Answer:
(a) 5.46 C
(b) 3.4125 x 10^19
Explanation:
q' = 1.08 C, q'' = 6.74 C, q''' = 4.61 C, q'''' = 9.41 C
When the charges are in contact to each other.
(a) So, the net charge is
[tex]q = \frac{q' + q'' + q''' + q''''}{4}[/tex]
[tex]q = \frac{1.08+6.74+4.61+9.41}{4}\\\\q = 5.46 C[/tex]
(b) As the charge is positive in nature, so the protons are there. The number of protons is
[tex]n = \frac{q}{e}\\\\n = \frac{5.46}{1.6\times 10^{-19}}\\\\n = 3.4125\times 10^{19}[/tex]
state any 3 properties of an ideal gas as assumed by the kinetic theory.
Answer:
The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the walls of the container, but otherwise do not interact; and (3) the transfer of kinetic energy between molecules is heat.
A physics student likes to study while listening to loud music. If electricity costs 12.00$/kWh (kilowatt-hour), how much would it cost the student to run a 220 W stereo system 8.0 hours per day for 10 days of studying?
Answer:
the cost of running the stereo is $211.2
Explanation:
Given;
cost of electricity, = 12.00$/kWh
power consumed by the stereo system, P = 220 W
duration of the power consumption, t = 8 hours
number of days, = 10 days
total time of the power consumption = 8 hours x 10 = 80 hours
power consumed in kW = 220 W / 1000 = 0.22 kW
Energy consumed = 0.22 kW x 80 h = 17.6 kWh
The cost of using 17.6 kWh
= 17.6 x $12
= $211.2
Therefore, the cost of running the stereo is $211.2
An object is acted upon by two and only two forces that are equal magnitude and oppositely directed. Is the objected necessarily in static equilibrium? Explain. You can draw a picture if that helps explain.
Answer:
the body is subjected to a continuous rotation and the body is not in rotational equilibrium
Explanation:
For an object to have a static equilibrium, it must meet two relationships
∑ F = 0
∑ τ =0
force acting on a body fulfills the relation of
sum F = F - F = 0
when two forces do not move from position.
To find the torque we assume that the counterclockwise rotations are positive
Σ τ = - F r - F r
Στ = -2 Fr <> 0
consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium
the air pressure at the base of the mountain is 75.0cm of mercury while at the top is 60cm of mercury. Given that theaverage density of air is 1.25kg/m³ and the density of mercury is 13 600 kg/m³and gravity-10N/kg, calculate the height height of the mountain
Answer:
질문?
Explanation:
평균 공기 밀도가 1.25kg/m³이고 수은 밀도가 13600kg/m³이고 g=10N/kg인 경우 산 기슭의 기압은 수은의 75.0cm이고 정상의 수은은 60cm입니다. 산의 높이를 계산?
68.6 bags are required to fill the specified volume of 10"x20"x20".
It is 40 lbs/1.9 gallons, assuming US gallons, for the topsoil.
US gallons equal 3785 ml. 1 lb = 1/2.2 = 0.454 kg or 454 grammes since 1 kilogramme equals 2.2 lbs. 3785 x 1.9 = 7192 ml is equal to 1.9 gallons. 40 lbs = 18160 g. Therefore, the topsoil density is 2.52 g/cc (18160/7192).
The volume of the peat bag is 14x20x30 inches, which is equal to 35.6x50.8x76.2 cm (1 inch = 2/54 cm)=137806 ccs.
In other words, for the 40 lb again, 18160 grams/137806 equals 0.13 g/cc. The peat is therefore significantly lighter than topsoil.
The volume of the latter volume is 120"x240"x20" or 576,000 cubic inches, and the volume of a bag is 14"x20"x30" or 8400 cubic inches, hence 576,000/8400 = 68.6 bags are required to fill the specified volume of 10"x20"x20".
