Final answer:
The crate has an acceleration of 0.43 m/s^2. It travels 21.5 m in 10.0 s and has a speed of 4.3 m/s at the end of 10.0 s.
Explanation:
To solve this problem, we can use Newton's second law, which states that force is equal to mass times acceleration (F = ma).
(a) We are given that the mass of the crate is 32.5 kg and the net horizontal force acting on it is 14.0 N. Plugging these values into the equation, we get:
F = ma
14.0 N = 32.5 kg * a
a = 14.0 N / 32.5 kg
a = 0.43 m/s^2
So, the acceleration produced is 0.43 m/s^2.
(b) To find the distance traveled by the crate in 10.0 s, we can use the equation of motion: distance = initial velocity * time + (1/2) * acceleration * time^2.
Since the crate starts at rest, the initial velocity is 0 m/s:
distance = 0 * 10.0 s + (1/2) * 0.43 m/s^2 * (10.0 s)^2
distance = 0 + (1/2) * 0.43 m/s^2 * 100.0 s^2
distance = 21.5 m
So, the crate travels 21.5 m in 10.0 s.
(c) To find the speed of the crate at the end of 10.0 s, we can use the equation of motion: final velocity = initial velocity + acceleration * time.
Since the crate starts at rest, the initial velocity is 0 m/s:
final velocity = 0 + 0.43 m/s^2 * 10.0 s
final velocity = 4.3 m/s
So, the speed of the crate at the end of 10.0 s is 4.3 m/s.
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.
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.
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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
While troubleshooting a network problem, a technician realized it could be resolved by opening a port on a firewall. The technician opened the port and verified the system was now working. However, an attacker accessed this port and launched a successful attack. What could have prevented this problem?A. Patch management processes
B. Vulnerability management processes
C. Configuration management processes
D. Change management processes
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!
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.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.
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
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.
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 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.
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.
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
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.
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.
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
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.
Describe how the water cycle is driven by the energy of the sun
Explanation:
Sun is the is the cause of entire water cycle because it is responsible for for its major two components that are
1. Condensation
2. Evaporation
The Sun's heat cause the evaporation( water converting into vapor) of water from the earth. This water ends up in atmosphere as water vapor. It cools and rise becoming cloud, and this eventually condenses to water droplets in form of rain or dwe.
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
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.
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]
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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.
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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
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.
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.
what's your current physics grade
Answer: grade 8 (uk grading system)
A string is 9.00 m long with a mass of 7.00 g. The string is held taut with a tension of 200.00 N applied to the string. A pulse is sent down the string. How long (in ms) does it take the pulse to travel the 9.00 m of the string?
Answer:
It takes the pulse 25.1 ms to travel the 9.00 m of the string
Explanation:
Force (F) = Mass( m) × Acceleration.( a)
F = ma............... equation 1
Making a the subject of equation 1 above,
a = F/m .............. equation 2
Where F = T = 200 N, m = 7.00 g = 0.007 kg,
Substituting these values into equation 2,
a = 200/0.007 = 28571.43 m/s².
Using one of the equation of motion,
S = ut + 1/2at²................... equation 2
Where S = distance, u = initial velocity, t = time, a = acceleration.
u = 0.
Therefore, S = 1/2at²................. equation 3
Making t the subject of equation 3,
t = √ (2S/a).................. equation 4
Where S = 9.00 m , a = 28571.43 m/s²
Substituting this values into equation 4,
t = √{(2×9)/28571.43}
t = √(18/28571.43
t = √0.00063
t = 0.0251 s
t = 25. 1 ms
Therefore it takes the pulse 25.1 ms to travel the 9.00 m of the string.
a parallel RC circuit has a capacitive reactance of 962 ohms and a resistance of 1,200 ohms what is the impedance of this circuit
A. 751 ohm
B. 653 ohms
C. 968 ohm
D, 1,254 ohm
Answer:
A. 751 ohm
Explanation:
Impedance: This is the total opposition to the flow of current in an a.c circuit by any or all of the three circuit elements ( R, L, C). The unit of impedance is Ohms (Ω). The impedance in a parallel circuit is gives a s
Z = RXₐ/√(Xₐ² + R²)............................... Equation 1
Where Z = The impedance of the a.c circuit, Xₐ = capacitive reactance, R = resistance.
Given: Xₐ = 962 Ω, R = 1200 Ω
Substituting these values into equation 1,
Z = 962×1200/√(962² + 1200²)
Z = 1154400/√(925444 + 1440000)
Z = 1154400/√(925444+1440000
Z = 1154400/1538
Z = 750.59 Ω
Z≈ 751 Ω
Therefore the impedance of the circuit = 751 Ω
The right option is A. 751 ohm
A box slides downwards at a constant velocity on an inclined surface that has a coefficient of friction uK = 0.58, find the angle of the incline, in degrees?
Answer:
30.11°
Explanation:
[tex]\mu[/tex] = Coefficient of friction = 0.58
g = Acceleration due to gravity = 9.81 m/s²
[tex]\theta[/tex] = Angle of incline
As the forces of the system are conserved we have
[tex]mgsin\theta-\mu mgcos\theta=0\\\Rightarrow mgsin\theta=\mu mgcos\theta\\\Rightarrow sin\theta=\mu cos\theta\\\Rightarrow \mu=\dfrac{sin\theta}{cos\theta}\\\Rightarrow \mu=tan\theta\\\Rightarrow \theta=tan^{-1}\mu\\\Rightarrow \theta=tan^{-1}0.58\\\Rightarrow \theta=30.11^{\circ}[/tex]
The angle of incline is 30.11°
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
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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.