction on W₂, N₂ = T/(μ × cos(θ))
∴ N₂ = 183.27/(0.25 × 5/6) = 879.7
N₂ ≈ 879.7 N
The friction force, [tex]F_{f2}[/tex] = N₂ × μ
∴ [tex]F_{f2}[/tex] = 879.7 N × 0.25 = 219.925 N
The total normal reaction on the ground, [tex]\mathbf{N_T}[/tex] = W₁ + N₂
[tex]N_T[/tex] = 3,924 N + 879.7 N = 4,803.7 N
The friction force, on the ground [tex]\mathbf{F_T}[/tex] = [tex]\mathbf{N_T}[/tex] × μ
∴ [tex]F_T[/tex] = 4,803.7 N × 0.25 = 1,200.925 N
The horizontal force applied to the lower block, P = [tex]\mathbf{F_T}[/tex] + [tex]\mathbf{F_{f2}}[/tex]
Therefore;
P = 1,200.925 N + 219.925 N = 1,420.85 N
The horizontal force applied to the lower block, P ≈ 1,420.85 N
What is the submarine's maximum safe depth in sea water? The pressure inside the submarine is maintained at 1.0 atm
Answer:
The submarine's maximum safe depth in sea water is 801.678 m.
Explanation:
P=Po+(rho)*g*h
Max Pressure = Initial Pressure + (Water Density)(Gravity)(Max Depth)
Area of Window = Pi*(Diameter/2)^2 = Pi*(.4m/2)^2 = 0.125664 m^2
Max Pressure= (1.0*10^6 N)/(0.125664 m^2)= 7.95775-E6 Pa
Initial Pressure= 1atm= 101.3kPa= 101300Pa
Water Density (rho) = 1000kg/m^3
Gravity= 9.8m/s^2
So rearranging for h= (P-Po)/((rho)*g)
h=((7.95775-E6Pa)-(101300Pa))/((1000kg/m^3)(9.8m/s^2))= 801.678 m
. A patient on the infectious disease floor takes 10 mL of levofloxacin syrup bid. If the product is only
available as a 5 mL unit-dose oral syringe, how many syringes will the technician prepare for a 24-
hour supply?
Answer:
The technician will need to prepare 48 syringes for a 24 hour supply since the patient needs double the available dose.
Explanation:
brainliest plz . . .
A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 2.5 kg box sliding across a frictionless floor hits the end of the spring and compresses it 6.5 cm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring
As the box compresses the spring, the spring performs
-1/2 (85 N/m) (0.065 m)² ≈ -0.18 J
of work on the box. By the work energy theorem, the total work performed on the box (which is done only by the spring since there's no friction) is equal to the change in the box's kinetic energy. At full compression, the box has zero instantaneous speed, so
W = ∆K ==> -0.18 J = 0 - 1/2 (2.5 kg) v ²
where v is the box's speed when it first comes into contact with the spring. Solve for v :
v ² ≈ 0.14 m²/s² ==> v ≈ 0.38 m/s
What would happen if you changed the position of the screen, but kept the other factors the same?
Answer:
I wish I could help but iam srry
RATIO of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is:
1) 7/29
2) 9/31
3) 5/27
4) 5/23
Answer:
[tex]5/27[/tex]
Explanation:
wavelengths for Lyman series
[tex]\lambda=\frac{1}{R(1-\frac{1}{4} })=\frac{4}{3R}[/tex]
wavelengths for Balmer series
[tex]\lambda_B=\frac{1}{R(\frac{1}{4}-\frac{1}{9}) } =\frac{1}{R(\frac{5}{36}) } =\frac{36}{5R}[/tex]
[tex]\frac{ \lambda_L}{ \lambda_B} =\frac{4}{3R} \times\frac{5R}{36} =5/27[/tex]
OAmalOHopeO
The ratio of longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is 5/27. The correct option is 3.
What is Lyman and Balmer series?
Lyman and Balmer series are sets of spectral lines in the emission spectrum of hydrogen, which result from the transitions of the electron from higher energy levels to lower energy levels.
The Lyman series consists of spectral lines that are produced by transitions of the electron from higher energy levels to the n=1 energy level. These transitions release energy in the form of ultraviolet photons. The lowest energy level in hydrogen is the n=1 energy level, which is also called the ground state. Therefore, the Lyman series includes the transition of the electron from any energy level greater than or equal to n=2 to the ground state.
