Answer:
4.44 ohms
Explanation:
To find the resistance of a circuit, you can use Ohm's law, which states that the resistance of a circuit is equal to the voltage across the circuit divided by the current through the circuit. So in this case, the resistance would be 12 volts / 2.7 amps = 4.44 ohms.
An astronaut's pack weighs 19.0 N when she is on earth but only 3.84 N when she is at the surface of moon. Part A What is the acceleration due to gravity on this moon? Express your answer with the appropriate units.
The weight of an astronaut's pack is 19.0 N on earth and 3.84 N on moon. The acceleration due to gravity on the moon is 1.98 m/s²
According to the Newton's second law of motion, the relation between force acted on an object with its acceleration is given by:
F = ma
Where:
F = force acted on the object
m = mass
a = acceleration
Weight is another name for the force due to gravity. If an object is brought from the earth to the moon, its weight will be different because the gravitational accelerations are different. However, its mass remains the same.
m = F / a
In case a = g = acceleration due to gravity:
m = w/g
Let
g = acceleration due to gravity on the earth = 9.8 m/s²
g₂ = acceleration due to gravity on the moon.
Hence,
19/g = 3.84/ g₂
g₂ = 3.84/19 x 9.8 = 1.98 m/s²
Hence, the acceleration due to gravity on the moon is 1.98 m/s²
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which of these variable stars would be classified as a rr lyrae? which of these variable stars would be classified as a rr lyrae? an f giant with a period of 14 hours a pulsar with a period of 0.14 seconds a k giant with a period of 14 days an m giant with a period of 140 days a b supergiant with a period of 0.14 days
An F giant with a period of 14 hours these variable stars would be classified as a RR Lyrae.
All the stars we see in the night sky are in our very own Milky way Galaxy. Our galaxy is referred to as the Milky way because it seems as a milky band of light inside the sky whilst you see it in a certainly darkish place. it's miles very difficult to be counted the range of stars in the Milky way from our position inside the galaxy.
The period of time at some stage in which an activity happens or a circumstance remains. it may be measured both in seconds or in thousands and thousands of years, depending upon the character of the pastime of situation being taken into consideration.
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the top of the saturated zone, as marked by the dashed line in the illustration, is called the water .
The dashed line in the figure denotes the water, which is at the top of the saturated zone. The underground region is included in the saturated zone.
The geologic medium's whole network of connecting apertures is fully submerged in water. This zone is frequently divided into the phreatic zone and the capillary fringe by hydrogeologists. The area of ground directly below the water table is known as the zone of saturation. Water has seeped into the pores and cracks of the earth and rocks. Over the water table, the saturated zone is more corrosive than the saturated zone. One extreme of the area's moisture content and the other extreme of the soil's dryness. a saturated area. The unsaturated zone can contain discrete saturated zones.
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The complete question is-
The top of the saturated zone, as marked by the dashed line in the illustration, is called ?
what is the period if a runner completes 5 laps around a circular track in 450s
Answer:
90 seconds to complete one lap around the track
Explanation:
To find the period of a runner's laps, we need to first find the length of time it takes for the runner to complete one lap around the track. We can do this by dividing the total time it takes to complete 5 laps by the number of laps:
period = 450 seconds / 5 laps
period = 90 seconds/lap
This means that it takes the runner 90 seconds to complete one lap around the track. This is the period of the runner's laps.
two wires cross at a 400 without electrical contact. find the magnitude and direction of the magnetic field at point p. point p is 10 cm from the wire intersection and equally distant from both wires.
