Answer:
Visibility conditions.
Explanation:
Safe speed is a speed at which the operator of the boat can take effective action to avoid stops within the distance. To calculate the safe speed visibility conditions (fog, rain, mist, and darkness) should be included. We should also include some other factors given below:
i) Traffic density
ii) Type of vessels in the area
iii) Wind
iv) Water conditions
The Navigation Rules dictate that a safe speed is determined by factors such as visibility, traffic density, vessel maneuverability, weather conditions, and proximity to navigational hazards.
Explanation:According to the Navigation Rules, several factors should be taken into account in determining a safe speed. These include the visibility, the traffic density, the maneuverability of the vessel in immediate circumstances, the state of wind, sea, and current, and the proximity of navigational hazards. For example, on a clear day with sparse traffic and calm seas, higher speeds might be safe. However, in heavy traffic or poor visibility, the navigation rules would call for lower speeds to ensure safety.
Learn more about Navigation Rules here:https://brainly.com/question/32157646
#SPJ11
The following images show the four terrestrial planets in our solar system (not to scale). Rank these planets from left to right based on the atmospheric pressure at the surface, from highest to lowest.
Answer:
Venus, Earth, Mars, Mercury
Explanation:
The atmospheric pressure is determined by the gravitational attraction of the planet over the atmosphere.
Since the gravitational force is defined as:
[tex]F = G\frac{m1m2}{r^{2}}[/tex] (1)
Notice from equation 1, how the gravitational force depends on the masses of the two objects that are interacting and the distance to the square between them.
For this particular case, m1 is the mass of the planet and m2 is the mass of the gasses in the atmosphere. While the r is the distance between those two.
Venus has a more dense atmosphere and than the Earth, so it has a higher atmospheric pressure. The same case can be seen for the Earth and Mars.
Mercury is the one with the lower atmospheric pressure since it has a thin atmosphere as a consequence of his small size.
The Atmospheric pressure on terrestrial planets, highest to lowest, are generally Venus, Earth, Mars, and Mercury. These rankings can change based on the specific positioning of the planets in the provided images.
The terrestrial planets in our solar system, from Mercury through Mars, have significant differences in atmospheric pressure due to several factors like their distance from the sun, their composition, size, and presence of a magnetic field. Based on the data we have:
Venus has the highest atmospheric pressure among terrestrial planets, which is about 92 times the pressure of Earth's atmosphere at sea level.
Earth comes next. Its atmospheric pressure, also known as barometric pressure, varies depending on altitude but averages about 1 bar at sea level.
Mars has a very thin atmosphere with a surface pressure less than 1% of Earth's.
Mercury, being the smallest and closest to the Sun, has virtually no atmosphere and thus has the lowest surface atmospheric pressure among the terrestrial planets.
Please note that the ranking could change if the images in question show the planets placed differently from the order mentioned.
For more such questions on Atmospheric pressure on terrestrial planets, click on:
https://brainly.com/question/32828430
#SPJ3
A degausser creates a magnetic field that erases data from magnetic storage media?
Explanation:
Degausser is a device used by automatically manipulating the alignment of magnetic domains on the medium to avoid data stored on computer and laptop hard disks, floppy disks, and magnetic cassette. There was a mistake. Therefore, a degausser is used to remove all audio, video and data signals from magnetic store media completely.
The most likely place to find abundant manganese nodules is on the: a. abyssal plain far from a continent. b. continental rise. continental shelf. c. crest of a mid-ocean ridge. d. All of the above locations contain manganese nodules.
Answer:
a. abyssal plain far from a continent
Explanation:
The sediments of the abyssal plains contain of the abundant mineral resources and notably polymetallic nodules and re also called the manganese nodules and these are on the shape of potatoes and distributed at the depths of about 4000 meters. And they lie at the international line of the pacific oceans stretching in an area of 118°–157°, and from 9°–16°N, that is 3 million m square is currently under the noddle exploration and discovery.The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass 40.0 kg resting on the output plunger. The piston and plunger are nearly at the same height, and each has a negligible mass. By how much is the spring compressed from its unstrained position?
The spring in the hydraulic chamber is compressed by 24.5 cm from its unstrained position when a 40.0 kg rock is resting on the output plunger. This is calculated using Hooke's Law.
