Answer:
Explanation:
yes
The following coplanar forces pull on a ring 200N at 30 degrees and 500N at 80 degrees and 300N at 240 degrees and an unknown force. Find the magnitude and direction of the unknown force of the ring is to be in equilibrium
MAGNITUDE AND DIRECTION OF A VECTOR
Given a position vector →v=⟨a,b⟩,the magnitude is located by |v|=√a2+b2. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tanθ=(ba)⇒θ=tan−1(ba)
What is the formula of magnitude?
The formula to determine the extent of a vector (in two dimensional space) v = (x, y) is: |v| =√(x2 + y2). This formula is derived from the Pythagorean theorem. the procedure to determine the magnitude of a vector (in three dimensional space) V = (x, y, z) is: |V| = √(x2 + y2 + z2)
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main answer is given below,
A solid cylinder of uniform density of 0.85 g/cm3 floats in a glass of water tinted light blue by food coloring.
Its circular surfaces are horizontal. What effect will the following changes, each made to the initial system, have on X, the height of the upper surface above the water? The liquids added do not mix with the water, and the cylinder never hits the bottom.
1. The cylinder is replaced with one that has the same density and diameter, but with half the height.
2. Some of the water is removed from the glass.
3. A liquid with a density of 1.06 g/cm3 is poured into the glass.
4. The cylinder is replaced with one that has the same height and diameter, but with density of 0.83 g/cm3.
5. A liquid with a density of 0.76 g/cm3 is poured into the glass.
6. The cylinder is replaced with one that has the same density and height, but 1.5× the diameter.
Options are: Increase, Decrease, No change
Each side has to have at least 44 horses
F61160 N. This is further explained below.
What is the force?Generally, We are only interested in the component that operates horizontally since the vertical components all cancel each other out. The pressure difference works on the hemisphere to generate a normal force all over the surface, but we are only concerned with that force's horizontal component. This may be determined by supposing the hemispheres to be two flat circular plates of the same radius as the hemispheres that have been forced together.
Therefore, force is equal to pressure multiplied by area, which is
F= (970 -15 )( * (0.45 m)2)
F=60754 N for each side.
Therefore, each side has to have at least 44 horses
44* 1390 = 61160 N
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When the current through a circular loop is 5.7 A, the magnetic field at its center is 3.9 ✕ 10−4 T. What is the radius (in m) of the loop?
Radius of the circular loop is 0.0091m.
What is magnetic field?Magnetic field is the area around a magnet where the magnetism influence is felt .
What is the magnetic field at the centre of a circular loop?The formula for magnetic field at the centre of a loop isB =(μ)I/2r
where B= Magnetic field at the centre of a circular loopμ= Magnetic permeability =4(π)*10^(-7)
I= current flowing through the loop
r= radius of the loop
Thus, radius of the loop =(4(π)×10^(-7)×5.7)/(2×3.9×10^(-4))=0.0091m
Thus, we can conclude that the radius of the loop is 0.0091m .
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A vector in the xy plane has components -14.0 units in the x-direction and 30.0 units in the y-direction. What is the magnitude of the vector? What is the angle between the vector and the positive x-axis?
[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]
We would calculate the magnitude by applying pythagorean theorem:
[tex]\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}[/tex]
[tex]\longrightarrow \sf{Magnitude = 33.12}[/tex]
[tex]\longrightarrow \sf{The \: vector \: is \: (- 14, 30)}[/tex]
The angle between two vectors is given by the formula:
[tex]\sf{\longrightarrow \small \cos \emptyset = \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }[/tex]
In two dimensional, the x axis of vector form is:
[tex]\small\sf{\longrightarrow (b1, b2) = (1, 0) }[/tex]
[tex]\sf{\longrightarrow \small \cos \: \emptyset = \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }[/tex]
[tex]\small\longrightarrow \sf{ \dfrac{14}{33.12} }[/tex]
[tex]\small\longrightarrow \sf{\emptyset \: = arcCos (\dfrac{ - 14}{33.12} )}[/tex]
[tex]\small\longrightarrow \sf{\emptyset= 115^\circ}[/tex]
[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]
[tex] \small\bm{The \: angle \: between \: the \: vector \: }[/tex]
[tex]\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}[/tex]
What is meant by significant figures
Answer:
The term significant figures refers to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation . The number of significant figures in an expression indicates the confidence or precision with which an engineer or scientist states a quantity.