The Balmer series consists of spectral lines that are produced by transitions of the electron from higher energy levels to the n=2 energy level. These transitions release energy in the form of visible photons. The lowest energy level in the Balmer series is the n=2 energy level. Therefore, the Balmer series includes the transition of the electron from any energy level greater than or equal to n=3 to the n=2 energy level.
Lyman and Balmer's series are named after the scientists who discovered them. The Lyman series is named after Theodore Lyman, an American physicist who discovered the series in 1906. The Balmer series is named after Johann Balmer, a Swiss mathematician who discovered the series in 1885.
Here in the Question,
The longest wavelength in the Lyman series of the hydrogen spectrum corresponds to the transition from the n = 2 energy level to the n = 1 energy level, while the longest wavelength in the Balmer series corresponds to the transition from the n = 3 energy level to the n = 2 energy level.
The wavelengths of these transitions can be calculated using the Rydberg formula:
1/λ = R(1/n1^2 - 1/n2^2)
where λ is the wavelength of the photon emitted, R is the Rydberg constant (1.097 × 10^7 m^-1), and n1 and n2 are the initial and final energy levels of the electron.
For the longest wavelength in the Lyman series, we have n1 = 2 and n2 = 1, so:
1/λ_lyman = R(1/2^2 - 1/1^2) = 3R/4
For the longest wavelength in the Balmer series, we have n1 = 3 and n2 = 2, so:
1/λ_balmer = R(1/3^2 - 1/2^2) = 5R/36
Therefore, the ratio of the longest wavelengths in the Lyman and Balmer series is:
λ_lyman/λ_balmer = (3R/4)/(5R/36) = 27/20
Simplifying this ratio gives:
λ_lyman/λ_balmer = 27/20
Multiplying both the numerator and denominator by 1/3R, we get:
λ_lyman/λ_balmer = (1/2)/(1/3) = 3/2
Therefore, the ratio of the longest wavelengths in the Lyman and Balmer series is 3:2, or 3/5 in fractional form. Simplifying this ratio gives:
λ_lyman/λ_balmer = 5/3
Taking the reciprocal of both sides, we get:
λ_balmer/λ_lyman = 3/5
Therefore, the correct answer is (3) 5/27.
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14. If you are asked to analyze an issue for an essay assignment, what should you do? a. Nb. C. Divide the issue into its main parts and discuss each part. Express your opinion about the issue. Explain what you think the issue means and how you came to that interpretation Describe the main ideas or points about the issue. d.
Answer:
Option C
Explanation:
Divide the issue into its main parts and discuss each part.
10. A mass weighing 10 pounds stretches a spring foot. This mass is removed and replaced with a mass of 1.6 slugs, which is initially released from a point foot above the equilibrium position with a downward velocity of . (a) Express the equation of motion in the form given in (6). (b) Express the equation of motion in the form given in (6) (c) Use one of the solutions obtained in parts (a) and (b) to determine the times the mass attains a displacement below the equilibrium position numerically equal to the amplitude of motion.
Answer:
Hah yyytttdvooophhfdbbb hhhctuionmm hhdtjjjov
If an object with constant mass is accelerating, what does Newton's second
law imply?
A. It will continue to accelerate until it meets an opposing force.
B. The object is exerting an opposite but equal force.
C. A force must be acting on the object.
D. The object will be difficult to decelerate.
Answer:
C. A force must be acting on the object.
Explanation:
This is due to the action of its momentum direction.
[tex].[/tex]
2. The vector sums of and the Ark witar must se rue our the directions and maintedes at an Bit CB? What meast le tue about the directions and magnitudes and it cor
Check attached photo
Check attached photo
physics approach to study macromoelcues at nanoscales
in detail plx
Answer:
Abstracto
Los ácidos nucleicos y las proteínas comprenden una red de biomacromoléculas que almacenan y transmiten información que sustenta la vida de la célula. El estudio de estos mecanismos es un campo llamado biología molecular. El desarrollo de esta ciencia siempre ha ido acompañado de avances técnicos que permiten romper barreras metodológicas para probar hipótesis novedosas. Entre los métodos disponibles para los biólogos moleculares, destacan cinco: electroforesis, secuenciación, clonación, transferencia y reacción en cadena de la polimerasa. Su impacto llega a la genética, la medicina y la biotecnología. Aquí, se revisan la relevancia histórica, los fundamentos técnicos y las tendencias actuales de estos cinco métodos esenciales. La revisión pretende ser útil tanto para estudiantes como para científicos profesionales que buscan adquirir conocimientos avanzados sobre el valor de estos métodos para investigar los mecanismos moleculares que sostienen la vida.
what is 5.02 Properties of Light Lab Report
Answer:
I Don't know
Explanation:
Because I'm 5 grade student
Flag question
Consider the pressure and force acting on the
dam retaining a reservoir of water. Suppose the
dam is 500-m wide
and the water is 80.0-m
deep at the dam, as illustrated below. What is
the average pressure on the dam due to the
water?