Both wire carrying current in same direction with the same magnitude A current-carrying wire produces a magnetic field and the Bio-Savart law enables us to calculate the magnitude and direction of this magnetic field. at any point where the magnetic field due to the segment of current-carrying wire is given by equation
[tex]B=\frac{\mu_0}{2 \pi} \frac{l}{r}[/tex]
where r is the distance between the wire and the point and I is the current of the wire. At point , the distance between it and the intersection is . So, its distance from the horizontal wire is B
[tex]r_1=(4 \mathrm{~cm}) \sin 75^{\circ}=3.86 \mathrm{~cm}[/tex]
This the same distance from the second wire as we are given
[tex]r_1=r_2=3.86 \mathrm{~cm}[/tex]
if we apply the right-hand rule, we find that the magnetic field at point of each wire has direction out of the magnetic field at point is the summation of both magnetic fields
[tex]B_1=\frac{\mu_0}{2 \pi} \frac{1}{r_2}+\frac{\mu_0}{2 \pi} \frac{1}{r_2}=2\left(\frac{\mu_0}{2 \pi} \frac{1}{r_1}\right) \text { (2) }[/tex]
Now, we can use the values for to get the magnetic field at point by
[tex]B_1 & =2\left(\frac{\mu_0}{2 \pi} \frac{1}{r_1}\right) \\& =2\left(\frac{\left(4 \pi \times 10^{-7} T \cdot m / A\right)(5 \mathrm{~A})}{2 \pi\left(3.86 \times 10^{-2} \mathrm{~m}\right)}\right) \\& =5.2 \times 10^{-5} \mathrm{~T}\end{aligned}[/tex]
At point , the horizontal while applies a magnetic field with direction out of the page while the other wire applies a magnetic feild with direction into . As both wires exert the same magnetic field at point is zero
At point , the horizontal while applies a magnetic field with direction out of the page while the other wire applies a magnetic feild with direction into the page. As both wires exert the same magnetic field at point is zero.
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If 1 cm on the globe represents 555 km in the real world, how many kilometers would you have traveled in 16.8 cm? (round)
9324 km
On travelling 16.8 cm, the distance travelled in kilometres is 9324 kilometres.
As per the known fact, the ratio of map distance will be equal. So, we will write the two ratio and equate it. Let us assume the distance travelled in 16.8 cm be x. Representing the information of equation form -
1 : 555 = 16.8 : x
x = 16.8 × 555
Performing multiplication on Right Hand Side of the equation to find the value of x (Denominator of 1 can be ignored)
x = 9324 kilometres
Thus, the distance travelled in 9324 kilometres.
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during a knee extension exercise there are two forces that are producing torque at the knee joint. one is the quadriceps muscle and the other is the weight at the ankle joint. the quad is capable of producing 500 newtons of force, while the weight at the ankle is 10 kg. the distance from the knee joint to the quad force is 0.05 meters and the distance from the knee joint to the weight is 0.38 meters (these are not the moment arms). the knee is flexed to 125 degrees. draw and calculate the moment arms for each force on the picture below and decide if the joint flexes or extends.
The Moment arm for equal force is 0.0287m, and the torque is -7.041Nm.
What is force ?
Everyday activities like walking, setting something down on a surface, tossing something into the air, and even the tides' regular variations all include the application of force. The result of the interaction between two or more things, a force is a push or a pull.
What is torque ?
The force that may cause an item to revolve along an axis is measured by torque. Similar to how force drives linear kinematics acceleration, torque drives angular acceleration. A vector quantity is torque.
a1 = a cos 55 = (0.38m) cos 55 = 0.218 m
Moment arm for equal force = q1 = q cos 55
= (0.05) cos 55
= 0.0287m
Now Tnet Fq - qT/counter clockwise – Wa a3/ clockwise
Tnet = (500 N) (0.207m) – (98N) (0.218m)
= 14.35Nm – 21.364 Nm
= -7.041Nm
As net torque is negative is weight will pull down the log joint will flex not extend.
Therefore, Moment arm for equal force is 0.0287m, and the torque is -7.041Nm.
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consider an earthquake to be a three-dimensional wave traveling through the earth's crust. the intensity of the earthquake passing through the earth is measured to be 11000 j/(m2*sec) at a distance 35 km from the source. the earthquake lasts for 14 seconds. ef picup part description answer save status a. determine the intensity at 1.3 km from the source. (include units with answer) format check click here to check your answer 6.25 pts.75% 2% try penalty 1 hint available
The Energy that can be captured is 8.267*10^7 KwH, Energy produced by earth quake in 11 seconds is 1.86*10^15 J.
What is intensity?
Intensity is the amount of energy a wave carries over a surface of a particular area in a unit of time, and it is the same as the energy density times the wave speed. The most popular unit of measurement is watts per square meter. The strength and size of a wave affect its frequency.
