Answer:
a) W = 2635.56 J
b) Wf = 423.27 J
c) c) The Sign of the work done by the frictional force (Wf) is negative (-)
d) W=0
Explanation:
Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .
The formula for calculate the work is :
W = F*d*cosα
Where:
W : work in Joules (J)
F : force in Newtons (N)
d: displacement in meters (m)
α :angle that form the force (F) and displacement (d)
Known data
m = 18.8 kg : mass of the block
F= 156 N,acting at an angle θ = 31.9◦°: angle above the horizontal
μk= 0.209 : coefficient of kinetic friction between the cart and the surface
g = 9.8 m/s²: acceleration due to gravity
d = 19.9 m : displacement of the block
Forces acting on the block
We define the x-axis in the direction parallel to the movement of the cart on the floor and the y-axis in the direction perpendicular to it.
W: Weight of the cart : In vertical direction downaward
N : Normal force : In vertical direction the upaward
F : Force applied to the block
f : Friction force: In horizontal direction
Calculated of the weight of the block
W= m*g = ( 18.8 kg)*(9.8 m/s²)= 184.24 N
x-y components of the force F
Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N
Fy = Fsinθ = 156 N*sin(31.9)° = 82.44 n
Calculated of the Normal force
Newton's second law for the block in y direction :
∑Fy = m*ay ay = 0
N-W+Fy= 0
N-184.24+82,44= 0
N = 184.24-82,44
N = 101.8 N
Calculated of the kinetic friction force (fk):
fk = μk*N = (0.209)*( 101.8)
fk = 21.27 N
a) Work done by the F=156N.
W = (Fx) *d *cosα
W = (132.44 )*(19.9)(cos0°) (N*m)
W = 2635.56 J
b) Work done by the force of friction
Wf = (fk) *d *cos(180°)
Wf = (21.27 )*(19.9) (-1) (N*m)
Wf = - 423.27 J
Wf = 423.27 J :magnitude
c) The Sign of the work done by the frictional force is negative (-)
d) Work done by the Normal force
W = (N) *d *cos(90°)
W = (101.8 )*(19.9) (0) (N*m)
W = 0
The work done by the 156 N force is 2630.77 J. The magnitude of the work done by the friction force is 743.14 J. The work done by the normal force is zero.
Explanation:a. The work done by a force is calculated by multiplying the force applied by the displacement in the direction of the force. In this case, the force is applied at an angle of 31.9 degrees above the horizontal, so we need to find the component of the force in the horizontal direction.
Fx = F * cos(θ) = 156 N * cos(31.9°) = 132.3 N
The work done is given by W = Fx * d = 132.3 N * 19.9 m = 2630.77 J
b. The magnitude of the work done by the force of friction can be calculated using the formula:
Wfriction = friction force * displacement = μ * m * g * d, where μ is the coefficient of kinetic friction, m is the mass of the block, g is the acceleration due to gravity, and d is the displacement. Substituting the given values:
Wfriction = 0.209 * 18.8 kg * 9.8 m/s2 * 19.9 m = 743.14 J
c. The work done by the frictional force is negative, indicating that it acts against the motion of the block.
d. The work done by the normal force is zero because the displacement of the block is perpendicular to the direction of the normal force.
Learn more about Work and Energy here:https://brainly.com/question/16987255
#SPJ11
The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.56 m and 1.00 m, respectively. The respective rotational speeds are 453 rev/min and 4137 rev/min. Calculate the speeds of the tips of both rotors.
Answer:
Explanation:
Given
Diameter of main rotor [tex]d_1=7.56 m[/tex]
Tail rotor [tex]d_2=1 m[/tex]
[tex]N_1=453 rev/min[/tex]
[tex]N_2=4137 rev/min[/tex]
[tex]\omega =\frac{2\pi N}{60}[/tex]
[tex]\omega _1=\frac{2\pi 453}{60}=47.44 rad/s[/tex]
[tex]\omega _2=\frac{2\pi 4137}{60}=433.28 rad/s[/tex]
Speed of the tip of main rotor[tex]=\omega _1\times r_1=47.44\times \frac{7.56}{2}=179.32 m/s[/tex]
Speed of tail rotor[tex]=\omega _2\times r_2=433.28\times \frac{1}{2}=216.64 m/s[/tex]
The electric motor of a model train accelerates the train from rest to 0.740 m/s in 30.0 ms. The total mass of the train is 560 g. Find the average power delivered to the train during its acceleration.
Answer:
= 5.1 W
Explanation:
time (t) = 30 ms = 0.03 s
mass (m) = 560 g = 0.56 kg
initial velocity (U) = 0 m/s
final velocity (V) = 0.74 m/s
power = \frac{work done}{t} = \frac{f x d}{t} = f x v = m x a x v
m x a x v = m x \frac{V-U}{t} x \frac{V + U}{2}
m x \frac{V-U}{t} x \frac{V + U}{2} = 0.56 x \frac{0.74 - 0}{0.03} x \frac{0.74+0}{2}
= 5.1 W
A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with an angular speed of 4.0 rad/s. neglecting friction and air resistance, find (a) the rod’s kinetic energy at its lowest position, (b) and how far above that position the center of mass rises.
Final answer:
The rod's kinetic energy at its lowest position is 1.26 J, calculated using the formula for rotational kinetic energy. The center of mass of the rod rises to a height of approximately 30.6 cm above the lowest point, using the principle of conservation of mechanical energy to equate potential and kinetic energies.
