A process filling small bottles with baby formula has a target of 3 ounces ±0.10 ounces. In other words, a bottle must contain between 2.9 and 3.1 ounces of baby formula. As part of a periodic process capability study, two hundred bottles from the process were sampled. The results showed the average amount of formula placed in the bottles to be 3.042 ounces with a standard deviation of 0.033 ounces.
a. Would you consider this process capable of meeting the required specifications? Answer by calculating an appropriate measure of capability.
b. Assuming that the quantity of formula in a bottle is normally distributed, out of a batch of a million bottles how many do you think would be unacceptable – i.e., outside the specification limits?
c. Currently the process is not centered. If the process is adjusted so that it becomes centered, what is the impact on the process capability measure that you calculated in (a)? d. Assume that the process is centered. What should the process standard deviation be in order for the process to be considered Six-Sigma compliant (i.e., have Cp or Cpk of 2.0)?
Answers
a. This process is not capable of meeting the required specifications.
b. Less than 0.3% of the bottles in a batch of a million would be outside the specification limits.
c. The process standard deviation must be 0.0054 ounces or less for the process to be considered Six-Sigma compliant
a. To determine if the process is capable of meeting the required specifications, we can calculate the process capability index, Cp, using the formula:
Cp = (USL - LSL) / (6 * standard deviation)
where USL is the upper specification limit, LSL is the lower specification limit, and standard deviation is the sample standard deviation.
In this case, USL = 3.1 ounces, LSL = 2.9 ounces, and standard deviation = 0.033 ounces. Therefore,
Cp = (3.1 - 2.9) / (6 * 0.033) = 0.505
Since Cp < 1, the process is not considered capable of meeting the required specifications.
b. Assuming the quantity of formula in a bottle is normally distributed, we can use the empirical rule to estimate the percentage of bottles outside the specification limits. The empirical rule states that for a normal distribution, approximately:
68% of the data falls within 1 standard deviation of the mean
95% of the data falls within 2 standard deviations of the mean
99.7% of the data falls within 3 standard deviations of the mean
Using the mean of 3.042 ounces and the standard deviation of 0.033 ounces, we can calculate the number of standard deviations away from the mean that correspond to the specification limits:
Upper specification limit: (3.1 - 3.042) / 0.033 = 1.76 standard deviations above the mean
Lower specification limit: (2.9 - 3.042) / 0.033 = 4.30 standard deviations below the mean
Therefore, using the empirical rule, we can estimate that less than 0.3% of the bottles in a batch of a million would be outside the specification limits.
c. If the process is adjusted so that it becomes centered, the process capability measure Cp will increase. The new Cp can be calculated using the same formula as in part (a), but with the mean shifted to the center of the specification limits, which is (2.9 + 3.1) / 2 = 3.
New Cp = (3.1 - 2.9) / (6 * 0.033) = 0.759
d. In order for the process to be considered Six-Sigma compliant, it must have a Cp or Cpk of 2.0 or higher. If the process is centered at 3 ounces, then the formula for Cpk is:
Cpk = min[(USL - mean) / 3 sigma, (mean - LSL) / 3 sigma]
where sigma is the process standard deviation. We want to find the value of sigma that gives Cpk = 2.0.
Substituting the values of USL, LSL, and mean, we get:
2.0 = min[(3.1 - 3) / (3 * sigma), (3 - 2.9) / (3 * sigma)]
Simplifying, we get:
2.0 = min[0.033 / (3 * sigma), 0.033 / (3 * sigma - 0.1)]
Since the minimum value occurs when the denominator of the second term is smallest, we can solve for sigma as:
0.033 / (3 * sigma - 0.1) = 2.0
Simplifying, we get:
sigma = 0.0054 ounces
Therefore, the process standard deviation must be 0.0054 ounces or less for the process to be considered Six-Sigma compliant.
Learn more about sigma compliant:
brainly.com/question/28659811
#SPJ11
Related Questions
a gardener mounts a pot plant on the wall using a metal ring and a chain the ring exerts a horizontal force of 20n on the pot plant and the weight of the pot plant is 30n determine the magnitude of the chain by calculation and construction
Answers
The magnitude of the force exerted by the chain is 36.1 N.