What is earth quake ?
An earthquake is the abrupt release of energy from the earth's crust that causes the earth's surface to shake. Seismic waves, also called S waves, are consequently produced.
I = 11000 J/(m^2 s)
r = 35 km
t = 11 s
a) from the given data
Energy produced by earth quake in 11 seconds
E = I*A*time
= I*4*pi*r^2*t
= 11000*4*pi*(35*10^3)^2*11
= 1.86*10^15 J
using,
log(E) = 4.8 + 1.5*M
log(1.86*10^15) = 4.8 + 1.5*M
==> M = 6.98
b) Energy that can be captured, E_captured = 16% of E
= 0.16*1.86*10^15
= 2.976*10^14
= 2.976*10^14/(3.6*10^6) KwH
= 8.267*10^7 KwH
the number of houses the earthquake could power for a year, N = E/10000
= 8.267*10^7/10000
= 8267
Therefore, Energy that can be captured is 8.267*10^7 KwH, Energy produced by earth quake in 11 seconds is 1.86*10^15 J.
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A vertical plate is submerged in water and has the indicated shape. Find a Riemann sum that describes the hydrostatic force against one side of the plate. Then, write the sum as an integral and evaluate it.The pressure of a fluid that is exerted on an object due to its gravity is called hydrostatic pressure. Mathematically hydrostatic pressure is represented as P=rhogh. The hydrostatic force is calculated by multiplying the hydrostatic pressure with the area.
The hydrostatic force is calculated by multiplying the hydrostatic pressure with the area is 6000lb.
What is hydrostatic force ?
A hydrostatic force is the outcome of a liquid's pressure loading when it acts on submerged surfaces. Calculating the hydrostatic force and determining the Centre of pressure are two important concepts in fluid mechanics.
What is gravity ?
A planet or other body's gravity draws objects toward its centre. The force of gravity keeps all the planets in their orbits around the sun.
The area of the strip is calculated as follows
Astrip = 6dy
Pstrip = pgy
Here, the density of water is p, the acceleration due to gravity is g and the vertical height of the strip from free surface of water is y.
The weight density of water is
Pg= 62.5 lb/ft3
The hydrostatic force again one side of the plate is calculated as follows.
F hydrostatic = ∫ P strip A strip
Substitute all the values in the above equation and integrate.
F hydrostatic = ∫ (pgy) (6dy)
F hydrostatic = ∫6pg ydy
F hydrostatic = 6pg y²/2
F hydrostatic = 6pg (6²/2 – 2²/2)
F hydrostatic = 6 X 62.5 (6²/2 – 2²/2)
F hydrostatic= 6000 lb
Therefore, hydrostatic force is calculated by multiplying the hydrostatic pressure with the area is 6000lb.
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a 345 pf capacitor is charged to 145 v and then quickly connected to a 165 mh inductor. frequency of oscillation
A 165 mh inductor was attached right after that. The frequency of the oscillation is f=7957.74 Hz.
What is motion-induced oscillation?Periodic or oscillatory motion is defined as a motion that repeats itself. A restoring force or torque causes the object in this motion to oscillate about its equilibrium position.
For Class 8, what is an oscillation?Oscillation is a revolving motion between two states or locations. The side-to-side swing of a pendulum or the up-and-down motion of a spring with a weight are two examples of periodic motions that repeat themselves in a regular cycle and are considered oscillations.
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1. suppose a computer using direct mapped cache has 2^20 bytes of byte-addressable main memory, and a cache of 32 blocks, where each cache block contains 16 bytes.
Solution:
The block size = 16 words
Tag Block Offset
11 bits 5 bits 4 bits
The memory address 0DB63 will map to the 22nd block and 3 bytes of a block.
cache advantage is, when the cache block is made larger or bigger then, there are fewer misses. The disadvantage is that if the data is not used before the cache block is removed from the cache, then it is no longer useful.
After analyzing the question, the solution is:
As given The Cache of 32 blocks and the memory is word addressable:
The main memory size.
The block size.