Explanation:The key to solving this problem is understanding that the force exerted by a spring follows Hooke's Law, which states that F = -kx, where F is the force, k is the spring constant, and x is the displacement of the spring from its equilibrium position. Given that the spring constant k = 1600 N/m and the force exerted by the rock of mass 40.0 kg is F = mg = 40.0 kg * 9.81 m/s² = 392.4 N, we can find the displacement as follows:
x = F/k = 392.4 N / 1600 N/m = 0.245 m or 24.5 cm
Therefore, the spring is compressed by 24.5 cm from its unstrained position when the 40.0 kg rock is resting on the output plunger.
Learn more about Hooke's Law and Displacement here:https://brainly.com/question/27988275
#SPJ12
The spring is compressed by 24.5 cm from its unstrained position.
Explanation:The amount by which the spring is compressed from its unstrained position can be calculated using Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. In this case, the force exerted by the spring is equal to the weight of the rock.
Using the equation F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring, we can rearrange the equation to solve for x:
x = F / k = (m * g) / k
Plugging in the values, we have x = (40.0 kg * 9.8 m/s²) / 1600 N/m.
Simplifying the equation, we get x = 0.245 m = 24.5 cm.
Learn more about Calculating spring compression here:https://brainly.com/question/32197840
#SPJ11
What is Icm, the moment of inertia of the disk around its center of mass? You should know this formula well.
[tex]I_{cm} =mr^{2}[/tex] is the moment of inertia of rigid body along its centre of mass.
The moment of inertia of a body is the rotational analog of mass in linear motion. If we consider a rigid body of mass m, assuming its mass concentrated at its center of mass (cm) which is at a distance of r from the axis of rotation of the body, then the moment of inertia of the rotating body is given by:
[tex]I_{cm} =mr^{2}[/tex]
Learn more about moment of inertia:
https://brainly.com/question/13891007
The moment of inertia (Icm) of a disk about an axis through its center is 0.5 * mass * radius^2. It plays a significant role in the study of rotational motion.
Explanation:The moment of inertia of a disk about an axis passing through its center of mass, perpendicular to the plane of the disk (also known as Icm) is calculated using the formula:
Icm = 0.5 * mass * radius^2
where the mass is how much matter is in the disk, and the radius is the distance from the center of the disk to its edge. This formula is important as it helps determine the rotational motion of an object.
Learn more about moment of inertia here:https://brainly.com/question/34228994
#SPJ3
When you shoulder your shotgun, what part of your body should fit snugly against the stock?
Answer:
Cheek
Explanation:
Typically the scope is a bead on the gun's edge. Our eye must be in alignment with the muzzle, so that the proper placement of our head on the stock is crucial. The stock will fit snugly to our cheek as we put the pistol to our face with our head on other side just above gun's center line.
The correct answer is cheek.
The combined mass of the asteroids in the Solar System is comparable to the mass of__________________.
Answer:
25 times less than the Moon's mass
Explanation:
The asteroids are small, rocky objects that orbit the Sun. Although asteroids orbit the Sun like planets, they are much smaller than planets.
There are millions of boulder-size and larger rocks that orbit the Sun, most of them between the orbits of Mars and Jupiter.
The largest asteroid is Ceres with a diameter of 1000 kilometers. Pallas and Vesta have diameters of about 500 kilometers and about 15 others have diameters larger than 250 kilometers
Although in the Solar System there are millions the asteroids, the combined mass of all of the asteroids is about 25 times less than the Moon's mass.
Then the answer for your question is:
25 times less than the Moon's mass
bibliographic source: www.astronomynotes.com
A stone is thrown vertically upward from the roof of a building.Does the position of the stone depend on the location chosen for the origin of the coordinate system?
Displacement is the difference between two co-ordinates so the origin doesn't matter.
Explanation:
Displacement is a physical quantity that depicts the change of position in a particle, this further is measured by the difference in position vectors limiting the duration under these two limits.Displacement = Velocity × TimeInstantaneous velocity = d/dt (position vector)
d = vt + 1/2 at² is the displacement acceleration equation.
d = u + at is the velocity displacement equation.
Displacement always has continuity with velocity and Acceleration.The average velocity is 0 if the displacement is 0.
How fast must a 3000 kg elephant move to have the same kinetic energy as a 65 spinter running at 10ms?
Q. : How fast must a 3000 kg elephant move to have the same kinetic energy as a 65.0 kg sprinter running at 10m/s?
Answer:
The elephant must be 1.47 m/s fast to have the same kinetic energy with the sprinter
Explanation:
Kinetic Energy: This is the energy of a body in motion. It can be expressed mathematically as
Ek = 1/2mv²
Since the kinetic energy the elephant = the kinetic energy of the sprinter.