Answer:
each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first nonzero digit.
Explanation:
Most towns use a water tower to store water and provide pressure in the pipes that deliver water to customers. The figure below shows a spherical water tank that holds 5.80 105 kg of water when full. Note that the tank is vented to the atmosphere at the top and that the pipe delivering water to customer Smith is a height h = 3.75 m above the level of the pipe delivering water to customer Jones. Determine the gauge pressure of the water at the faucet of each house when the tank if full.
There is a spherical water-filled chamber with a vent on top and underneath it is a pipe 18 m long that leads down to the ground. At ground level a horizontal pipe connects the faucet in the Jones house to the water supply. The faucet in the Smith house is connected to the water supply with a pipe at a height h above ground level.
(a) Jones house
Pa
(b) Smith house
Pa
The effective height of the water for Smith's house will be 24.61m.
How to calculate the height?Based on the information given, the volume of the water in sphere will be:
= 4/3πr³ = (5.80 × 10^5)/1000
= 4.18r³ = 580
r³ = 138.7
r = 5.18m
The effective height of the water will be:
= 18.0 + 2(5.18)
= 28.36
The gauge pressure at Faucet of Jones house will be:
= pgh
= 1000(9.8)(28.36)
= 277.9kPa
The effective height of the water for Smith's house will be:
= 18.0 + 2(5.18) - 3.75
= 24.61m
The gauge pressure at Faucet of Jones house will be:
= 1000 × 9.8 × 24.61
= 241.2kPa
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Only using dimensional analysis, answer the following question:
Convert 65 miles per hour to kilometers per second. Show your work and explain how you got it please!
Brainliest goes to most detailed response
Answer:
[tex]2.9055\times 10^-2 km/s \approx 0.03 km/s[/tex]
Explanation:
One (statute) mile is defined as [tex]1\ 609.344 m = 1.609\ 334 km[/tex] or, cutting the last significant digits, [tex]1.6 km[/tex]. An annoying European would drop some kind of remark here like "but what about nautical miles?", but let's assume it's a statute mile.
One hour is 3600 seconds (60 minutes in a hour, times 60 seconds in a minute), no complains across the pond here.
Now we're ready to crunch numbers.
[tex]65\frac{mi}h = 65\frac{1.6km}{3600s} \approx 0.03 \frac {km}s[/tex]
You can use the exact definition of mile and you will get the exact value of [tex]0.029055 \frac{km}s = 2.9055\times10^-^2 km/s[/tex] with an error of about a meter.
Answer:
See below
Explanation:
Since you want detailed dimensional analysis here is a lot of details:
mile / hr / (60 min / hr * 60 sec / min) * 1760 yds/mile * 36 in/yd * 1m /39.37 in * 1 km / 1000 m
After cancelling everything out (dimensional analysis) , you are left with:
= mile/ hr * 1/3600 * 1760 * 36 /39.37 * 1/1000
= mile /hr * .00044704089 km -hr/ sec-mile = km/s
the underlined portion is the conversion factor for miles/hr to km/s
65 * .00044704089 = .029057 km / sec
PLEASE HELP!!!!
Compared to driving at 20 mph, about how much longer will it take for you to stop at 60 mph?
about nine times as far
about six times as far
about twice as far
Answer:
The braking distance would be about nine times as long (assuming that acceleration during braking stays the same.)
Explanation:
Let [tex]u[/tex] denote the initial velocity of the vehicle ([tex]20\; \text{mph}[/tex] or [tex]60\; \text{mph}[/tex]) and let [tex]v[/tex] denote the velocity of the vehicle after braking ([tex]0\; \text{mph}[/tex]). Let [tex]x[/tex] denote the braking distance.
Assume that the acceleration during braking are both constantly [tex]a[/tex] in both scenarios. The SUVAT equations would apply. In particular:
[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a}\end{aligned}[/tex].
Since [tex]v = 0[/tex] (the vehicle has completely stopped), the equation becomes [tex]x = (-u^{2}) / (2\, a)[/tex].
Assuming that [tex]a[/tex] (braking acceleration) stays the same, the braking distance [tex]x[/tex] would be proportional to [tex]u^{2}[/tex], the square of the initial velocity.