Learn more about density herebrainly.com/question/4970627
SPJ1
A light spectrum is formed on the screen using a diffraction grating. The entire apparatus made up of laser, grating and the screen is now immersed in a liquid with refractive index 1.33. Do the bright spots on the screen get closer together, farther apart, remain the same or disappear entirely? Explain
Answer:
the points are closer to each other
Explanation:
The expression for the diffraction of a grating is
d sin θ = m λ
sin θ = m λ / d (1)
where d is the distance between slits and m is the order of diffraction, the most general is to work in the order m = 1, the angle te is the angle of diffraction
When we immerse the apparatus in a medium with refractive index n = 1.33, the light emitted by the laser must comply
v = λ f
where v is the speed of light in the medium, the frequency remains constant
velocity and refractive index are related
n = c / v
v = c / n
we substitute
c / n = λf
λ = [tex]\frac{c}{f} \ \frac{1}{n}[/tex]
λ = λ₀ / m
where λ₀ is the wavelength in vacuum
we substitute is equation 1
d sin θ = m λ₀ / n
sin θ = λ₀/ n d
sin θ = [tex]\frac{1}{n}[/tex] sin θ₀
we can see that the value of the sine is redueced since the refractive index is greater than 1,
consequently the points are closer to each other
The velocity of a body is given by the equation v=a+bx, where 'x' is displacement. The unit of b is
Answer:
2bsnsnsnns181991oiwiw
The propeller on a boat motor is initially rotating at 8 revolutions per second. As the boat captain reduces the boat speed, the propeller SLOWS at a steady rate of 0.9 revolutions per second per second. After 17 revolutions, how fast is the propeller spinning in revolutions per second
Answer: [tex]5.77\ rps[/tex]
Explanation:
Given
Initial angular velocity is [tex]\omega_i=8\ rps[/tex]
rate of reduction [tex]\alpha=0.9 rev/s^2[/tex]
after 17 revolution i.e. [tex]\theta =17\ rev[/tex]
using [tex]\Rightarrow \omega_f^2-\omega_i^2=2\alpha\theta[/tex]
Insert the values
[tex]\Rightarrow \omega_f^2=8^2-2\times (0.9)\times17\\\Rightarrow \omega_f^2=33.4\\\Rightarrow \omega_f=5.77\ rps[/tex]
During normal beating, the heart creates a maximum 4.10-mV potential across 0.350 m of a person's chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? V/m (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? T (c) What is the wavelength of the electromagnetic wave?
Explanation:
Given that,
Maximum potential, V = 4. mV
Distance, d = 0.350 m
Frequency of the wave, f = 100 Hz
(a) The maximum electric field strength created is given by:
[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]
(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :
[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]
(c) The wavelength of the electromagnetic wave can be calculated as :
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]
So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].
5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?
Answer:
gshshs
Explanation:
hshsksksksbsbbshd
An object undergoing simple harmonic motion takes 0.40 s to travel from one point of zero velocity to the next such point. The distance between those points is 50 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.
Answer:
a) [tex]P=0.80[/tex]
b) [tex]1.25Hz[/tex]
c) [tex]A=25cm[/tex]
Explanation:
From the question we are told that:
Travel Time [tex]T=0.40s[/tex]
Distance [tex]d=50cm[/tex]
a)
Period
Time taken to complete one oscillation
Therefore
[tex]P=2*T\\\\P=2*0.40[/tex]
[tex]P=0.80[/tex]
b)
Frequency is
[tex]F=\frac{1}{T}\\\\F=\frac{1}{0.80}[/tex]
[tex]1.25Hz[/tex]
c)
Amplitude:the distance between the mean and extreme position
[tex]A=\frac{50}{2}[/tex]
[tex]A=25cm[/tex]
Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10^-4 m. One laser emits light of wavelength 480 nm and the other is 640 nm. A screen to view the light intensity pattern is 2.0 m behind the barrier. The distance from the center of the pattern to the nearest completely dark spot (dark for both colors) is ____ cm. (include 2 digits after the decimal point)
Answer:
a) y = 2.4 x 10⁻³ m = 0.24 cm
b) y = 3.2 x 10⁻³ m = 0.32 cm
Explanation:
The formula of Young's Double Slit experiment will be used here:
[tex]y = \frac{\lambda L}{d}\\\\[/tex]
where,
y = distance between dark spots = ?