1/2m₁v₁² = 1/2m₂v₂².......................... Equation 1
Making v₁ the subject of the equation,
v₁ = √(m₂v₂²/m₁)................................. Equation 2
Where v₁ = speed of the elephant, m₁ = mass of the elephant, v₂ = speed of the sprinter, m₂ = mass of the sprinter.
Given: m₁ = 3000 kg, m₂ 65 kg, v₂ = 10 m/s
Substituting these values into equation 2,
v₁ =√(65×10²/3000)
v₁ = [tex]\sqrt{(6500)/3000[/tex]
v₁ =[tex]\sqrt{2.167}[/tex]
v = 1.47 m/s
Therefore The elephant must be 1.47 m/s fast to have the same kinetic energy with the sprinter
A particle moves along the x-axis with velocity given by v(t)=3t2+6t for time tâ¥0. If the particle at position x=2 at time t=0, what is the position of the particle at time t=1?
a) 4
b) 6
c) 0
d) 11
e) 12
Answer:
b) 6
Explanation:
Given
v(t)=3t²+6t
X(0) = 2
X(1) = ?
Knowing that
v(t)=3t²+6t = dX/dt
⇒ ∫dX = ∫(3t²+6t)dt
⇒ X - X₀ = t³ + 3t²
⇒ X(t) = X₀ + t³ + 3t²
If X(0) = 2
⇒ X(0) = X₀ + (0)³ + 3(0)² = 2
⇒ X₀ = 2
then we have
X(t) = t³ + 3t² + 2
when
t = 1
X(1) = (1)³ + 3(1)² + 2
X(1) = 6
After a time, t =1, the position of the particle has been 6. Thus, the correct option is b.
The velocity of the particle moving along the x-axis has been given by:
[tex]\rm v(t)\;=\;3t^2\;+\;6t[/tex]
The differentiation of v(t) in terms of x will be:
[tex]\rm \dfrac{dx}{dt}\;=\;3t^2\;+\;6t[/tex]
[tex]\rm \int dx\;=\;\int (3t^2\;+\;6t)\;dt[/tex]
differentiation with the limits if X be x and [tex]\rm x_0[/tex]:
x - [tex]\rm x_0[/tex] = [tex]\rm t^3\;+\;3t^2[/tex]
In the given question, the value of [tex]\rm x_0[/tex] = 2:
At time t = 0
x = [tex]\rm t^3\;+\;3t^2\;+\;x_0[/tex]
x = [tex]\rm (0)^3\;+\;3(0)^2\;+\;2[/tex]
x = 2.
To find the position of the particle at time =1, given, [tex]\rm x_0[/tex] = 2.
x = [tex]\rm (1)^3\;+\;3(1)^2\;+2[/tex]
x = 1 + 3 + 2
x = 6.
Thus, after time, t =1, the position of the particle has been 6. Thus, the correct option is b.
For more information about the position of the particle, refer to the link:
https://brainly.com/question/18328170
When entering a bicycle lane to make a right turn, within how many feet must you enter the lane before making the turn?
Answer:
200 feet
Explanation:
Most streets in US with bike lane, especially San Francisco, when entering a bicycle lane to make a right turn, you must enter 200 feet to the lane before making the turn.
Typically, when making a right turn, you should enter the bicycle lane no more than 200 feet before the turn. Always yield to cyclists before merging into the lane. These rules can vary by location.
Explanation:Regarding the query about bicycle lanes and right turns, each city or state can have different regulations, but a common rule in many regions is that a driver intending to make a right turn must enter the bicycle lane no more than 200 feet before making the turn. This allows for a safe and orderly flow of both motorized and non-motorized traffic. It's important to check if this rule applies specifically to your region though, as variations can and do exist. Always make sure to yield to any cyclists in the bike lane before merging into it.
Learn more about Road Laws here:https://brainly.com/question/31848272
#SPJ11
You have just been appointed as director of your company's corporate training division. The CEO of your company has been displeased with your company's prior training programs, so you are tasked with rehauling the entire training division. You convene a meeting of all training division managers to decide on the types of training that the division will implement. One of your managers is a firm supporter of e-training programs for employees in your company's international offices. He touts the benefits of e-programs by stressing that ________.
Answer:
He touts the benefits of e-programs by stressing that:
E-training are highly adjustable and flexible as well as employees can complete their training according to their ease.
Explanation:
The concept of E-training is becoming more and more popular now-a-days as it has the following advantages:
You can train the people all over the world with the help of internet.You can complete the training by remaining in your comfort zone.You can get the feedback of the people in a structured manner in less time.You can evaluate the performance in the real time.Due to these reasons, the manager wants the company to do the e-training programs.