Hence, increasing the initial speed from [tex]20\; \text{mph}[/tex] to [tex]60\; \text{mph}[/tex] would increase the braking distance by a factor of [tex]3^{2} = 9[/tex].
Answer:
9 times
Explanation:
A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and
400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the
steam is cooled. Determine the compression work if the final state is
a) 1.0 MPa and 250°C and
b) 500 kPa.
c) Also determine the temperature at the final state in part b.
A piston-work cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops corresponds to 40 percent of the initial volume. Now the steam is cooled. compression work is 44.32 KJ,
The amount of labor put into the piston to cause its reciprocating motion is known as the piston work. It is calculated by multiplying the piston's displacement by the net force.
An expanding gas cylinder's force output is transferred by pistons to the crankshaft, which then drives the flywheel's rotation. A reciprocating engine is a device like this.
Piston work is the effort made by the piston to make its reciprocating motion. The piston's displacement is calculated by multiplying it by the net force.
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An internal explosion breaks an object, initially at rest, into two pieces: A and B. Piece A has 1.9 times the mass of piece B. The energy of 7900 J is released in the explosion.
a)Determine the kinetic energy of piece A after the explosion.
Express your answer to two significant figures and include the appropriate units.
b)Determine the kinetic energy of piece B after the explosion.
Express your answer to two significant figures and include the appropriate units.
Kinetic energy of pieces A and B are 2724 Joule and 5176 Joule respectively.
What is the relation between the masses of A and B?Let mass of piece A = MaMass of piece B = Mb
Velocities of pieces A and B are Va and Vb respectively.As per conservation of momentum,Ma×Va = Mb×Vb
Here, Ma=1.9MbSo, 1.9Mb × Va = Mb×Vb
=> 1.9Va = Vb
What are the kinetic energy of piece A and B?Expression of kinetic energy of piece A = 1/2 × Ma × Va²Kinetic energy of piece B = 1/2 × Mb × Vb²Total kinetic energy= 7900J=>1/2 × Ma × Va² + 1/2 × Mb × Vb² = 7900
=> 1/2 × Ma × Va² + 1/2 × (Ma/1.9) × (1.9Va)² = 7900
=> 1/2 × Ma × Va² ×(1+1.9) = 7900 j
=> 1/2 × Ma × Va² = 7900/2.9 = 2724 Joule
Kinetic energy of piece B = 7900 - 2724 = 5176 JouleThus, we can conclude that the kinetic energy of piece A and B are 2724 Joule and 5176 Joule respectively.
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define universal vibrations ?
Answer:
The second universal law, the law of vibration, posits that everything (every atom, object, and living thing) is in constant motion, vibrating at a specific frequency.
Explanation:
Please mark brainly
The second universal law defines this.
The second universal law, also known as the Law of Vibration, The Law of Vibration states that everything in the universe is in a constant state of movement. We refer to these movements as vibration, and the speed or rate at which something vibrates is called its frequency. The only difference between one object and another is its vibration rate.
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A 5 Ω and a 5 Ω resistor are in parallel. What is their total resistance?
A. 10 Ω
B. 2.5 Ω
C. 25 Ω
D. 5 Ω
Answer:
B. 2.5 Ω
Explanation:
[tex]\frac{1}{5} +\frac{1}{5}=0.4[/tex]
Put a one on top of the 0.4
[tex]\frac{1}{0.4} =2.5[/tex]
is a pot plant a closed system?
Answer:
No because a pot plant system must interact with its surroundings (air, water, etc) and these are not closed systems.
Only if the pot plant were completely enclosed, say in a large glass jar would the system be closed.
Maya made this picture to represent a chemical reaction:
Which of the following statements best explains the type of chemical reaction represented by Maya's picture?
A. It is neither a synthesis reaction nor a decomposition reaction because the total mass of the products is less than the total mass of the reactants.
B. It is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.
C.It represents a synthesis reaction because the same atoms are present in the reactants and products.
D. It represents a decomposition reaction because two reactants break apart and form two products.
The statement that best explains the type of chemical reaction represented by Maya's picture is that it is neither a synthesis reaction nor a decomposition reaction because two reactants form two products. That is option B.
What is a chemical reaction?A chemical reaction is the combination of two elements to yield a new product through the formation of bonds.