Resultant forcesThe magnitude of the force exerted by the chain must balance the force exerted by the ring and weight of the pot.
The magnitude of the force exerted by the chain is calculated by applying Pythagoras theorem;
\(F_3^2 = F_1^2 + F_2^2\\\\F_3 = \sqrt{F_1^2 + F_2^2} \\\\F_3 = \sqrt{20^2 + 30^2} \\\\F_3 = 36.1 \ N\)
Thus, the magnitude of the force exerted by the chain is 36.1 N.
Learn more about resultant force here: https://brainly.com/question/25239010
how is charged particles related to electric current, electric circuits, and resistance
Answers
Charged particles are fundamental to the behavior of electric currents, electric circuits, and resistance. An electric current is the flow of charged particles, typically electrons, through a conductor.
The flow of charged particles generates an electric field that induces a potential difference, or voltage, across the conductor.Electric circuits are constructed by connecting conductors and electrical components, such as resistors, capacitors, and inductors, in a specific configuration. The arrangement of the components determines how the current flows through the circuit.
The flow of current through the circuit depends on the resistance offered by the components in the circuit and the potential difference across the circuit.Resistance is the property of a conductor that opposes the flow of current. The resistance of a conductor is proportional to the number of charged particles in the conductor, the length of the conductor, and the cross-sectional area of the conductor. The resistance can also be affected by the temperature of the conductor and its material properties.
In summary, charged particles are responsible for generating electric currents that flow through electrical circuits. The behavior of the currents is determined by the arrangement of the components in the circuit and the resistance offered by the conductors and components. Resistance is a fundamental property of a conductor that opposes the flow of charged particles and can be affected by various factors.
For more such questions on resistance visit:
https://brainly.com/question/7669692
#SPJ11
?course_assessment_id=_192597_ Remaining Time: 23 minutes, 45 seconds. * Question Completion Status: Moving to the next question prevents changes to this answ
Answers
The relationship between the speed of light and the refractive index of a medium is essential in understanding the propagation of light in different materials.
The refractive index (n) of a medium is defined as the ratio of the speed of light in vacuum (c) to the speed of light in the medium (v). Mathematically, n = c/v. When light passes through a medium, it slows down due to interactions with atoms or molecules in the material, resulting in a decrease in speed compared to its velocity in a vacuum. The refractive index determines how much light is bent or refracted as it enters a different medium, impacting phenomena like refraction, reflection, and dispersion. This relationship plays a crucial role in various applications, such as optics, telecommunications, and the study of wave behavior in different media.
To know more about refractive index, here
brainly.com/question/30761100
#SPJ4
--The complete Question is, What is the relationship between the speed of light and the refractive index of a medium, and how does it affect the propagation of light in different materials?--
dry suits become almost essential in water temperatures below
Answers
Dry suits become almost essential in water temperatures below approximately 50 degrees Fahrenheit (10 degrees Celsius). Below this temperature, the risk of hypothermia and cold-water shock increases significantly, making it dangerous to enter the water without adequate protection.
The primary function of dry suits is to provide comprehensive insulation and shield the wearer from water exposure. Unlike wetsuits, which allow a small amount of water to enter and then retain and warm it against the body, dry suits are completely sealed to prevent water from penetrating. This ensures the wearer stays dry and creates a layer of air between the body and the suit, which acts as insulation.In colder water temperatures, the body loses heat at an accelerated rate, increasing the likelihood of rapid heat loss and hypothermia upon immersion. By wearing a dry suit, the risk is minimized as it offers thermal protection and prevents direct contact between the body and the cold water.
However, it's crucial to understand that relying solely on a dry suit may not guarantee safety in extremely cold water. Additional precautions include proper insulation underneath the dry suit, appropriate safety gear, and familiarity with cold-water immersion techniques. Additionally, obtaining training and experience in cold-water environments is highly recommended to ensure personal safety.
Remember to seek guidance from local experts, such as diving instructors or experienced individuals familiar with cold-water conditions, as they can provide specific advice based on the local environment and your intended activities in cold water.
To learn more about temperatures visit: https://brainly.com/question/27944554
#SPJ11
Charge q = + 15 nC is uniformly distributed on a spherical shell that has a radius of 120 mm.