Small-plane pilots regularly compete in "message drop" competitions, dropping heavy weights (for which air resistance can be ignored) from their low-flying planes and scoring points for having the weights land close to a target. A plane 65 m above the ground is flying directly toward a target at 46 m/s .
A.) At what distance from the target should the pilot drop the weight?B.) The pilot looks down at the weight after she drops it. Where is the plane located at the instant the weight hits the ground?not yet over the targetpast the targetdirectly over the targetnot enough information to determine
Answer:
a)165,6 m
b)Directly over the target
Explanation:
a)To calculate the time that ball hits the ground after released, we will use the free fall formula below:
[tex]h=(1/2)*g*t^2[/tex]
h=65m
g=9,8m/s^2
[tex]65=0,5*9,8*t^2\\t^2=65/(0,5*9,8)\\t=3,6 sec[/tex]
Plane has the speed of 46m/s. Due to there is no air resistance weight remains the same speed while free falling. So the distance will be:
[tex]x=46*3,6=165,6 m[/tex]
b) Due to no air resistance plane and weight will be at the same speed on the air. Therefore, when the weight hits the ground plane will be directly over the target.
The pilot should drop the weight 165.6 m prior to the target, and the plane will have passed the target by the time the weight lands.
Explanation:
In Physics, we know that the motion of the dropped weight is governed by the equations of free fall, specifically that the time t for an object to fall from rest under gravity g, ignoring air resistance, from a height h is given by t = sqrt(2h/g). In this case, g = 9.81 m/s2 (earth's acceleration due to gravity), and h= 65 m. Therefore, if we put these values into the formula, we get t = sqrt(2*65/9.81) ≈ 3.6s.
For part A, the distance from the target where the pilot should drop the weight is the same as the distance the plane will travel in the time it takes for the weight to hit the ground, which is the product of the plane's speed (46m/s) and the time of flight t of the weight. Therefore, the distance is 46*3.6 = 165.6 m.
The answer for part B is that the plane will have passed the target when the weight hits the ground as it continues to move forward at 46 m/s.
Learn more about Free Fall here:https://brainly.com/question/35159043
#SPJ3
Which friction-reducing technologies are used in the Variable Compression Turbo Engine?
Friction-reducing technologies used in the Variable Compression Turbo Engine are Diamond-like coating on valve lifters, micro finishing on crankshaft and camshaft and mirror bore coating on cylinder wall.
Explanation:
Variable compression is a technology to adjust the compression of an internal combustion engine while the engine is in operation. At this time friction may occur that need to be reduced. To reduce this friction some technologies are used like
Diamond-like coating on valve liftersMicro finishing on crankshaft and camshaftMirror bore coating on cylinder wallA hydrogen free diamond like carbon coating is applied to an engine valve lifter to reduce mechanical loss. Micro finishing on crankshaft and camshaft achieves improvement in geometric parameters such as roundness. Mirror bore coating on cylinder wall raises energy efficiency by reducing the friction inside the engine.
Friction-reducing technologies in Variable Compression Turbo engines include gasoline direct injection, variable valve timing, and multi-valve configurations, all enhanced by advanced computer controls. These technologies help to optimize engine efficiency by minimizing frictional losses.
Explanation:The Variable Compression Turbo (VCT) Engine employs several friction-reducing technologies to enhance performance and efficiency. One key technology used in such engines is gasoline direct injection, which enables precise control over the fuel-air mixture and improves combustion efficiency. Another technology is variable valve timing, which optimizes the opening and closing of the engine's valves to match the engine's speed and load, reducing mechanical friction and improving efficiency.
In addition to these, the VCT engines may use multi-valve configurations that increase the engine's ability to breathe by allowing for more intake and exhaust flow, further reducing frictional losses. Enhanced computer controls also play a critical role in adjusting the compression ratio and monitoring engine performance to minimize friction. The use of advanced materials and engineering solutions contribute to reducing heat transfer into the environment, although it cannot be eliminated entirely due to the second law of thermodynamics.
A huge cannon is assembled on an airless planet (ignore any effects due to the planet's rotation). The planet has a radius of 5.00 × 106 m and a mass of 3.95 × 1023 kg. The cannon fires a projectile straight up at 2000 m/s. An observation satellite orbits the planet at a height of 1000 km. What is the projectile's speed as it passes the satellite?
Answer:
The projectile's speed as it passes the satellite is 1497.8 m/s.