The total number of blocks = main memory size/ block size:
Says it is 1 M words in size and it requires 20 bits. Now, from which 32 cache block requires 2^20, 20 bits, and block size means offset requires 2^4, 4 bits. Now the sizes of the bag, and, block and word fields are following:
Tag Block Offset
11 bits 5 bits 4 bits
The hexadecimal address value is 0DB63, its binary equivalent is
The tag = ( first 11 bits) = 0000 1101 101
block = ( next 5 bits ) = 10110
offset = ( next 4 bits ) = 0011
Therefore, the memory address 0DB63 will map to the 22nd block and 3 bytes of a block.
The main favored position or circumstance of the cache happens that, when the cache block is created best or considerably therefore, there happen hardly any misses. this takes place when the information in visible form in the block exists secondhand.
One of the loss exists that if the data exist not secondhand before the cache block exist detached from the cache, then it exist not any more valuable. in this place there exist an addition to the best miss punishment.
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which of the following is not a strong argument for the theory that some large elliptical galaxies formed as the result of galaxy collisions?
The majority of galaxies in the universe are elliptical galaxies, and galaxy collisions are frequent.
Mergers produce large elliptical galaxies. There is proof that galaxies develop through mergers and collisions. Star formation is triggered by galaxic collisions. All of the gas is transformed into stars before a disc can develop because the higher gas density produces stars more quickly. High redshift elliptical galaxies lack young, blue stars. Old red stars with erratic orbits in several planes are abundant in elliptical galaxies, which have a small amount of gas and dust. The largest galaxies that we are currently aware of are enormous elliptical galaxies. The majority of galaxies found now are small elliptical galaxies. Elliptical galaxies often have very little cold gas.
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complete question: Which of the following is not a strong argument for the theory that some large elliptical galaxies formed as the result of galaxy collisions?
A) Elliptical galaxies dominate the population in dense galaxy clusters.
B) Some ellipticals have stars and gas that rotate opposite to the rest of the galaxy.
C) Some elliptical galaxies are surrounded by shells of stars.
D) Computer simulations predict that the product of a galaxy collision is generally an elliptical galaxy.
E) Galaxy collisions are common and most galaxies in the universe are elliptical.
determine the displacement of a plate of length l pinned at its ends with a concentrated load vs applied at its center. This problem is illustrated in Figure 3–9.
The radius of gyration will be k = (h2/12)1/2 , or k2 = h2/12.
What is radius?
The radius of a circle is the distance a circle's center from any point along its perimeter. Usually, "R" or "r" is used to indicate it.
What is displacement ?
The term "displacement" describes the shortest path an object takes to travel from one spot to another. It is understood to be the modification of the object's position, to put it simply. Its magnitude and direction make it a vector quantity.
Radius of Gyration is given by, k = (I/M)1/2; Where I = Moment of Inertia of the body and M is its mass.
Now, for a plate of rectangular cross-section of height h and width b, the area moment of inertial about horizontal midline is given by, I = Mh2/12 .
Hence, the radius of gyration will be k = (h2/12)1/2 , or k2 = h2/12.
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How can a dilute solution of salt in water be made more concentrated?
Answer:
Heating it
Explanation:
Helps the salt dissolve faster and is more effective when heated up :)
The answer is A. by heating it
Consider the broadcasting circuit for an AM radio station which broadcasts at a frequency of 1400 kHz. The free electrons in such a circuit are moving back and forth in simple harmonic motion. (a) How long does it take for the free electrons in this circuit to go back and forth once? Give your answer in microseconds. μs (b) Assuming the average speed of the free electrons is 100 μm/s, what is the range of motion of the electrons as they go back and forth in the wires of the circuit. (HINT: As a free electron goes back and forth, it travels its full range every half cycle.) Give your answer in nanometers. nm (c) What is the wavelength of the electromagnetic radio waves emitted by this broadcasting circuit? Give your answer in meters. m
a) Time period is given by 0.751 b) Displacement is 0.0751 nm c) The wavelength of the electromagnetic radio waves emitted is 225.56 m
T = 1/f
T = 1/(1330*10^3 Hz)
T = 751.88 ns
Time period T = 0.751 us
b) displacement is given by:
x = velocity * Time
x = 100*10^-6 m/sec*0.751*10^-6 sec
x = 0.0751 nm
c) lambda = c/f
lambda = 3*10^8/(1330*10^3 Hz)
lambda = 225.56 m
Displacement, commonly known as length or distance and denoted by the letters d or s, is a one-dimensional variable that represents the separation between two defined points. In the International System of Units (SI), the meter is the common unit of displacement (m). Transferring unfavorable emotions from one thing or person to another is referred to as displacement and is a defensive tactic. To "take out" their rage on a family member, for instance, a person can yell at them if they are upset with their job. Replace, supersede, and supplant are some frequent alternatives to the word "displace."