1. What is the magnitude of the electric field just outside the shell? (Express your answer with the appropriate units.)
2. What is the magnitude of the electric field just inside the shell? (Express your answer with the appropriate units.)
3. What is the direction of the electric field just outside and just inside the shell?
a. radially inward just outside the shell, radially outward just inside the shell
b. radially inward just outside the shell, zero just inside the shell
c. radially outward just outside the shell, radially inward just inside the shell
d. zero just outside the shell, radially inward just inside the shell
e. zero just outside the shell, radially outward just inside the shell
f. radially outward just outside the shell, zero just inside the shell
4. What is the electrostatic potential just outside the shell, relative to zero at infinity? (Express your answer with the appropriate units.)
5. What is the electrostatic potential just inside the shell, relative to zero at infinity? (Express your answer with the appropriate units.)
6. What is the electrostatic potential at the shell's center? (Express your answer with the appropriate units.)
7. What is the electric field magnitude at the shell's center? (Express your answer with the appropriate units.)
Answers
1.The magnitude of the electric field just outside the shell is approximately 46,875 N/C.
2.The electric field just inside the shell is zero.
3.The correct option is (c)
4.The electrostatic potential just outside the shell is V = (8.99 x 10^9 N*m^2/C^2) * (15 x 10^-9 C) / (0.12 m) ≈ 1.12 x 10^6 V.
5.The electrostatic potential just inside the shell is the same as the electrostatic potential just outside the shell.
6.The electric potential at the center of the shell is V = (8.99 x 10^9 N*m^2/C^2) * (15 x 10^-9 C) / (0.12 m) ≈ 1.12 x 10^7 V.
7.The electric field magnitude at the shell's center is zero .
1. The electric field just outside the shell can be calculated using the formula for electric field due to a uniformly charged sphere:
E = kq/r^2
where k is the Coulomb's constant, q is the charge on the sphere, and r is the distance from the center of the sphere to the point where the electric field is being calculated.
Here, q = +15 nC, r = 120 mm = 0.12 m, and k ≈ 8.99 x 10^9 N*m^2/C^2. Thus, substituting these values, we get:
E = (8.99 x 10^9 N*m^2/C^2) * (15 x 10^-9 C) / (0.12 m)^2 ≈ 46,875 N/C
Therefore, the magnitude of the electric field just outside the shell is approximately 46,875 N/C.
2. The electric field just inside the shell is zero because the net electric field at any point inside a uniformly charged spherical shell is zero due to the symmetry of the distribution.
3. The direction of the electric field just outside the shell is radially outward, and the direction of the electric field just inside the shell is radially inward. The correct option is (c).
4. The electrostatic potential just outside the shell can be calculated using the formula V = kq/r, where V is the electric potential at a distance r from the center of the sphere. Therefore, the electrostatic potential just outside the shell is V = (8.99 x 10^9 N*m^2/C^2) * (15 x 10^-9 C) / (0.12 m) ≈ 1.12 x 10^6 V.
5. The electrostatic potential just inside the shell is the same as the electrostatic potential just outside the shell, since the potential at any point just inside or on the surface of a charged spherical shell is constant.
6. The electric potential at the shell's center can be calculated using the formula V = kq/R, where R is the radius of the spherical shell. Therefore, the electric potential at the center of the shell is V = (8.99 x 10^9 N*m^2/C^2) * (15 x 10^-9 C) / (0.12 m) ≈ 1.12 x 10^7 V.
7. The electric field magnitude at the shell's center is zero since the net electric field at the center of a uniformly charged spherical shell is zero due to the symmetry of the distribution.
For such more questions on magnitude
https://brainly.com/question/24468862
#SPJ11
At what point in this sequence of events is potential energy
being converted to kinetic energy?
A. When the spring is recoiling B. When the spring is fully stretched. C. When the spring is fully recoiled. D. When the spring is stretching
Answers
Answer:
your answer is C
Explanation:
the spring when fully recoiled is at maximum PE and will then convert to KE
The total magnification of an image is determined by.
Answers
Answer:
Magnification of the objective lens used and the magnification of the ocular lens.