A chemical reaction is said to be a synthesis reaction when when two different atoms or molecules interact to form a different molecule or compound.
A chemical reaction is said to be a decomposition reaction when one reactant breaks down into two or more products.
Therefore, from the picture, the chemical reaction is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.
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Is it possible for an object to be in equilibrium (no net force), if only one force is acting on it ?
A weight lifter lifts a mass of 250 kg with a force of 5000N to a height of 5m. a. What is the workdone by the weight lifter? b. What is the workdone by the gravity? c. What is the net workdone on the object?
hii friends
Answers are:
a. The work done by the weight lifter W = 25000 Joule
b. The work done by the gravity W = 12250 Joule
c. The net work done on the object W = 12750 Joule
What is Work ?
Work is the product of force and distance in the direction of the force applied. That is, W = F x h
The S.I unit is Joule.
Given that a weight lifter lifts a mass of 250 kg with a force of 5000N to a height of 5m.
The given parameters are;
Mass m = 250 kgForce F = 5000 NHeight h = 5 ma. The work done by the weight lifter will be
W = F x h
W = 5000 x 5
W = 25000 Joule
b. The work done by the gravity will be
W = mg x h
W = 250 x 9.8 x 5
W = 12250 Joule
c. The net work done on the object will be
W = 25000 - 12250
W = 12750 Joule
Since the work done by the gravity is in opposite direction to the work done by the weight lifter, the net work done on the object will be the difference between the two.
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a ball rolls off a platform that is 12 m above ground. The ball's horizontal velocity as it leaves the platform is 5 m/s. How much time it takes for the ball to hit the ground
The time it takes for the ball to hit the ground from the platform is determined as 1.56 seconds.
Time of motion of the ball
The time of motion of the ball is calculated as follows;
h = vt + ¹/₂gt²
where;
v is initial vertical velocity = 0t is time of motionh is height of fall of the ballh = 0 + ¹/₂gt²
h = ¹/₂gt²
t = √(2h/g)
t = √(2 x 12 / 9.8)
t = 1.56 seconds
Thus, the time it takes for the ball to hit the ground from the platform is determined as 1.56 seconds.
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If the internal energy of a system is decreased, which of the following is impossible?
a
Work done by the system is larger than heat
released.
b
Work done on the system is smaller than
heat released.
C
Work done by the system is larger than heat
absorbed.
d
Work done on the system is smaller than
heat absorbed.
Work done on the system is smaller than heat absorbed.
What happens when internal energy decreases?A cell's internal energy drops when it does work or expels heat. There won't be a net change in internal energy if the work performed by a cell matches the energy transferred in by heat or if the work performed on a cell matches the energy transported out by the heat.The energy within remains constant. The ideal gas law states that the temperature decreases according to the volume when a gas is compressed while maintaining a constant pressure. In this instance, more energy is lost as heat from the system is gained through work. Internal energy levels drop.Ideal gases' internal energy and enthalpy depend solely on temperature; neither volume nor pressure play a role. Using property relations, we may demonstrate these characteristics of ideal gases.
If the internal energy of a system is decreased, which of the following is impossible:
C) Work done on the system is smaller than heat absorbed.
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A grinding wheel 0.35 m in diameter rotates at 2600 rpm .
a) Calculate its angular velocity in rad/s .
Express your answer using three significant figures.
b)What is the linear speed of a point on the edge of the grinding wheel?
Express your answer to two significant figures and include the appropriate units.
c)What is the acceleration of a point on the edge of the grinding wheel?
Express your answer to two significant figures and include the appropriate units.
It is calculated that a) The angular velocity of the wheel is 272.13 rad/s,
b) On the edge of the grinding wheel, the linear speed is 47.62 m/s,
and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².
Calculation of angular velocity, linear speed & acceleration:
Provided that,
the diameter of the wheel = 0.35 m
So, the radius, r = 0.35/2 = 0.175 m
As 1 revolution = 2π rad
(a) the angular velocity, ω = 2600 rpm = [tex]\frac{2600 * 2\pi }{60}[/tex] rad/s
⇒ω = 272.13 rad/s
So, the angular velocity is 272.13 rad/s.