λ = wavelength
L = distance of screen = 2 m
d = slit width = 4 x 10⁻⁴ m
a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:
[tex]y = \frac{(4.8\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}[/tex]
y = 2.4 x 10⁻³ m = 0.24 cm
a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:
[tex]y = \frac{(6.4\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}[/tex]
y = 3.2 x 10⁻³ m = 0.32 cm
Solve numerical problem. Please give me step - step explanation Help me out plz
Answer:
You should multiply 60 kg*9.8 and answer will come.
Hope this will help you.
Answer:
yes she is right you should multiple 60*9.8
have a great day God bless you
what is the energy of an electromagnetic wave that has a frequency of 8.0 x 10^15 Hz? Use the equation...
(C)
Explanation:
[tex]E = hf = (6.626×10^{-34}\:\text{J•s})(8.0×10^{15}\:\text{Hz})[/tex]
[tex]= 5.3×10^{-18}\:\text{J}[/tex]
Answer:
It's D
Explanation:
It's from alvs
An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X is stationary. An observer on Planet X sees rocket 2 travel toward Earth at 0.4c. What is the speed of rocket 1 according to an observer on rocket 2?
Answer:
0.625 c
Explanation:
Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.
In the context,
The relative speed of body 2 with respect to body 1 can be expressed as :
[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]
Speed of rocket 1 with respect to rocket 2 :
[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]
[tex]$u' = \frac{0.7 c}{1.12}$[/tex]
[tex]u'=0.625 c[/tex]
Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c
Relative to a stationary observer, a moving clock Group of answer choices can do any of the above. It depends on the relative energy between the observer and the clock. always runs faster than normal. can do any of the above. It depends on the relative velocity between the observer and the clock. always runs slower than normal. keeps its normal time.
Answer:
always runs slower than normal.
Explanation:
The basic concept of theory of relativity was given famous scientist, Albert Einstein. The relativity theory provides the theory of space and time, which are the two aspects of spacetime.
According to the theory of relativity, the laws of physics are same for all the non-accelerating observers.
In the context, according to the theory of relativity, a moving clock relative tot a stationary observer always runs slower than the normal time.
define nortons theorem
Answer:
In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.
1. A sequence of potential differences v is applied accross a wire (diameter =0.32 mm length = 11 cm and the resulting current I are measured as follows: V 0.1 0.2 0.3 0.4 0.5 I (MA) 72 144 216 288 360 2) a) plot a graph of v against I.
b) determine the wire's resistence , R.
c) State ohm's law and try to relate it . your results.
Answer:
a. Find the graph in the attachment
b. 720 kΩ
c. The ratio V/I gives us our resistance which is 720 kΩ
Explanation:
a) plot a graph of V against I.
To plot the graph of V against I, we plot the corresponding points against each other. With the voltage V measured in volts and the current I measured in mA, the plotted graph is in the attachment.
b) Determine the wire's resistance , R.
The resistance of the wire is determined as the gradient of the graph.
R = ΔV/ΔI = (V₂ - V₁)/(I₂ - I₁)
Taking the first two corresponding measurements. V₁ = 72 V, I₁ = 0.1 mA, V₂ = 144 V and I₂ = 0.2 mA
R = (144 V - 72 V)/(0.2 - 0.1) mA
R = 72 V/0.1 mA
R = 72 V/(0.1 × 10⁻³ A)
R = 720 × 10³ V/A
R = 720 kΩ
c) State ohm's law and try to relate it your results.
Ohm's law states that the current flowing through a conductor is directly proportional to the voltage across it provided the temperature and all other physical conditions remain constant.
Mathematically, V ∝ I
V = kI
V/I = k = R
Since the ratio V/I = constant, from our results, the ratio of V/I for each reading gives us the resistance. Since we have a linear relationship between V and I, the gradient of the graph is constant and for each value of V and I, the ratio V/I is constant. So, the ratio V/I gives us our resistance which is 720 kΩ.
Since V/I is constant, we thus verify Ohm's law.
A(n) _______________ absorbs energy and then emits electromagnetic radiation based on its _______________. Classical physics predicted that at a high enough temperature, _______________ light would be emitted. Instead, white light was emitted, resulting in the ultraviolet _______________. The photoelectric effect occurs when light shining on a metal creates a(n) _______________. However, only light of a certain minimum _______________ causes electrons to flow. Gas atoms excited by an electric current emit bands of colors of light in a(n) _______________. Each narrow band of light is associated with _______________ of a specific energy.