Explanation: I hope you have/had an amazing day today<3
For an Ultrasonic Vaporizer Us Duty Rate 4.2%, valued in USD$750, transported via Air and no valid CUSMA certificate available calculate: - Duty Charge - MPF - HMF
2. Same Ultrasonic Vaporizer Us Duty Rate 4.2%, valued in USD$750, transported via Ocean and no valid CUSMA certificate available calculate:
Duty Charge
MPF
HMF
Answers
1. The duty charge, MPF and HMF for the Ultrasonic Vaporizer transported via air are USD$31.50, USD$2.60 and USD$0.94, respectively.
2. The duty charge, MPF and HMF for the Ultrasonic Vaporizer transported via ocean are USD$31.50, USD$0.94 and USD$0.94, respectively.
How to calculate duty charge, MPF and HMFFor the Ultrasonic Vaporizer transported via air
Duty Charge:
Given information;
Duty rate = 4.2%
Value of goods = USD$750
Duty charge = Value of goods x Duty rate
Duty charge = USD$750 x 4.2%
Duty charge = USD$31.50
Merchandise Processing Fee (MPF)
MPF rate = 0.3464% (as of 2021)
Value of goods = USD$750
MPF = Value of goods x MPF rate
MPF = USD$750 x 0.3464%
MPF = USD$2.60
Harbor Maintenance Fee (HMF)
HMF rate = 0.125% (as of 2021)
Value of goods = USD$750
HMF = Value of goods x HMF rate
HMF = USD$750 x 0.125%
HMF = USD$0.94
Therefore, the total import fees for the Ultrasonic Vaporizer transported via air are:
Duty charge = USD$31.50
MPF = USD$2.60
HMF = USD$0.94
For the Ultrasonic Vaporizer transported via ocean
Duty Charge:
Given information:
Duty rate = 4.2%
Value of goods = USD$750
Duty charge = Value of goods x Duty rate
Duty charge = USD$750 x 4.2%
Duty charge = USD$31.50
Merchandise Processing Fee (MPF)
MPF rate = 0.125% (as of 2021)
Value of goods = USD$750
MPF = Value of goods x MPF rate
MPF = USD$750 x 0.125%
MPF = USD$0.94
Harbor Maintenance Fee (HMF):
HMF rate = 0.125% (as of 2021)
Value of goods = USD$750
HMF = Value of goods x HMF rate
HMF = USD$750 x 0.125%
HMF = USD$0.94
Therefore, the total import fees for the Ultrasonic Vaporizer transported via ocean are:
Duty charge = USD$31.50
MPF = USD$0.94
HMF = USD$0.94
Learn more on Duty charge on https://brainly.com/question/29786125
#SPJ4
Duty Charge=$31.50
The MPF will be $27.23.
The HMF rate is $0.125 per metric ton.
Ultrasonic Vaporizer transported via Air:
Duty Charge: $31.50
MPF: $26.79
HMF: Not applicable
Ultrasonic Vaporizer transported via Ocean:
Duty Charge: $31.50
MPF: $27.23
HMF: HMF rate and weight of the vaporizer are needed to calculate the HMF.
Ultrasonic Vaporizer transported via Air:
Duty Charge: The duty charge is calculated by multiplying the value of the product ($750) by the duty rate (4.2%): Duty Charge = \(\$750 * 4.2\%\)
= \(\$31.50.\)
MPF: The MPF for air transport is based on the value of the merchandise. As the value of $750 is below the maximum, the MPF will be $26.79.
HMF: The HMF is not applicable for air transport.
Ultrasonic Vaporizer transported via Ocean:
Duty Charge: The duty charge is calculated in the same way as for air transport: Duty Charge = \(\$750 * 4.2\%\)
= \(\$31.50\).
MPF: The MPF for ocean transport is also based on the value of the merchandise. As the value of $750 is below the maximum, the MPF will be $27.23.
HMF: The HMF is applicable for ocean transport. The HMF rate is $0.125 per metric ton. The weight of the Ultrasonic Vaporizer is needed to calculate the HMF.
Learn more about Duty Charge from the given link:
https://brainly.com/question/24128124
#SPJ11.