Answer:
P = density (p) * g * h
P = 1000 kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2
since kg m / s^2 = Newtons
The average pressure is 1/2 (pressure at 0m + pressure 80 m) for liquid of uniform density
What is a benefit of joining the YMCA instead of a private health club?
The YMCA has nicer equipment and buildings.The YMCA has nicer equipment and buildings. , ,
The YMCA is likely to be in an upscale neighborhood.The YMCA is likely to be in an upscale neighborhood. , ,
The YMCA is much more affordable.The YMCA is much more affordable. , ,
The YMCA has restaurants and spa services.
Answer:
Explanation:
This depends on where you are.
The YMCA where I grew up was affordable. Very. Or my mother would not have let me go.
Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.
Compute the PE when the spring is compressed by 0.50 m.
Answer
E - 1/2 K x^2 potential energy of compressed spring
E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m
Which item will be shipped third?
—-
Answer:
I know it's groceries
Explanation:
electronics ship before clothing
electronics ship after groceries
urgent items are first so
order:
1.) A/Electronics
2.) Clothing/B
3.) Groceries(since groceries aren't urgent)
thing is it's C or D I'm leaning to D since it says it ships last but i dont know so if I'm wrong sorry.
A/An is a type of blood cell that's also called a red blood cell. a) Jeukocyte O b) thrombocyte c) plasma d) erythrocyte
Answer:
red blood cell, also called erythrocyte
Explanation:
Hope it helps
Mark me as Brainliest plz
:DD
A 4kg block is attached to a vertical spring with a spring constant of 800N/m. How much elastic potential energy is stored in the system?
E= [tex]\frac{1}{2}[/tex]×k×x² = 1J
given k=800n/m x=0.05m
Current is the rate at which charge is flowing.
a. True
b. Fals
Answer:
A. True
Explanation:
The pressure of sea water increases by 1.0atm for each 10m increase in the depth, by what percentage is the density of water increased in the deepest ocean of water of 12km. Compressibility is 5.0×10^-5 atm
The percentage by which the water density increased is 4.1[tex]\mathbf{\overline 6}[/tex] %
The known values are;
The increase in pressure per 10 meter increase in depth = 1.0 atm
The depth of the deepest ocean = 12 km = 12,000 m
The compressibility of the ocean = 5.0 × 10⁻⁵ 1/atm
The unknown
The percentage the density of water increased in the deepest ocean
Strategy;
Find the pressure at the deepest point of the deepest ocean and apply the compressibility
We have;
[tex]\mathbf{Compressibility = \dfrac{1}{V} \times \dfrac{\partial V}{\partial p}}[/tex]
The change in pressure, [tex]\partial p[/tex] = (12,000 m/(10 m)) × 1.0 atm = 1,200 atm
Therefore, we have for one cubic meter of water
[tex]\mathbf{5.0 \times 10^{-5} \ atm^{-1} = \dfrac{1}{1 \, m^3} \times \dfrac{\partial V}{1,200 \, atm}}[/tex]
Therefore;
[tex]\mathbf{\partial}[/tex]V = 5.0 × 10⁻⁵ atm⁻¹ × 1 m³ × 1,200 atm = 0.06 m³
The new volume = V - [tex]\mathbf{\partial}[/tex]V
∴ The new volume = 1 m³ - 0.06 m³ = 0.94 m³
The initial density = mass/(1 m³)
The new density = mass/(0.96 m³)
The percentage increase in density, [tex]\partial[/tex]ρ%, is given as follows;
[tex]\mathbf{\partial p \% = \dfrac{ \dfrac{Mass}{0.96 \ m^3} - \dfrac{Mass}{1 \ m^3} }{ \dfrac{Mass}{1 \ m^3}} \times 100 = \dfrac{25}{6} \% = 4.1 \overline 6 \%}[/tex]
∴ [tex]\mathbf{\partial}[/tex]ρ% = 4.1[tex]\mathbf {\overline 6}[/tex] %
The percentage by which the water density increased, [tex]\partial[/tex]ρ% = 4.1[tex]\mathbf{\overline 6}[/tex] %
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Answer:
solving for: velocity
equation: velocity = distance / time
substitution: velocity = 1425 km / 12.5 hrs
answer: 114 km/hr
A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.
a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?
b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?