Explanation:
Given that,
Radius of planet [tex]r=5.00\times10^{6}\ m[/tex]
Mass of planet [tex]m=3.95\times10^{23}\ kg[/tex]
Speed = 2000 m/s
Height = 1000 km
We need to calculate the projectile's speed as it passes the satellite
Using conservation of energy
[tex]E_{1}=E_{2}[/tex]
[tex]\dfrac{1}{2}mv_{1}^2+\dfrac{GmM}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{GmM}{R+h}[/tex]
[tex]\dfrac{v_{1}^2}{2}+\dfrac{GM}{r_{1}}=\dfrac{v_{2}^2}{2}+\dfrac{GM}{R+h}[/tex]
[tex]-\dfrac{v_{2}^2}{2}=-(\dfrac{GM}{R}-\dfrac{GM}{R+h}-\dfrac{v_{1}^2}{2})[/tex]
[tex]v_{2}^2=v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})[/tex]
[tex]v_{2}=\sqrt{v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})}[/tex]
Put the value into the formula
[tex]v_{2}=\sqrt{2000^2+2\times6.67\times10^{-11}\times3.95\times10^{23}(\dfrac{1}{5.00\times10^{6}+1000\times10^{3}}-\dfrac{1}{5.00\times10^{6}})}[/tex]
[tex]v_{2}=1497.8\ m/s[/tex]
Hence, The projectile's speed as it passes the satellite is 1497.8 m/s.
2 Physics Questions WILL GIVE BRAINLIEST!!
A book is being pushed along a table at a constant speed. The coefficient of static friction is 0.75, and the coefficient of kinetic friction is 0.30. The book has a mass of 0.50 kg. What is the force of friction?
0.15 N
1.5 N
3.7 N
0.38 N
Suppose the mass of each of two spherical objects is doubled, and the distance between their centers is three times as large. How is the magnitude of the force of gravity between them affected?
It is 4/9 as large
It is 3/2 as large
It is 4/3 as large
It is 2/3 as large
It is 2/9 as large
Answer:
a) The force of friction is 1.5 N.
b) The new force is 4/9 times as large as the original force.
Explanation:
It is given that the book is moving on the table.
∴ The friction acting on the block should be kinetic.
∴ Coeffecient of kinetic friction , k = 0.3
The force of friction = kN , where 'N' is the normal force acting on the surface where friction is acting.
In this case , N = mg , where 'm' is the mass of the book and 'g' is the acceleration due to gravity.
∴ Force of friction , f = kmg = 0.3×0.5×10 = 1.5N
b) We know that ,
Gravitational force , F ∝ m , where 'm' is the mass of the body.
F ∝ [tex]x^{2}[/tex] where 'x' is the distance between the two bodies.
Since mass of both the bodies is doubled , the force will be 4 times as large compared to the original force F from the first proportionality equation.
However , as the distance is 3 times as large , the force will be 9 times as small as compared to the original force F
∴ The new force is 4/9 times as large as the old force
A door 1.00 m wide and 2.00 m high weighs 330 N and is supported by two hinges, one 0.50 m from the top and the other 0.50 m from the bottom. Each hinge supports half the total weight of the door. Assuming that the door’s center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.
Each hin-ge exerts a horizontal force of 165 N on the door.
Explanation:To find the horizontal components of the force exerted on the door by each hin-ge, we need to divide the weight of the door evenly between the two hing-es. The weight of the door is 330 N, so each hin-ge will support 330 N / 2 = 165 N.
Next, we need to determine the horizontal component of the force exerted by each hin-ge. Since the door's center of gravity is at its center and the hin-ge is 0.50 m from the top and 0.50 m from the bottom, the horizontal distance between the hin-ge and the center of gravity is the same for both hin-ges.
Using the formula for torque, τ = force × distance, we can calculate the torque exerted by the weight of the door about each hin-ge. Since the torque is equal to the force times the perpendicular distance from the axis of rotation to the line of action of the force, and the horizontal component of force is perpendicular to the distance, we can set up the following equation: torque = force × distance
Solving for the horizontal component of force, we have: force = torque / distance
By plugging in the torque value for each hin-ge and the distance value, we get:
force = 165 N × 0.50 m / 0.50 m = 165 N
Therefore, the horizontal components of force exerted on the door by each hin-ge are both 165 N.
1. A 21.3-kg child positions himself on an inner-tube which is suspended by a 7.28-m long rope attached to a strong tree limb. The child and tube is drawn back until it makes a 17.4-degree angle with the vertical. The child is released and allowed to swing to and from. Assuming negligible friction, determine the child's speed at his lowest point in the trajectory.