Answer:
Blackbody radiator, temperature, ultraviolet, catastrophe, electric current, frequency, spectrum, photons
Explanation:
# a p e x
1 and 2 ) A blackbody radiator is an object that absorbs energy, then emits electromagnetic radiation based on the temperature of the object. This comes directly from the definition in the passages.
3 and 4 ) Ultraviolet catastrophe describes when old physicists assumed as frequency increased the waves would go from visible to ultraviolet because that is what comes next on the spectrum. Instead of this happening, the light became white light and it was an apparent 'catastrophe'
Appropriate words for blank position shown below,
Blackbody radiatortemperature ultravioletcatastropheelectric currentfrequency spectrum photonsA blackbody radiator is defined as an object that absorbs all electromagnetic radiation that falls on it at all frequencies over all angles of incidence.
Ultraviolet light is a type of electromagnetic radiation that makes black-light posters glow
A movement of positive or negative electric particles produce current.
Frequency is defined as the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.
Photons are particles which transmit light.
Learn more:
https://brainly.com/question/18533748
(a) Calculate the force needed to bring a 800 kg car to rest from a speed of 85.0 km/h in a distance of 115 m (a fairly typical distance for a nonpanic stop).
N
(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a), i.e. find the ratio of the force in part(b) to the force in part(a).
(force in part (b) / force in part (a))
Answer:
2Al + 2H2O + 2NaOH ⟶ 3H2 + 2NaAlO2
Chất rắn màu xám bạc của nhôm (Al) tan dần trong dung dịch, sủi bọt khí là hidro (H2).
Explanation:
Two long, straight wires are separated by 0.120 m. The wires carry currents of 11 A in opposite directions, as the drawing indicates. Find the magnitude of the net magnetic field.
Answer:
The magnitude of the magnetic field is 1.83 x [tex]10^{-5}[/tex] T.
Explanation:
The flow of an electric current in a straight wire induces magnetic field around the wire. When current is flowing through two wires in the same direction, a force of attraction exists between the wires. But if the current flows in opposite directions, the force of repulsion is felt by the wires.
In the given question, the direction of flow of current through the wires is opposite, thus both wires applies the same field on each other. The result to repulsion between them.
The magnetic field (B) between the given wires can be determined by:
B = [tex]\frac{U_{o}I }{2\pi r}[/tex]
where: I is the current, r is the distance between the wires and [tex]U_{0}[/tex] is the magnetic field constant.
But, I = 11 A, r = 0.12 m and [tex]U_{0}[/tex] = 4[tex]\pi[/tex] x [tex]10^{-7}[/tex] Tm/A
So that;
B = [tex]\frac{4\pi *10^{-7}*11 }{2\pi *0.12}[/tex]
= 1.8333 x [tex]10^{-5}[/tex]
B = 1.83 x [tex]10^{-5}[/tex] T
calculate the pressure of water having density 1000 kilo per metre square at a depth of 20 m inside the water
Answer:
the pressure of the water at the given depth is 196,200 N/m².
Explanation:
Given;
density of the water, ρ = 1000 kg/m³
depth of the water, h = 20 m
acceleration due to gravity, g = 9.81 m/s²
The pressure at the given depth of the water is calculated as;
P = ρgh
P = 1000 x 9.81 x 20
P = 196,200 N/m²
Therefore, the pressure of the water at the given depth is 196,200 N/m².
Tres personas, A, B y C jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección horizontal y la persona B aplica a su vez 5 en dirección horizontal, ¿Cuál es el valor de la fuerza que debe ejercer la persona C, para que la caja esté en equilibrio físico?
Answer:
Un objeto se encuentra en equilibrio físico si la fuerza neta que se le aplica es igual a 0.
En este caso solo se aplican fuerzas en el eje horizontal, por lo que las podremos sumar directamente.
La persona A aplica una fuerza:
Fa = -3N
La persona B aplica una fuerza:
Fb = 5N
La persona C aplica una fuerza Fc, la cual aún no conocemos.
Pero sabemos que la caja está en equilibrio físico, por lo que:
Fa + Fb + Fc = 0N
reemplazando los valores que conocemos, obtenemos:
-3N + 5N + Fc = 0N
Ahora podemos resolver esto para Fc, la fuerza que aplica la persona C.