(b) The linear speed, v = r * ω
⇒v = 0.175 * 272.13
⇒v= 47.62 m/s
(c) The angular acceleration, [tex]a=\frac{v^{2} }{r}[/tex]
[tex]a = \frac{(47.62)^{2} }{0.175}[/tex]
⇒[tex]a[/tex] = 12958.08 m/s²
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The graph shows the US Department of Labor noise regulation for working without ear protection. A machinist is in an environment where the ambient sound level is of 85 dB, i.e., corresponding to the 8 Hours/day noise level. The machinist likes to listen to music, and plays a Boom Box at an average level of 86.0 dB.
1. Calculate the INCREASE in the sound level from the ambient work environment level (in dB).
2. A student at a rock concert has a seat close to the speaker system, where the average sound intensity corresponds to a sound level of 116 dB. By what factor does that sound intensity exceed the 1.5- Hours/day intensity limits from the graph?
(1) The INCREASE in the sound level from the ambient work environment level (in dB) is 1 dB.
(2) The factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
Increase in the sound levelThe INCREASE in the sound level from the ambient work environment level (in dB) is calculated as follows;
Increase in sound level = final sound level - original sound level
Increase in sound level = 86 dB - 85 dB = 1 dB
Factor of sound level increasefrom the graph at 1.5 hours/day, sound level = 97 dB
Increase in sound intensity = final sound level - original sound level
Increase in sound intensity = 116 dB - 97 dB = 19 dB
Factor increase = 19/97 = 0.196 = 19.6%
Thus, the factor that sound intensity exceed the 1.5- Hours/day intensity limits from the graph is 19.6 %.
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Solve this question please A.S.A.P ⁉️
No spamming ❗
Answer:
Here given is :-Velocity (u) = 10m/sGravity (g) = 10 m/s²Height (h) = ?Explanation:
v² - u² = 2gh0² - 10² = 2 × (-10) × h-100 = -20 h[tex]h = \frac{ - 100}{ - 20} = 5m[/tex]V = u + gt0 = 10 - 10t10t = 10[tex]t = \frac{10}{10} = 1 \: sec[/tex]Total time = 2 secAnswer:
See below
Explanation:
vf = vo + at when it reaches top vf = 0 vo = 10m/s a = -9.81 m/s^2
0 = 10 + (-9.81) t
t = 1.02 seconds to reach max
time up = time down so roundtrip = 2.04 seconds
Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
The minimum necessary height of the IV bag above the position of the needle is 0.37 m.
Minimum necessary heightThe minimum necessary height of the IV bag above the position of the needle is calculated as follows;
P = ρgh
where;
ρ is density = 1.308 kg/L = 1308 kg/m³g is acceleration due to gravity = 9.8 m/s²p is pressure = 35.7 mmHg = 4759.609 Pah is height, (m) = ?Substitute the given parameters and solve for minimum height
h = P/ρg
h = (4759.609) / (9.8 x 1308)
h = 0.37 m
Thus, the minimum necessary height of the IV bag above the position of the needle is 0.37 m.
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According to current evidence, what happens when gays or lesbians rear a child?
a. the child is more likely than average to develop depression or anxiety disorders.
O b. the child develops about the same as other children.
O c. the child is more likely than average to become gay or lesbian.
O d. the child does poorly in school.
Answer: The answer is A.
Explanation:
Just because u have gay parents doesnt mean theyre bad lol XD
A thin-walled hollow sphere has a radius 4cm from the center
of the sphere, the eletric field points radially inward and
has a magnitude of 1.5 × 10^4NC^-1 How much charge is on the surface
A thin-walled hollow sphere has a radius of 4 cm from the center
of the sphere, the electric field points radially inward and
has a magnitude of 1.5 × 10⁴ N/C then the charge on the surface would be 2.6690×10⁻⁹ C
What is an electric charge?Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge.
The electric field inside a spherical shell is given by the formula
E = q/4πεr²
where q is the charge on the surface
r is the radius of the sphere
ε is the electrical permeability
By substituting the respective values in the given formula
1.5 × 10⁴ = q/4π(8.85✕ 10⁻⁻¹²)(.04²)
q = 2.6690×10⁻⁹ C
Thus, the charge on the surface would be2.6690×10⁻⁹ C
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The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to
PU is ... PQ
VU is ... 2VT
PR is ... PU.