In a model AC generator, a 502 turn rectangular coil 8.0 cm by 23 cm rotates at 120 rev/min in a uniform magnetic field of 0.61 T. (a) What is the maximum emf induced in the coil
Answers
Answer:
9896
Explanation:
multiply them all together
Energy cannot be created or destroyed only transferred is the law of conservation of
Answers
The electric potential at a point in space is 235V. If a 0.0 485C charge is placed there what will its potential energy u be
Answers
Answer:
PE (Energy) = V / Q * Q where V is the electric potential at that point ( the energy required to bring a charge of 1 C to that point)
PE = 235 J/C * .0485 C = 11.4 J
a force acts on a 4.8 kg mobile object that moves from an initial position of to a final position in 4.30s find the work done on the object
Answers
The work done on the 4.8 kg mobile object by the force acting on it is 350 J.
The work done on a 4.8 kg mobile object by a force acting on it, which moves from an initial position to a final position in 4.30 s, needs to be calculated.
The work done on an object is equal to the force applied to it multiplied by the distance it moves in the direction of the force. The formula for work is W = Fd, where W is work, F is force, and d is distance. If the force is constant, the work done can be calculated as W = Fdcosθ, where θ is the angle between the force and the direction of motion.
In this case, the force and the distance are not given, but the time taken to travel the distance is given. However, we can use the formula for average velocity to find the distance. The formula for average velocity is v = Δd/Δt, where v is velocity, Δd is the change in distance, and Δt is the change in time.
We can rearrange this formula to find the distance traveled: Δd = vΔt. Since the initial velocity is zero, the final velocity is equal to the average velocity. Therefore, the distance traveled is given by Δd = (vf+vi)/2 * t, where vf is the final velocity and vi is the initial velocity.
Next, we need to find the force applied to the object. We can use the formula for acceleration to find the force. The formula for acceleration is a = F/m, where a is acceleration, F is force, and m is mass. Rearranging this formula, we get F = ma.
We can use the formula for average velocity to find the final velocity. The formula for average velocity is v = Δd/Δt, where v is velocity, Δd is the change in distance, and Δt is the change in time. We can rearrange this formula to find the final velocity: vf = Δd/Δt.
Given: m = 4.8 kg, t = 4.30 s
Assume initial velocity, vi = 0 m/s
Assume final position, xf = 25.0 m
Using v = Δd/Δt, we can find the average velocity, vave:
vave = (xf - xi) / t = (25 - 0) / 4.30 = 5.81 m/s
Using vf = (vi + vave) / 2, we can find the final velocity, vf:
vf = (0 + 5.81) / 2 = 2.91 m/s
Using F = ma, we can find the force, F:
F = ma = (4.8 kg) * (2.91 m/s²) = 14 N
Using W = Fd, we can find the work done on the object:
W = Fdcosθ = Fdcos0 = Fd = (14 N) * (25.0 m) = 350 J
Therefore, the work done on the 4.8 kg mobile object by the force acting on it is 350 J.
To know more about force refer here:
https://brainly.com/question/24115409#
#SPJ11
9.6 a 2-m long center-fed dipole antenna operates in the am broadcast band at 1 mhz. the dipole is made of copper wire with a radius of 1 mm. (a) determine the radiation efficiency of the antenna
Answers
The radiation efficiency of the antenna is: 11.45Ω
The radiation efficiency of a 2-m long center-fed dipole antenna operating in the AM broadcast band at 1 MHz, made of copper wire with a radius of 1 mm, can be determined by calculating the antenna's radiation resistance, or Rrad. Rrad is calculated by dividing the radiation power by the square of the voltage at the antenna terminals.
Once Rrad is known, the radiation efficiency is simply the ratio of Rrad to the antenna's total resistance. Calculating Rrad of a dipole antenna requires knowledge of the antenna's reactance, which is a function of the antenna's length and frequency. A dipole antenna of length l will have a reactance of Xd = -j75Ω/l, where l is measured in meters.
In this case, the reactance is Xd = -j37.5Ω. Once the reactance is known, Rrad is calculated using the following formula: Rrad = [75 - (Xd)^2]/[75 + (Xd)^2]. Plugging in the values from the problem, the radiation resistance is Rrad = 7.86Ω.