2. A baseball player catches a 163-gram baseball which is moving horizontally at a speed of 39.8 m/s. Determine the force which she must apply to the baseball if her mitt recoils a horizontal distance of 25.1 cm.
Answer
1) given,
mass of child = 21.3 Kg
length of the rope = L = 7.28 m
angle made with vertical = 17.4°
speed of the child at the bottom most point
using conservation energy
[tex]m g h = \dfrac{1}{2}mv^2[/tex]
[tex] v = \sqrt{2gh}[/tex]
H = L cos θ
H = 7.28 x cos 17.4°
H = 6.94
where
L = H + h
h = L - H
h = 7.28 - 6.94 = 0.34 m
now,
[tex] v = \sqrt{2\times 9.8 \times 0.34}[/tex]
v = 2.58 m/s
2) given,
mass of baseball = 163 g = 0.163 kg
initial speed = 39.8 m/s
final speed = 0
horizontal distance = 25.1 cm = 0.251 m
Force = ?
using equation of motion
v = u + at
0 = 39.8 + at
at = -39.8 m/s
using equation
[tex]s = ut + \dfrac{1}{2}at^2[/tex]
[tex]0.251= 39.8 t - 0.5 \times 39.8 t [/tex]
t = 0.0126 s
using formula of impulse
I = F x t
I = m(v - u)
F x 0.0126 = 0.163 x (-39.8)
F = 514.87 N
Final answer:
To determine the child's speed at his lowest point in the trajectory, we can use the conservation of mechanical energy by equating the gravitational potential energy to the sum of the kinetic and elastic potential energy. By using the equations for kinetic energy and tension, we can calculate the child's speed and the force of tension in the rope at the lowest point.
Explanation:
To determine the child's speed at his lowest point in the trajectory, we can use the conservation of mechanical energy. At the highest point, the gravitational potential energy is equal to the sum of the kinetic energy and the elastic potential energy of the rope. At the lowest point, the gravitational potential energy is zero, so the kinetic energy is equal to the elastic potential energy. Using this information, we can calculate the child's speed at the lowest point using the equation:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the child, and v is the child's velocity.
Since we are given the mass of the child and the length of the rope, we can also calculate the force of tension in the rope at the lowest point using the equation:
T = mg + (mv^2)/L
where T is the tension, m is the mass of the child, g is the acceleration due to gravity, v is the child's velocity, and L is the length of the rope.
By using these equations, we can determine the child's speed at his lowest point in the trajectory.
What BEST completes the following sentence.The dust and gas that escapes from a comet creates a/an _____________________________.
Answer:
Explanation:
The dust and gases that escape from a comet creates a coma.
Coma can also be defined as an unclear envelope around the comet, formed when the comet passes near the sun. Sun temperature melts the comet ice and thus give comet a fuzzy appearance when viewed telescope and distinguishes it from a star.
This technique can be useful to find the size of the comet of different size
Federal Law mandates that the steering or helm area of a power boat less than 20 feet in length must have which of the following?a. Latex gloves.b. Capacity Plate.c. Ladder stand.d. Rear
Answer:Federal Law mandates that the steering or helm area of a power boat less than 20 feet in length must have Capacity Plate. The correct option is B.
Explanation: The capacity Plate indicates the maximum weight capacity and/or the maximum number of people that the boat can carry safely in good weather. Information on the capacity Plate includes the maximum number of adult persons, the maximum gross load, and the maximum size of engine, in horsepower,
Answer:
b. Capacity Plate.
Explanation:
According to the Federal Law, boats of less than 20 feet in length need to have a capacity plate that has to be in the steering area all the time. Also, this capacity plate has to show the maximum people capacity, weight and horsepower recommended. Because of this, the answer capacity plate.
A particle P is projected from the origin O so that it moves along the x-axis. At time t after projection, the velocity of the particle v is given by v = 3t^2 -24t +45. ) Find an expression for the displacement of P in the first 3 seconds of its motion using calculus(integration).
Answer:
x = t³ − 12t² + 45t
Explanation:
Position is the integral of velocity.
x = ∫ v dt
x = ∫ (3t² − 24t + 45) dt
x = t³ − 12t² + 45t + C
The particle is originally at the origin, so at t = 0, x = 0.
0 = (0)³ − 12(0)² + 45(0) + C
0 = C
x = t³ − 12t² + 45t
You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 13.2 cm. When the cylinder is rotating at 1.63 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall?