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A tank whose bottom is a mirror is filled with water to a depth of 19.3 cm. A small fish floats motionless 7.50 cm under the surface of the water. What is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at normal incidence?
The fish's perceived depth is 5.3 cm, while the fish's picture is 24.8 cm. when observed at normal incidence, the perceived depth of a fish's reflect in the tank's bottom.
What do you deem an incident to be?
A "incidence" is the appearance of new cases of illness or injury within a population over a specific period of time. The amount of new cases in a population is how some public health experts define incidence, however
What distinguishes incidence from prevalence?
In contrast to prevalence, which is the percentage of a population with a particular characteristic, incidence is the number of new cases of a characteristic that appear in a population during a given time frame.
Briefing:
apparent depth=(Na/Nw)*s
=(1.00/1.33)*33cm
=24.8cm
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construction worker Bob IS standing at the top of a slanted rooftop when he accidentally drops & orange from his lunch bag; The orange rolls down the rool, which i> angled at 0 28,58 relative t0 Ihe harizontal as shown the figure below The bottom edge of the roofus meters above Ihe ground. Ihe orarge leaves the right edge ol Ihe rooftop with velocity 0l Ti 4.471,how fal, Arto the right of the roof edge does his 8 meter tall fellow consiruction warker Werdy have sland on the ground 50 that the orange just barely passes over her head? [Figure not diawn t0 scalel Imapa *x4: ALhae
Wendy must stand 1.19m to the left of the edge of the roof so that the orange just barely passes over her head.
Bob's orange will travel down the slanted roof with a velocity of 4.47 m/s, and it will hit the ground 3.0 m below the edge of the roof. The height of the orange above the ground at the edge of the roof is 1.8 m, and the angle of the roof is 28.5° relative to the horizontal.
To calculate the distance Ar that Wendy has to stand from the edge of the roof, we must first calculate the time it takes for the orange to reach the ground. Using the equation for the vertical displacement of an object with an initial velocity, we can solve for the time, t:
d = v*t - (1/2)*g*t^2
0 = 4.47*t - (1/2)*9.8*t^2
t = 0.938s
Next, we must calculate the horizontal displacement of the orange from the time, t. Using the equation for the horizontal displacement of an object with an initial velocity and a constant acceleration we can solve for the displacement:
x = v*t + (1/2)*a*t^2
x = 4.47*0.938 + (1/2)*0*0.938^2
x = 4.19m
To calculate the distance Ar that Wendy must stand from the edge of the roof, we must subtract the horizontal displacement of the orange from the height of the edge of the roof:
Ar = 3.0 - 4.19
Ar = -1.19m
Therefore, Wendy must stand 1.19m to the left of the edge of the roof so that the orange just barely passes over her head.
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Mary weighs 525 N and she walks down a flight of stairs to a level 6.5 m below her starting point. What is the change in Mary’s potential energy? Answer in units of J.
The change in Mary’s potential energy is 3412.5 Joule.
What is potential energy?
Potential energy is a form of stored energy that is dependent on the relationship between different system components. If a steel ball is raised above the ground as opposed to falling to the ground, it has more potential energy. It is capable of performing more work when raised.
Given that: Mary weighs 525 N and she walks down a flight of stairs to a level 6.5 m below her starting point.
Hence, change in potential energy is = weight × change in height
= 525 N × 6.5 m
= 3412.5 Joule.