The total resistance of a dipole antenna is calculated using the formula Rt = Rr + Xd^2/Rr. This equation yields a total resistance of Rt = 11.45Ω. Dividing Rrad by Rt yields a radiation efficiency of 0.69 for this antenna.
To know more about radiation refer here:
https://brainly.com/question/13934832#
#SPJ11
Explanation of the physics involved i the northern lights(aurora borealis) why can't it occur in mire southerns latitudes?
Answers
Answer:
The northern lights come from the interaction of electrically charged particles given off by the Sun surface during solar storms and the Earth's magnetic field which is pointed in the northern to southern poles of the Earth. Therefore, the charged particles that travel through millions of miles from the Sun to the Earth accelerate towards the magnetic poles and appear as the northern (and southern) lights
Boreal is an adjective word for the north or northern regions. In the southern latitudes, in the southern hemisphere, the phenomenon is called aurora australis (australis is the adjective for 'of the south')
Explanation:
A soil with modest CEC of 15 cmol(+)/kg soil has a pH of 6.0. If you took 100 pounds of soil and mixed it with 100 pounds of KCl, the resulting soil would: have a pH near 7.0 have a pH less than 6 have a pH remaining at 6.0 be a moderately alkaline soil (pH 7.5 to 8.5) be calcareous
Answers
When you mix 100 pounds of soil with 100 pounds of KCl, the resulting soil would have a pH near 7.0.
This is because KCl is a neutral salt and does not affect the pH of the soil. The modest CEC of the soil means that it has a relatively low capacity to retain positively charged ions such as K+. Therefore, adding KCl would not significantly affect the pH of the soil. The resulting soil would have a similar pH to the original soil, but it may have a slightly higher pH due to the dilution effect of adding more material. However, the pH increase would be minimal and would not make the soil moderately alkaline (pH 7.5 to 8.5) or calcareous.
More on pH: https://brainly.com/question/491373
#SPJ11
If a train going 60 m/s hits the brakes, and it takes the train 1 minute 30 seconds to stop, what is
the train's acceleration?
Answers
Explanation:
u=60
t=90s
a=
v=0
v=u+at
0=60+90a
90a=-60
a=-3/2
Table salt is made up of one sodium atom and one chlorine atom. Is table salt an element
Answers
An enemy ship is on the east side of a mountain island, as shown in the figure. The enemy ship has maneuvered to within d1= 2156 m of the h = 1840 m high mountain peak and can shoot projectiles with an initial speed of vi= 240 m/s. If the western shoreline is horizontally d2 = 245 m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?
Answers
If d2 ≤ 4593.6 m, the safe distance is d2.
If d2 > 4593.6 m, the safe distance is d2 - 4593.6 m.
Depending on the specific value of d2, you can calculate the corresponding safe distance using the above conditions.
To determine the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship, we need to consider the projectile motion and the range of the projectiles.
Given:
Distance from the enemy ship to the mountain peak: d1 = 2156 m
Height of the mountain peak: h = 1840 m
The initial speed of the projectiles: vi = 240 m/s
Distance from the western shoreline to the mountain peak: d2 = 245 m
First, we can calculate the time it takes for a projectile to reach the mountain peak at height h. We can use the equation of motion for vertical displacement:
h = vit + (1/2)gt^2,
where h is the vertical displacement (h = 1840 m), vi is the initial vertical velocity (0 m/s since the projectile is launched horizontally), g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time of flight.
Plugging in the values, we have:
1840 = 0 + (1/2)(9.8)t^2.
Simplifying the equation, we find:
4.9t^2 = 1840.
Solving for t, we get:
t = sqrt(1840 / 4.9) ≈ 19.14 s.
Now, we can calculate the horizontal distance covered by the projectiles during this time. Since the initial horizontal velocity (vi) remains constant, the horizontal distance is given by:
d = vi * t.
Substituting the values, we have:
d = 240 m/s * 19.14 s ≈ 4593.6 m.
Therefore, the projectiles fired from the enemy ship can reach a horizontal distance of approximately 4593.6 m.