Answer:
centripetal acceleration[tex](a_{c})=13.8m/s^{2}[/tex]
Explanation:
convert 1.63rev/sec to rpm by multiplying by 60
= 1.63*60=97.8rpm
Convert this to rad/sec
1rpm =π/30 rad/sec
97.8rpm = 97.8 * (π/30 rad/sec)
=10.25rad/sec
linear velocity= angular velocity *radius
radius =13.2cm=13.2/100=0.132m
v=rω
v= 0.132*10.25
v=1.35m/s
centripetal acceleration = [tex]\frac{v^{2} }{r}[/tex]
[tex]a_{c}=\frac{1.35^{2} }{0.132}[/tex]
[tex]a_{c}=13.8m/s^{2}[/tex]
Answer:
[tex]a_{c}[/tex] = 13.8 m/s².
Explanation:
The acceleration centripetal [tex]a_{c}[/tex] is given by:
[tex] a_{c} = \frac{v^{2}}{r} [/tex] (1)
where v: is the tangential speed and r: is the container radius
The tangential speed is equal to:
[tex] v = \omega \cdot r [/tex] (2)
where ω: is the angular velocity
Since 1 revolution is equal to 2π rad, the velocity (equation 2) is:
[tex] v = 1.63 \frac{rev}{s} \cdot \frac{2\pi rad}{1rev} \cdot 0.132m = 1.35m/s [/tex]
Now, by entering the velocity value calculated into equation (1) we can find the acceleration centripetal:
[tex] a_{c} = \frac{(1.35m/s)^{2}}{0.132m} = 13.8m/s^{2} [/tex]
I hope it helps you!
If we found an Earth-sized planet orbiting very close to another star (causing it to be very hot), which planet would its surface resemble the most?
Answer:
Venus is definitely the answer
Explanation:
If we found an Earth-sized planet orbiting very close to another star (causing it to be very hot), which planet would its surface resemble the most?
Venus is known as the earth twin because it is similar in shape to the earth. Although, it is not the closest to the sun, it traps the suns heat in its atmosphere making it the hottest planet in the solar system. the temperature is as high as 465 degrees Celsius. Venus is known to be the second brightest body in the solar system, the moon being the first. A day in venus could be longer than a year on the earth. It revolves in the opposite direction around the sun compared to other planets. It has an atmospheric pressure which ninety times grater than what we have here on earth.
An amplifier has an input power of 5 mW. The output power is 100 mW. What is the decibel gain to the nearest decibel?
Answer:
The decibel gain to the nearest decibel = 13 db
Explanation:
Gain of an amplifier: This is defined as the measure of the ability of an amplifier to increase the power of a signal from the input to the output port by adding energy converted from some power supply to the signal.
It is represented mathematically as.
db = 10×log (p₂/p₁)............................. Equation 1
Where dp = decibel gain of the amplifier, p₁ = input power of the amplifier, p₂ = output power of the amplifier.
Given: p₁ = 5 mW, p₂ = 100 mW.
Substituting these values into equation 1
db = 10×log(100/5)
db = 10×log(20)
dp = 10 ×1.301
dp = 13 db gain
Therefore the decibel gain to the nearest decibel = 13 db
At an age of 380,000 years, the temperature of the universe had fallen to 3000 K, and electrons could then combine with protons to produce hydrogen gas instead of roaming freely through space.
What major transition occurred as a consequence of this change in the universe at this time?
a) The present laws of physics were applicable to the properties of the universe time.
b) The universe became transparent to light for the first time.
c) Nuclear fusion no longer occurs below this temperature, and so, general fusion throughout the universe would have ceased.
d) The universe would have lost its electrical charge suddenly to become elect.
Answer:
The major transition occurred as a consequence of this change in the universe at this time is that b)The universe became transparent to light for the first time.
Explanation:
For the first 380,000 years or so, the universe was essentially too hot for light to shine. The heat of creation smashed atoms together with enough force to break them up into a dense plasma, an opaque soup of protons, neutrons and electrons that scattered light like fog. Then 380,000 years after the Big Bang, matter cooled enough for atoms to form during the era of recombination, resulting in a transparent, electrically neutral gas.
This set loose the initial flash of light created during the Big Bang, which is detectable today as cosmic microwave background radiation. However, after this point, the universe was plunged into darkness, since no stars or any other bright objects had formed yet.
What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.
Incomplete question.The complete question is
A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm. What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.
Answer:
Total gauge pressure= 3724Pa or 3.724kPa
Explanation:
Pressure (water alone) = (1,000 x 9.8 x 0.2m) = 1,960Pa.