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A 150-kg electromagnet is at rest and is holding 100 kg of scrap steel when the current is turned off dropping the steel. - 2- Knowing that the cable and the supporting crane have a total stiffness equivalent to a spring of constant k 60 kN/m and a damper c = 300 Ns/m, determine (a) the frequency of oscillation (b) the maximum upward displacement І В
The frequency of the oscillation is determined to be 0.123 Hz.
The number of waves passing through a fixed point in unit time is said to be frequency.
Given that, Mass of the scrap steel = 100 kg
Spring constant k = 60 kN/m
The formula for time period is as follows,
T = 2π √(m/k)
And the relation between time period and frequency is, T = 1/f
f = 1 / [2π √(m/k)] = 1/ [ 2π √(100/60) = 1/ [ 2π * 1.29] = 1/8.105 = 0.123 Hz
Thus, the frequency of oscillation is 0.123 Hz.
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difference between constant and
variable work
Work done on an object by a constant force is known as Constant Work, whereas the work done on an object by variable force is known as Variable Work.
In constant work or work done by constant force, the magnitude and direction of the force are constant or they do not change. Therefore, constant work is simply calculated by force acting on the object multiplied by displacement of the object.
But in case of variable work, things are not so easy. In this scenario, the force's magnitude and direction may alter at any point while the job is being done. The majority of the work we do on a daily basis is an instance of variable force work. The same calculation involves integration and is fairly difficult.
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(Figure 1) shows a thin liquid film bounded on the right side by a sliding wire that is attached to a spring with spring constant 0.50 N/m. The spring is stretched by 1.4 cm. Figure 1 of 1 < > 0.50 N/m 6.0 cm Mai W Part A What is the liquid's surface tension in mN/m? Express your answer in millinewtons per meter. || ΑΣφ ? 7= mN/m
The liquid's surface tension in mN/m is 58.33.
Surface tension is the tendency of the surface of a liquid at rest to contract into the smallest possible surface area. Surface tension allows objects denser than water, such as razor blades and insects, to float on the surface without being partially submerged.
Given:
Spring constant of the spring = k = 0.5 N/m
Distance by which the string is stretched = d = 1.4 cm = 0.014 m
Force of the spring on the wire = Fs
Fs = kd
Surface tension of the liquid = γ
Length of the sliding wire = L = 6 cm = 0.06 m
Surface tension force of the liquid on the wire = Ft
A liquid film has 2 surfaces therefore,
Ft = 2γL
The force of the spring on the wire and the surface tension force on the wire will be equal.
Fs = Ft
kd = 2γL
(0.5)(0.014) = 2γ(0.06)
0.007 = 0.12γ
γ = 0.05833 N/m
Converting the surface tension from Newton per meter to milli-Newton per meter, (1 N = 103 mN)
γ = (0.05833) x (103) mN/m
γ = 58.33 mN/m
Surface tension of the liquid = 58.33 mN/m
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Two pipes of identical diameter and material are connected in parallel. The length of pipe A is three times the length of pipe B. Assuming the flow is fully turbulent in both pipes and thus the friction factor is independent of the Reynolds number and disregarding minor losses, determine the ratio of the flow rates in the two pipes.
According to the question the ratio of the flow rates in the two pipes is [tex]\sqrt{3}[/tex] /1
What in physics is flow?Fluid Flow, a branch of fluid dynamics, is concerned with fluids. It involves the movement of a fluid under the influence of uneven forces. As soon as unbalanced pressures are applied, this motion will persist.
Briefing:Lets take
Length of pipe B = L
Length of pipe A = 3 L
Discharge in pipe A = Q₁
Discharge in pipe B = Q₂
[tex]h_{f}[/tex] = (FLQ²)/12.1.[tex]d^{5}[/tex]
F=Friction factor, Q=Discharge,L=length
d=Diameter of pipe
here all only Q and L is varying and all other quantity is constant
So we can say that
LQ²= Constant
L₁Q₁²=L₂Q²₂
By putting the values
3LQ₁²=LQ²₂
Therefore
[tex]Q_{1}[/tex]/ [tex]Q_{2}[/tex] =[tex]\sqrt{3}[/tex]/ 1
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the in situ moisture content of a soil is 15% with saturation level equal to 70% and void ratio of 0.9. the specific gravity of soil solids is 2.75. this soil is to be excavated and transported to a construction site for use in a compacted fill. if the specifications call for the soil to be compacted
This soil is to be excavated and transported to a construction site for use in a compacted Volume fill is 8680.73 m³
The specific gravity of solids = 2.75
situ moisture content of a soil is 15% with saturation level equal to 70% and void ratio of 0.9.