To find safe distances from the western shore, we need to consider the following scenarios:
If the distance from the western shoreline to the mountain peak (d2) is less than 4593.6 m, any ship at or beyond the western shoreline will be safe from the bombardment. In this case, the safe distance is given by d2 itself.
If the distance from the western shoreline to the mountain peak (d2) is greater than 4593.6 m, the safe distance will be the difference between d2 and 4593.6 m. This ensures that the ship is outside the range of the projectiles.
In summary:
If d2 ≤ 4593.6 m, the safe distance is d2.
If d2 > 4593.6 m, the safe distance is d2 - 4593.6 m.
Therefore, depending on the specific value of d2, you can calculate the corresponding safe distance using the above conditions.
To learn more about projectile motion visit: https://brainly.com/question/24216590
#SPJ11
I need help with some graph.
Which graph shows acceleration?
Answers
Answer:
The first graph
Explanation:
Graph A shows acceleration.
I just know
A depiction of a famous scientific experiment is given. Consider how the beam changes when the magnet is off compared to when the magnet is on.
Answers
The beam used is a negatively charged electron beam with a velocity of
v = E / B where E is the electric field and B is a magnetic field.
Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced
Fe = Fm
q * E = q * v * B
v = E / B
this configuration is called speed selector.
The beam used is a negatively charged electron beam with a velocity of v = E / B.
For more information on magnets click on the link below:
https://brainly.com/question/14997726
#SPJ4
Although part of the question is missing the complete question is A depiction of a famous scientific experiment is given. Consider how the beam changes when the magnet is off compared to when the magnet is on. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned off. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned off, the beam path bends toward the positively charged plate and ends at the lower half of the wide end of the tube. A bell-shaped evacuated glass tube with a narrow end and a wide end is connected to a battery at the narrow end. In the center of the tube there is a negatively charged plate above the tube, a positively charged plate below the tube, and a magnet with the field turned n. A beam originating at the narrow end of the tube travels toward the wide end of the tube. With the magnetic field turned on, the beam path travels in a straight path to the center of the wide end of the tube. What type of beam was used in this experiment?
Consider a cylindrical specimen of a steel alloy 10.0 mm (0.39 in.) in diameter and 75 mm (3.0 in.) long that is pulled in tension. Determine its elongation when a load of 20,000 N (4,500 lbf) is applied.
Answers
the elongation of the cylindrical steel alloy specimen with a diameter of 10.0 mm and length of 75 mm, when subjected to a tension load of 20,000 N, is approximately 0.095 mm.
To determine the elongation of a cylindrical steel alloy specimen when a load is applied, we can use the formula:
Elongation (ΔL) = (Load (F) × Length (L₀)) / (Area (A) × Young's Modulus (E))
First, let's find the cross-sectional area of the cylindrical specimen:
A = π × (diameter / 2)²
A = π × (10.0 mm / 2)²
A = 78.54 mm²
Next, we need the Young's Modulus (E) of the steel alloy. This value is typically provided, but for this example, let's assume E = 200 GPa (200 x 10^3 MPa).
Now we can calculate the elongation:
ΔL = (20,000 N × 75 mm) / (78.54 mm² × 200,000 MPa)
ΔL = (1,500,000 N·mm) / (15,708,000 N)
ΔL ≈ 0.095 mm
So, the elongation of the cylindrical steel alloy specimen with a diameter of 10.0 mm and length of 75 mm, when subjected to a tension load of 20,000 N, is approximately 0.095 mm.
learn more about cylindrical here
https://brainly.com/question/28043239
#SPJ11
You are driving down the highway and a bug splatters on your windshield.
Which is greater: the force of the bug on the windshield, or the force of the windshield on the bug?
Answers
The force of the windshield on the bug is greater than the force of the bug on the windshield. This is due to Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
According to Newton's third law of motion, when two objects interact, they exert equal and opposite forces on each other. In the scenario of a bug splattering on a windshield, the bug exerts a force on the windshield upon impact. However, the windshield also exerts an equal and opposite force on the bug .The force of the windshield on the bug is greater because of the relative masses involved. The windshield is much larger and heavier compared to the bug, so it can exert a greater force.