Pressure (oil on top) = (900 x9.8 x 0.2m) = 1,764Pa.
Total gauge pressure at bottom = (1,960 + 1,764) = 3724Pa or 3.724kPa
A steel scale measures the length of a copper rod as 80cm when both are at 20°c , the caliberation temperature of the scale . what would the scale read for the length of the rod when both are at 40°c,
given coefficient of linear expansion of steel =1.1×10^-5°c^-1
coefficient of linear expansion of copper =1.7×10^-5°c^-1
a. 80.0096 cm
b. 80.0272 cm
c. 1 cm
d. 25.2 cm
Answer:
a) 80.0096 cm
Explanation:
This is an exercise of thermal expansion, we must calculate how much each material is dilated and the difference is the expansion of the measurement
The expression for linear dilation
ΔL = α L₀ ΔT
Let's start with the steel
ΔL / L₀ = 1.1 10⁻⁵ (40 -20)
ΔL / L₀ = 22 10⁻⁵
Copper
ΔL / L₀ = 1.7 10⁻⁵ (40-20)
ΔL / L₀ = 34.0 10⁻⁵
What made the copper bar high is the differentiated being read
ΔD = ΔL / L₀_copper - ΔL / L₀_Steel
The initial length of the steel balance is made equal to the length of the copper rod
ΔL = (34 - 22) 10⁻⁵ / 80
ΔL = 0.0096
For which the final length of the copper bar
[tex]L_{f}[/tex] = L₀ +ΔL
[tex]L_{f}[/tex] = 80 +0.0096
[tex]L_{f}[/tex] = 80.0096 cm
The correct answer is a
A train has a constant speed of 10 m/s around a track with a diameter of 45 m what is the centripal acceleration?
Answer:
[tex]a_{c}= 4.44\frac{m}{s}[/tex]
Explanation:
When an object goes on a circular movement, it has a centripetal acceleration that always points toward the center of the circle, it is the responsible of the change of direction in the movement of the object. and that centripetal acceleration is related with the speed in the next way:
[tex]a_{c}=\frac{v^{2}}{r} [/tex], with v the speed, r the radius of the track that is half of the diameter (22.5 m)
[tex]a_{c}=\frac{10^{2}}{22.5} [/tex]
[tex]a_{c}= 4.44\frac{m}{s}[/tex]
Which telescope has a better (smaller) angular resolution: a 2-m telescope observing visible light (wavelength 5.0×10-7 m) or a 10-m radio telescope observing radio waves (wavelength 2.1×10-2 m)?
The 2-m telescope observing visible light has a better (smaller) angular resolution than the 10-m radio telescope observing radio waves.
Explanation:In order to determine which telescope has a better (smaller) angular resolution, we need to calculate the angular resolution for each telescope. The formula for angular resolution is given by θ ≈ 1.22λ/D, where λ is the wavelength of light and D is the diameter of the telescope's aperture.
For the 2-m telescope observing visible light with a wavelength of 5.0×10-7 m, the angular resolution is approximately 1.22×(5.0×10-7)/(2) = 3.05×10-7 radians.
For the 10-m radio telescope observing radio waves with a wavelength of 2.1×10-2 m, the angular resolution is approximately 1.22×(2.1×10-2)/(10) = 2.54×10-3 radians.
Therefore, the 2-m telescope observing visible light has a better (smaller) angular resolution than the 10-m radio telescope observing radio waves.
Final answer:
The 2-m optical telescope observing visible light has a better (smaller) angular resolution compared to the 10-m radio telescope observing radio waves.
Explanation:
To determine which telescope has a better (smaller) angular resolution, we can use the formula for resolving power (θ), which is given by: θ ≈ λ / D, where λ is the wavelength and D is the diameter of the telescope's aperture. For the 2-m telescope observing visible light (λ = 5.0×10⁻⁷ m), the angular resolution would be θ ≈ 5.0×10⁻⁷ m / 2 m = 2.5×10⁻⁷ radians. For the 10-m radio telescope observing radio waves (λ = 2.1×10⁻² m), the angular resolution would be θ ≈ 2.1×10⁻² m / 10 m = 2.1×10⁻³ radians. Thus, the 2-m optical telescope has a smaller (better) angular resolution than the 10-m radio telescope.
The part of the building structure, typically below grade, upon which all other construction is built is known as:________
Answer:
horizontal structural member that supports a floor. Beams are typically wood, cold formed metal framing or steel.
Joists
Horizontal timbers, beams or bars supporting a floor.