VR is ... VS
VQ is ... VU
PR is ... PS
a.
i. PU is greater than PQ.ii. VU is Greater than 2VTb.
i. PR is Equal to PU.ii. VR is Equal to VSc.
i. VQ is Equal to VUii. PR is Greater than PS.What is pressure?Pressure is the force per unit area on a surface.
What is speed?Speed is the distance moved per unit time.
Pressure
Since pressure, P = hρg where
h = depth, ρ = density of liquid and g = acceleration due to gravity.Since ρ and g are constant
P ∝ h
So, we see that pressure is directly proportional to depth.
a. i. Pressure between R and USince U is lower than Q, Pressure at U is greater than pressure at Q.
So,PU is greater than PQ.
ii. Speed between U and TUsing the continuity equation
VUAU = VTAT where
VU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmVUT= speed at T, AT = cross-sectional area at T = π(dT)² where dT = diameter at T = 2.0 mmSo, VUAU = VTAT
VUπ(dU)² = VTπ(dT)²
VU = VT(dT)²/(dU)²
VU = VT(2.0)²/(1.0)²
VU = VT(4)
VU = 4VT
Since VU = 4VT,VU is Greater than 2VT
b i. Pressure between R and USince R is at the same depth as U, Pressure at R is equal to pressure at U.
So,PR is Equal to PU.
ii. Speed between R and SUsing the continuity equation
VRAR = VSAS where
VR = speed at R, AR = cross-sectional area at R = π(dR)² where dR = diameter at R = 2.0 mmVS= speed at S, AS = cross-sectional area at S = π(dS)² where dS = diameter at S = 2.0 mmSo, VRAR = VSAS
VRπ(dR)² = VSπ(dS)²
VR = VS(dS)²/(dS)²
VR = VS(2.0)²/(2.0)²
VR = VS(1)
VR = VS
Since VR = VS,VR is Equal to VS
c. i. Speed between Q and UUsing the continuity equation
VQAQ = VUAU where
VQ = speed at Q, AQ = cross-sectional area at Q = π(dQ)² where dQ = diameter at Q = 1.0 mmVU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmSo, VQAQ = VUAU
VQπ(dQ)² = VUπ(dU)²
VQ = VU(dU)²/(dQ)²
VQ = VU(1.0)²/(1.0)²
VQ = VU(1)
VQ = VU
Since VQ = VU, VQ is Equal to VU
Ii. Pressure between R and SSince R is lower than S, Pressure at R is greater than pressure at S.
So,PR is Greater than PS.
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Find the orbital speed of an ice cube in the rings of Saturn. The mass of Saturn is 5.68 x 1026 kg, and use an orbital radius of 3.00 x 105 km. (G = 6.67 × 10-11 N ∙ m2/kg2)
19.5 km/s
27.5 km/s
11.2 km/s
20.5 km/s
The orbital speed of an ice cube in the rings of Saturn is 355358.97m/s
Law of gravitationAccording to the gravitation law, the force of gravitation is directly proportional to the product of the masses and inversely proportional to the distance between them. Mathematically;
F = GMm/r²
where
m = mass of ice cube and
s = Gm1/r^2
Hence,
F = sm2
On rearranging,
s = m2/F
let V = orbital speed
centripetal acceleration = V^2/r
Such that;
V²/r = Gm/r²
V² = Gm/r
V = √Gm/r
Substitute the given parameters
V = √6.67×10^-11 * 5.68 x 10^26 / 3.00 x 10^5
V = 355358.97m/s
Hence the orbital speed of an ice cube in the rings of Saturn is 355358.97m/s
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Two uncharged spheres are separated by 1.70 m. If 2.40 ✕ 10¹² electrons are removed from one sphere and placed on the other, determine the magnitude of the Coulomb force (in N) on one of the spheres, treating the spheres as point charges.
_______N
**Hint** Find the net charge on each sphere and substitute values into Coulomb's law.
The magnitude of the Coulomb force (in N) on one of the spheres, given the data is 4.59×10⁻⁴ N
How to determine the charge on each spheresSphere 1 losses 2.40×10¹² electrons
But
1 electron = 1.6x10¯¹⁹ C
Thus,
Charge on sphere 1 = +1.6x10¯¹⁹ × 2.40×10¹² = +3.84×10¯⁷ C
Sphere 2 gains 2.40×10¹² electrons
But
1 electron = 1.6x10¯¹⁹ C
Thus,
Charge on sphere 2 = -1.6x10¯¹⁹ × 2.40×10¹² = -3.84×10¯⁷ C
How to determine the coulomb forceCharge on sphere 1 (q₁) = +3.84×10¯⁷ CCharge on sphere 2 (q₂) = 3.60 mC = -3.84×10¯⁷ CElectric constant (K) = 9×10⁹ Nm²/C²Distance apart (r) = 1.7 mForce (F) =?