However, the windshield remains intact due to its structural strength. Overall, while both the bug and the windshield experience forces during the collision, the force exerted by the windshield on the bug is greater .
Learn more about Newton's third law click here: brainly.com/question/974124
#SPJ11
if the rocket has mass 6.567 kg, if the circle has radius 15 km, and if the speed of the rocket is 5000 m/s in its path along the circle, then the force of the engines on the rocket (the thrust) is
Answers
The force of the engines on the rocket (thrust) is approximately 1.963 × 10^6 N.
The force of the engines on the rocket (thrust) can be calculated using the centripetal force equation. The thrust is found to be approximately 1.963 × 10^6 N.
In order to maintain a circular path, the rocket experiences a centripetal force directed towards the center of the circle. This force is provided by the engines and is equal to the product of the rocket's mass and the centripetal acceleration.
The centripetal acceleration can be calculated using the equation:
a = v^2 / r
where v is the velocity of the rocket and r is the radius of the circular path. Substituting the given values, we have:
a = (5000 m/s)^2 / 15000 m = 1666.67 m/s^2
The centripetal force is then given by:
F = m * a
Substituting the mass of the rocket (6.567 kg) and the calculated acceleration, we find:
F = (6.567 kg) * (1666.67 m/s^2) = 1.963 × 10^6 N
Therefore, the force of the engines on the rocket (thrust) is approximately 1.963 × 10^6 N.
Learn more about thrust : brainly.com/question/20211023
#SPJ11
A 1100 kg car accelerates from 48.0 km/hr to 59.0 km/he 100 m. What was the magnitude of the net force acting
on the car?
Answers
Answer:
Explanation:
Given:
m = 1100 kg
V₁ = 48.8 km / h = 48.8*1000 m / 3600 s ≈ 13.6 m/s
V₂ = 59.0 km / h = 59.0*1000 m / 3600 s ≈ 16.4 m/s
S = 100 m
_______________
F - ?
Acceleration:
S = (V₂² - V₁²) / (2*a)
a =(V₂² - V₁²) / (2*S) = (16.4² - 13.6²) / (2·100) ≈ 0.42 m/s²
Force:
F = m*a = 1100 * 0.42 = 462 N
Use dimensional analysis to determine how many centimeters are in 15 kilometers.A. 15,000 cmB. 1,500 cmC. 1,500,000 cmD. 150 cm
Answers
In order to find how many centimeters are in 15 kilometers, we can use the following conversion rate:
1 km = 100,000 cm.
So, for 15 kilometers, we have:
\(15\text{ km}=15\cdot(100,000\text{ cm})=(15\cdot100,000)\text{ cm}=1,500,000\text{ cm}\)Therefore the correct option is C.
5. Gravitational force gets weaker as ________ increases and gets stronger as the __________ of the object increases.
Answers
Answer:
Distance, Mass
hii please help i’ll give brainliest if you give a correct answer please
Answers
Answer:
they are equal and act in opposite directions
Explanation:
newton's third law of motion states that if object A exerts a force on object B, then object B will exert an equal but opposite in direction force.
hope it helps, I think my response was late,
Which sample of water has the most thermal energy?
O A. A 0.5 kg sample at 1°C
O B. A 1.5 kg sample at 10°C
O C. A 1 kg sample at 10°C
O D. A1 kg sample at 4°C
Answers
Answer:b
Explanation:
A light bulb is connected to two batteries. Each battery has 1.5V. If a student measures the current flowing through the light bulb to be 2 A, What is the resistance of the light bulb?
Answers
Answer:
R = 0.75 ohms
Explanation:
Given that,
The voltage of each battery, V = 1.5 V
Current through the light bulb, I = 2 A
We need to find the resistance of the light bulb. We know that,
Ohm's law, V = IR
Where
R is the resistance of the light bulb
Put all the values,
\(R=\dfrac{V}{I}\\\\R=\dfrac{1.5}{2}\\\\R=0.75\ \Omega\)
So, the resistance of the light bulb is equal to 0.75 ohms.
William took a 7 hour bicycle trip. In all, he traveled 112 miles. What was his average rate of speed in miles per hour?
Answers
Answer:
16 miles
Explanation:
112 miles/ 7 hours= 16 miles per hour