Dry weight = 103.5
Yd = 16.5/(1 + 18/100) = 14.34KN/m³
Volume of soil to be excavated to produce 7651m³
V = Yd compacted / Yd insite × compacted fill volume
V = 16.27/14.34 × 7651 = 8680.73 m³
No of trucks = V × Yinsity / 1.78 = 8680 × 16.5/178 = 826trucks.
Using the human body, such as by pinching or measuring an object's volume using the size of your hand, is the most traditional method of doing so. The human body, however, is incredibly unpredictable due to its differences. A more accurate technique to measure volume is to utilize fairly constant and long-lasting natural objects, like gourds, sheep or pig stomachs, and bladders. Small volumes are typically measured using standardized human-made containers today since metallurgy and glass manufacture have advanced through time. [3]: 393 By using a multiple or portion of the container, this approach is frequently used to measure small volumes of fluids or granular materials. To create a nearly flat surface, the container is shook or leveled off for granular items. The most exact approach to measure is not with this technique.
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For each of the following, create a DFA that recognizes exactly the language given. (a) [5 points) Binary strings with at least two ls. (b) [5 points) Binary strings that have at least one 1 and an even number of Os. (c) [5 points) Binary strings such that none of their runs of ls have odd length. A run of ls is a substring consisting of all 1s (i.e. "11..-1") that is either at the beginning of the string, or at the end of the string, or in the middle of the string, surrounded by Os. For example, the following strings are included in the language: "11","110","00001111", and "011001111011". But the following strings are not included in the language because they all contain at least one run of ls with odd length: "111", "010", and "00110111".
A binary string is a sequence of bytes. Unlike strings, which typically contain textual data, binary strings are used to contain non-traditional data such as images. The answer is in Images.
Binary string length is the number of bytes in the sequence. Binary strings have a CCSID of 65535. A binary string is a series of bytes. Unlike strings which usually contain text data binary strings are used to store data such as image sounds and mixed media.
A binary string is a series of octets. Binary strings are divided into two types of strings. First, binary strings specifically allow the storage of zero-valued octets and other non-printable octets. Each character in the string is defined by a number and that number is encoded in binary like an int. Contains a zero terminator.
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Figure shows three forces applied to a trunk that moves leftward by 3.00 m over a frictionless floor. The force magnitudes are F 1
​
=5.00 N,F 2
​
=9.00 N, and F 3
​
=3.00 N. and the indicated angle is θ=60.0 0
. During the displacement,
(a) what is the net work done on the trunk by the three forces?
(b) does the kinetic energy of the trunk increase or decrease?
The net work done on the trunk by the three forces is 1.50J and the kinetic energy of the trunk will increase
The data given in the question is:
F 1 =5.00N,
F2 =9.00N,
F3 =3.00N,
θ=60°
d=3m
Work done by each force can be found as
We know that, work done by a force is given by:
W =F d cosϕ,
where ϕ is the angle between Force and displacement vectors.
F₁ is in the direction of the displacement.
So,
W₁=F₁dcosϕ
=(5.00N)(3.00m)cos0°
=15.0J.
Force F₂ makes an angle of
ϕ ₂=180°−θ
=180° −60°
=120°
with the displacement i.e. in negative x direction.
So,
W₂=F₂dcosϕ₂
=(9.00N)(3.00m)cos120°
=−13.5J.
Force F₃ is perpendicular to the displacement.
So, W₃=F₃dcosϕ
=(3.00N)(3.00m) cos90°
=0J
Net Work done by three forces is
As work is a scalar quantity, so total work done will be the algebraic sum of all the individual works.
So, W=W₁+W₂+W₃
=15−13.5+0
=+1.50J.
According to Work energy theorem, total work done on a system equal the change in kinetic energy of the system.