Fc = 0N + 3N - 5N
Fc = -2N
Podemos concluir que la persona C aplica una fuerza horizontal de -2N
.. Solve: 91
Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of
a slit of width 12x10^-5 cm when the slit illuminated by monochromatic light of wave length
6000 A
[KUET’10-11)
(a) 30°
(b) 60°
(c) 15°
(d) None of these
Solution
Explanation:
bro I have no idea fam......
A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hanging vertically, the mass is pulled aside a small distance of 7.6 cm and released from rest. While the mass is swinging the cord exerts an almost-constant force on it. For this problem, assume the force is constant as the mass swings. How much work in J does the cord do to the mass as the mass swings a distance of 8.0 cm
Answer:
work done is -2.8 × 10⁻⁶ J
Explanation:
Given the data in the question;
mass of the pendulum m = 6 kg
Length of core = 1.7 m
Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad
In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.
Now, the required work done will be;
[tex]W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta[/tex]
[tex]W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta[/tex]
[tex]W = -mgl \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta[/tex]
W = [tex]-mgl[[/tex] -cosθ [tex]]^{0.047}_{0.045 }[/tex]
W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]
W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]
W = -2.8 × 10⁻⁶ J
Therefore, work done is -2.8 × 10⁻⁶ J
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is
The question is incomplete. The complete question is :
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s) and the dissolution rate (kg/s).
Solution :
From flow over sphere, the mass transfer equation can be written as :
[tex]$Sh = 2 + 0.6 Re^{1/2} Sc^{1/3}$[/tex]
where, Sherood number, [tex]$Sh = \frac{K_L d}{D_{eff}}$[/tex]
Reynolds number, [tex]$Re=\frac{Vd\rho}{\mu}$[/tex]
Schmid number, [tex]$Sc= \frac{\mu}{\rho D_{eff}}$[/tex]
So,
[tex]$\frac{K_L d}{D_{eff}}=2+0.6 \left( \frac{V d \rho}{\mu} \right)^{1/2} \ \left( \frac{\mu}{\rho D_{eff}} \right)^{1/3}$[/tex]
Diameter, d = 1 cm = [tex]$1 \times 10^{-2}$[/tex] m
V = 1 m/s
[tex]$\rho = 1000 \ kg/m^3$[/tex]
[tex]$\mu = 10^{-3} \ kg/m/s$[/tex]
[tex]$D_{eff} = 2 \times 10^{-9} \ m^2/s$[/tex]
[tex]$\frac{K_L \times 10^{-2}}{2 \times 10^{-9}}=2+0.6 \left( \frac{1 \times 10^{-2} \times 10^3}{10^{-3}} \right)^{1/2} \ \left( \frac{10^{-3}}{10^3 \times 2 \times 10^{-9}} \right)^{1/3}$[/tex]
[tex]$K_L \times 5 \times 10^6=478.22$[/tex]
[tex]$K_L=9.5644 \times 10^{-5}$[/tex] m/s
So the mass transfer coefficient is 9.5644 [tex]$\times 10^{-5}$[/tex] m/s. It is given solubility,
[tex]$\Delta C = 2 \ kg/m^3$[/tex]
[tex]$N = Md^2 \times \Delta C \times K_L$[/tex]
[tex]$N= M \times (10^{-2})^2 \times 2 \times 9.5644 \times 10^{-5}$[/tex]
[tex]$N= 6 \times 10^{-8}$[/tex] kg/s (dissolution rate)
Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be
Answer:
a) [tex]F_g=1.5*10^9Ibf[/tex]
b) [tex]F_t=12490Ibf/ft^2[/tex]
[tex]F_b=0[/tex]
Explanation:
From the question we are told that:
Height [tex]h=200ft[/tex]
Width [tex]w=1200ft[/tex]
a)
Generally the equation for Dam's Hydro static force is mathematically given by
[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]
Where
[tex]\rho=Density\ of\ water[/tex]
[tex]\rho=62.4Ibm/ft^3[/tex]
Therefore
[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]
[tex]F_g=1.5*10^9Ibf[/tex]
b)
Generally the equation for Dam's Force per unit area is mathematically given by
[tex]F=\rho*g*h[/tex]
For Top
[tex]F_t=\rho*g*h[/tex]
[tex]F_t=62.4*32.2*200[/tex]
[tex]F_t=12490Ibf/ft^2[/tex]
For bottom
[tex]Here \\H=0 zero[/tex]
Therefore
[tex]F_b=0[/tex]
The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].