Using the Coulomb's law equation, the force can be obtained as illustrated below:
F = Kq₁q₂ / r²
F = (9×10⁹ × 3.84×10¯⁷ × 3.84×10¯⁷) / (1.7)²
F = 4.59×10⁻⁴ N
Thus, the magnitude of the Coulomb's force is 4.59×10⁻⁴ N
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The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5 m. If the box has a mass of 1 kg, what is the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s?
The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
Acceleration of the box
The acceleration of the box is calculated as follows;
vf² = vi² + 2as
a = (vf² - vi²)/2s
a = (11.5² - 13²) / (2 x 8.5)
a = -2.16 m/s²
Time of motion of the boxThe time taken for the box to travel is calculated as follows;
a = (vf - vi)/t
t = (vf - vi) / a
t = (11.5 - 13) / (-2.16)
t = 0.69 s
Average power supplied by the frictionP = Fv
P = (ma)(vf - vi)
P = (1 x -2.16) x (11.5 - 13)
P = 3.24 W
Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
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a guage is connected to tank in which the pressure of the fluid is 305kpa above atmospheric. if the absolute presssure of the fluid remains unchanged but the gage is in chamber where the air pressure is reduce to a vaccum of 648mmHg what reading in psi will then be observed
The pressure reading in psi observed after the gauge is placed in the chamber with a vacuum of 648 mmHg will be approximately 31.67 psi.
When the gauge is placed in a chamber with a vacuum of 648 mmHg we need to convert the pressure reading to a compatible unit to determine the pressure measurement in PSI.
First, convert the initial pressure of the fluid from kPa to mmHg.
7.5 mmHg is equal to about 1 kPa. So, 305 kPa * 7.5 mmHg/kPa = 2287.5 mmHg is the initial pressure of the fluid in mmHg.
Reducing the original pressure by vacuum pressure.
1639.5 mmHg = new pressure = 2287.5 mmHg - 648 mmHg
Converting the mmHg of the new pressure to the psi unit.
0.01934 psi is equal to approximately 1 mmHg. As a result, the new pressure in psi is equal to 31.67 psi: 1639.5 mmHg * 0.01934 psi/mmHg
Therefore, the pressure reading in psi observed after the gauge is placed in the chamber with a vacuum of 648 mmHg will be approximately 31.67 psi.
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A -4.00 nC point charge is at the origin, and a second -6.00 nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x -axis at x = 0.200 m'
The net electric force that the two charges would exert on an electron placed at point on the x -axis is 1.68 x 10⁻¹⁶ N.
Force on electron due to charge 1
The force exerted on the electron due to the charge q1 placed at the origin is calculated as follows;
F = kq₀q₁/r²
where;
k is coulomb's constantq0 is charge at the originq1 is the charge at 0.2 m (electron)r is the distance between the chargesF(01) = (9 x 10⁹ x 4 x 10⁻⁹ x 1.6 x 10⁻¹⁹)/(0.2²)
F(01) = 1.44 x 10⁻¹⁶ N
Force on electron due to charge 2The force exerted on the electron due to the charge q1 placed at the origin is calculated as follows;
F = kq1q2/r²
where;
k is coulomb's constantq2 is charge at the 0.8 mq1 is the charge at 0.2 m (electron)r is the distance between the charges = 0.8 m - 0.2 m = 0.6 mF(12) = (9 x 10⁹ x 6 x 10⁻⁹ x 1.6 x 10⁻¹⁹)/(0.6²)
F(12) = 2.4 x 10⁻¹⁷ N
Net force on the electronF(net) = 2.4 x 10⁻¹⁷ N + 1.44 x 10⁻¹⁶ N
F(net) = 1.68 x 10⁻¹⁶ N
Thus, the net electric force that the two charges would exert on an electron placed at point on the x -axis is 1.68 x 10⁻¹⁶ N.
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