Here, a positive work of 1.5 Joules is done on the trunk, so its Kinetic Energy will increase by 1.5 Joules.
Figure was missing it is attached with the answer
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a soccer player kicks a soccer ball of mass 0.45 kg that is initially at rest. the player's foot is in contact with the ball for 1.90 * 10^-3 s, and the force of the kick is given by f(t)
We set the derivative of F with respect to time equal to zero and solve for t since J=F avg t. The outcome is t=1.5103 s. It is clear that the force, F(t), reaches its maximum at t=0.0015s, with Fmax equal to 4500N.
How do you determine the force required to kick a ball?The formula Force (F) = Mass (M) x Acceleration (A), where the units are F (Newton) = Mass (kg) x Acceleration (m/s2), is used to determine the force of a kick.
What are the forces at work while kicking a soccer ball?The force exerted by the foot on the football during a kick is referred to as the action force. On the other side, the response force is the pressure the ball places on the foot. Because the action force and reaction force are of equal magnitude, no force is stronger than the other.
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Which of these is NOT a vector quantity?
O rotational kinetic energy
tO ranslational momentum
O velocity
O angular momentum
O acceleration
Answer: Rotational kinetic energy is not a vector quantity
Explanation:
Vector A volume that requires both magnitude and direction to be fully defined is called a vector.
Kinetic Energy It's defined as the energy held by a body when it's in stir,i.e haste isn't 0. When haste is 0, Kinetic Energy is basically 0.
Kinetic Energy is given by
K = 1 2 m v 2
where m = mass, v = haste.
Mass is a scalar volume and the haste which is a vector is squared which makes it a scalar volume as well. thus, Kinetic Energy is a scalar.
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spine in this situation. we will model the spine and upper body as a horizontal rigid rod of uniform density with a length of 50.0 cm and a mass of 40.0 kg. assume that the person attempts to lift an object with their arms, which we will model as attached at the far end of the rod. support of the back in this position is provided primarily by the erector spinalis muscle which we will model as being attached at one end to the spine at a point 33.0 cm from the hip at an angle of 10 degrees; the other end of the muscle is attached to the lower body below the hip.
The amount of tension in the back muscle is 3270 N.
Mass of object = 18.2 Kg
Mass of person = 40 Kg
Length = r = 0.5 m
Length at the spine at required position = r' = 0.33 m
Angle = θ = 10 degrees
Tension is defined as the force transmitted through a rope, string or wire when pulled by forces acting from opposite sides. The tension force is directed over the length of the wire and pulls energy equally on the bodies at the ends. Every physical object which is in contact exerts some force on one another.
Torque = Tension = T =
= T = r'sin10 -rmgsin90
= T = (0.5 X 40 X 9.8) / (0.33 X 0.1736)
= T = 3270 N
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The complete question:
We will model the spine and upper body as a horizontal rigid rod or uniform density with a length of 50.0 cm and a mass of 40.0 kg. Assume that the person attempts to lift an object with their arms, which we will model as attached at the far end of the rod. Support of the back in this position is provided primarily by the erector spinalis muscle which we will model as being attached at one end to the spine at a point 33.0 cm from the hip at an angle of 10 degrees; the other end of the muscle is attached to the lower body below the hip. The object being lifted has a mass of 18.2 kg. Calculate the tension , in the back muscle for this scenario.
when there is a net force exerted on an object, the magnitude of the acceleration of the object is [a] proportional to the magnitude of the net force
When there is a net force exerted on an object, the magnitude of the acceleration of the object is proportional to the magnitude of the net force and has a magnitude that is inversely proportional to the mass.--The statement is true.
Newton's second law of motion states that the acceleration of an object depends upon two variables i.e the net force which is acting upon the object and the mass of the object. The acceleration of an object is dependent directly upon the net force which is acting upon the object, and inversely upon the mass of the object. If the force acting upon an object is increased, the acceleration of the object is increased. If the mass of an object is increased, the acceleration of the object is decreased.
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The complete question is:
State whether a given statement is True or False.
When a net external force acts on an object of mass m, the acceleration that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass.
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What is the average speed of the dog if it’s goes 40 meter in 8 second