The force per unit area near the top is 86.74 psi.
The force per unit area near the bottom is zero.
Hydrostatic force
The hydrostatic force on the dam is the force exerted on the dam by the column of the water.
[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]
Force per unit area near the topThe force per unit area is the pressure exerted near the top of the dam.
[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]
where;
P is pressure in PSI
ρ is density of water in lb/gal
h is the vertical height in ft
[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]
The pressure near the bottom is zero, become the vertical height is zero.
Learn more about hydrostatic pressure here: https://brainly.com/question/11681616
How can I solve this?
You have three capacitors of values 40 F, 10 F and 50 F. What would their equivalent capacitance (in F) be if they were connected in parallel with each other? Enter your answer as a number only, to one decimal place.
Explanation:
The equivalent capacitance of capacitors in parallel can be determined as
[tex]C_{eq} = C_1 + C_2 + C_3[/tex]
[tex]\:\:\:\:\:= 40\:\text{F} + 10\:\text{F} + 50\:\text{F} = 100\:\text{F}[/tex]
17- How much work is needed for a climber in order to climb 45 m height, where his weight is 70 kg. also, calculate the power required to climb the height in 30 minutes ? g= 9,8 m.sec
Answer:
Work Done= 3150J
Power= 1.75W
Explanation:
Work Done= Force x the distance travelled in the direction of the force (W= f x d)
Weight is a force, i think the qn. stated it wrongly, it should be 70N not 70kg.
Work Done= 70 x 45
=3150J
Power= Work Done/Time
=3150/(30x60)
*convert minutes to seconds since the S.I. unit of Power is joules/seconds(J/s) or watts(W)
=1.75W
15- A racehorse coming out of the gate accelerates from rest to a velocity f 15.0 m/s due west in 1.80 s. What is its average acceleration?
Answer: (15 - 0)/1.8 = 8. 33m/s^2
Explanation:
The acceleration of the racehorse is 8.33 m/s²
The given parameters;
initial velocity of the racehorse, u = 0
final velocity of the racehorse, v = 15 m/s
time of motion of the horse, t = 1.8 s
The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;
[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]
Thus, the acceleration of the racehorse is 8.33 m/s²
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A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of air is 1.29 kg/m^3. The skin and structure of the balloon has a mass of 910 kg. Neglect the buoyant force on the cargo volume itself.
Determine the largest mass of cargo the balloon can lift.
The largest mass of cargo the balloon can lift is 791.06 kg
First, we need to calculate the mass of helium.
Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.
The volume of the balloon also equals the volume of helium present.
Now, the mass of helium m = density of helium, ρ × volume of helium, V
m = ρV
Since ρ = 0.179 kg/m³
m = ρV
m = ρ4πr³/3.
m = 0.179 kg/m³ × 4π(7.15 m)³/3
m = 0.179 kg/m³ × 4π(365.525875 m³)/3
m = 0.179 kg/m³ × 1462.1035π m³/3
m = 261.7165265π/3 kg
m = 822.207/3 kg
m = 274.07 kg
Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.
The weight of this mass W = Mg where g = acceleration due to gravity.
The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.
W' = m'g
Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'
We know that the volume of air displaced, V' = volume of balloon, V
So, V' = V = 4πr³/3.
Since the density of air, ρ' = 1.29 kg/m³,
m' = ρ'V
m = 1.29 kg/m³ × 4π(7.15 m)³/3
m = 1.29 kg/m³ × 4π(365.525875 m³)/3
m = 1.29 kg/m³ × 1462.1035π m³/3
m = 1886.113515π/3 kg
m = 5925.4/3 kg
m = 1975.13 kg
So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.
So, the net mass m" = W"/g = 791.06g/g = 791.06 kg
This net mass is the largest mass of cargo that the balloon can lift.
Thus, the largest mass of cargo the balloon can lift is 791.06 kg
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