Answer: it would be daddy
Explanation:
Because I’m daddy
A 50.0 ohm and a 30.0 ohm resistor are connected in parallel. What is their equivalent resistance? Unit=Ohms
R(parallel) = product/ sum
50×30/50+30
1500/80
18,75 ohms
Answer: 18.75
above is right but you need to put a dot after the number 18
What is a measure of how hard it is to stop a moving object?
25.
A. gravity
B. weight
C.
inertia
D. momentum
Answer:
C. inertia
Explanation:
inertia describes an object’s resistance to change in motion (or to get in motion due to a lack of motion), and momentum describes how much motion it has.
both are connected, as inertia depends on the object's momentum, but the answer here is inertia.
The kinetic energy of a particle of mass 500g is 4.8j. Determine the velocity of the particle
Answer:
4.38 m/s
Explanation:
A 2.5 m segment of wire supplying current to the motor of a submerged submarine carries 1050 A and feels a 3.6 N repulsive force from a parallel wire 5.00 cm away. show answer No Attempt What is the magnitude of the current in the other wire in A
force between two parallel wire is
f/l = mueo*i1i2/2pir
f/l = 2*10^-7*i1i2/rl
i2 = f*r/2*10^-7*i1
i2 = 342.9 A
A sinusoidal voltage is displayed on an oscilloscope screen. The separation on the screen between two adjacent peaks is 5.5 divisions, and you notice that the sweep speed is set to 1 ms per division. What is the frequency of the sinusoidal voltage
Answer:
f = 1.8 10² Hz
Explanation:
With the readings of the oscilloscope screen we can calculate the period of the wave
T = #_divisions time_base
T = 5.5 1 10⁻³
T = 5.5 10⁻³ s
the period and frequency are related
f = 1 / T
f = 1 / 5.5 10⁻³
f = 1.8 10² Hz
Why do we use semiconductor instead of metal in thermopile?
Answer:
Metal or conductors , what they do is that they allow full flow of current that is conduction is due to free electrons only and there is literally no gap between valence and conduction band,so free electrons can easily jump into conduction band from valence band.
And now lets talk about insulators,what they do is that they don’t allow any current to flow i.e they act as strong dielectric,and gap between valence band and conduction band is so big that free electron can never come into conduction band from valence band ever if they try is for eternity…
But semiconductors can act both as an insulator as well as a conductor based on the voltage input. Hence, there is a possibility to control the current flow in semiconductors , so they don’t just relax and let the current pass by, they can control it, and that is why you can design logic circuits with it.
But as the temperature increases free electrons from valence band of insulators can jump to conduction band and can cause a little conductivity, and then the insulator will act as a semi-conductor.
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B
[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Explanation:
Given:
[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]
[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by
[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]
[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]
[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
1) Un objeto realiza un movimiento circular uniforme en una circunferencia de 10 metros de diámetro y efectúa 20 vueltas por minuto. Se pide hallar:
a) El periodo.
b) La frecuencia en Hertz.
c) La velocidad tangencial. d) La velocidad angular.
e) La aceleración centrípeta.
Answer:
A RECIPE NEEDS A COMBINED WEIGHT OF 720G OF FLOUR AND SUGAR. IF THE RECIPE NEEDS 5TIME FLOUR THAN SUGAR,HOW MUCH OF EACH IS NEEDED
7) A force of 500N exists between two identical point charges separated by a dis-
tance of 40 cm the magnitude of the two charges is
Answer:
q=9.43×10^-5C
Explanation:
F=kq^2/r^2
500= 9×10^9 × q^2/ (0.4)^2
500N×0.16m=9×10^9Nm^2C^2 × q^2
80/(9×10^9)= q^2
√(8.8889×10^9) = q
q= 9.43×10^-5C
since they are identical both charges, q=9.43×10^-5C
The magnitude of the two charges is 9.43×10^-5 C.
To calculate the magnitude of an electric force, it is necessary to use the following expression:
[tex]F = k \frac{q_{1} \times q_{2} }{d^{2}}[/tex]
Assuming that the constant is:
[tex]k = 9\times 10^{9}[/tex]
We can apply the values in the formula above, obtaining:
[tex]500 = 9 \times 10^{9} \times \frac{q^{2}}{0.4^{2}}[/tex]
[tex]q = \sqrt{0.88\times10^{-8}}[/tex]
[tex]q = 9.43 \times 10^{-5} C[/tex]
So, the magnitude of the two charges is 9.43×10^-5 C.
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