Explanation:
A thin rod acting as a pendulum is an example of physical principles related to rotational motion and conservation of energy. To find the rod's kinetic energy at its lowest position, we use the formula for rotational kinetic energy, K = (1/2)Iω2, where 'I' is the moment of inertia and 'ω' is the angular speed. For a rod rotating about one end, the moment of inertia, 'I', is given by (1/3)ml2, with 'm' being the mass of the rod and 'l' its length.
We can find the kinetic energy of the rod at its lowest position, where the angular speed is 4.0 rad/s: K = (1/2)Iω2 = (1/2)(1/3)(0.42 kg)(0.75 m)2(4.0 rad/s)2. After calculation, this gives K = 1.26 J (Joules).
To determine how far above the lowest position the center of mass rises, we have to consider the conservation of mechanical energy. At the highest point, all kinetic energy is converted into potential energy (assuming no energy losses), so we have mgh = K, where 'g' is the acceleration due to gravity, 'h' is the height gained, and 'm' is the mass of the rod. Therefore, h = K / (mg). Plugging in the values, we get h = 1.26 J / (0.42kg * 9.81 m/s2) = 0.306 m, or approximately 30.6 cm above the lowest position.
A molecule with a seesaw molecular geometry has a bond angle of
Answer
A molecule with seesaw molecular geometry has a bond angle of 90° and 120°.
Seesaw is a type of molecular geometry where there are four bonds attached to a central atom.
Bond is observed in the shape of playground seesaw that is why it is known as a seesaw bond.
Four bonds attached to the central bond formed results as tetrahedral or square planar geometry.
In see-saw, shape maximizes the bond angle of the lone pair and the other atoms in the molecule.
lone pair in the equatorial position offer 120 and 90-degree bond angles whereas bond angle will be 90-degree bond angles if placed at axial position.
A molecule with seesaw molecular geometry has bond angles that are less than 120 degrees for equatorial positions and less than 90 degrees for axial positions, due to lone pair repulsion. Hence the correct option is 1.
Bond Angles in Seesaw Molecular Geometry
A molecule with a seesaw molecular geometry has four nuclei and one lone pair of electrons. This molecular structure is based on a trigonal bipyramidal geometry but with one equatorial position occupied by a lone pair. As a result, the bond angles are adjusted due to electron pair repulsion.
The bond angles for a seesaw geometry are:
Less than 120 degrees for equatorial bonds, as the lone pair causes slight compression of these angles.Less than 90 degrees for axial bonds due to the same repulsive effects from the lone pair.Approximately 173 degrees between the two axial positions, slightly less than the ideal 180 degrees due to lone pair repulsion.The correct question is:
A molecule with a seesaw molecular geometry has a bond angle of
1. <120 for equatorial bonds and <90 for axial bonds.
2.180
3. <90
4. 120 for equatorial bonds and 90 for axial bonds.
5.120
A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a surface area of 7.70 m2. The total power it absorbs from the star is 3800 W.
Assuming the surface is an ideal absorber and radiator, calculate the equilibrium temperature of the asteroid (in K)
Answer:
Temperature will be 305 K
Explanation:
We have given The asteroid has a surface area [tex]A=7.70m^2[/tex]
Power absorbed P = 3800 watt
Boltzmann constant [tex]\sigma =5.67\times 10^{-8}Wm/K^4[/tex]
According to Boltzmann rule power radiated is given by
[tex]P=\sigma AT^4[/tex]
[tex]3800=5.67\times 10^{-8}\times 7.70\times T^4[/tex]
[tex]T^4=87.0381\times 10^8[/tex]
[tex]T=305K[/tex]
So temperature will be 305 K
When output is below potential and the policy rate has hit the floor of zero, the resulting fall in inflation leads to ________ real interest rates, which ________ output further, which causes inflation to fall further.
Answer:
high real rate of interest, depressing
Explanation:
If output is less than potential and the rate of policy has fall to the ground of zero, the subsequent decline in inflation results in high real rate of interest which further depressing output, causing inflation to even further decline.
If the actual output is less than potential output, demand or supply has fallen, resulting in job and output declines
A very loud train whistle has an acoustic power output of 100 W. If the sound energy spreads out spherically, what is the intensity level in dB at a distance of 100 meters from the train?
The intensity level in dB at a distance of 100 meters from the train is 98.46 dB.
Explanation:To calculate the intensity level in dB at a distance of 100 meters from the train, we need to first calculate the intensity of sound at that distance.
Since the sound energy spreads out spherically, the intensity decreases with the square of the distance from the source.
Using the formula for intensity, I = P/4πr², where P is the acoustic power output and r is the distance from the source, we can calculate the intensity at 100 meters:
I = 100W / (4π * (100m)²) = 0.0796 W/m².
Now, we can use the formula for the intensity level in dB, β = 10log(I/I₀), where I₀ is the reference intensity (10⁻¹² W/m²).
Substituting the values, β = 10log(0.0796 / 10⁻¹²) = 98.46 dB.
Learn more about sound intensity here:https://brainly.com/question/15089336
#SPJ3
Increasing the mass attached to a spring will increase the angular frequency of its vibrations. True or False
Answer: False
Explanation:
Using the angular frequency formula derived from Hooke's law we have :
[tex]\omega =\sqrt{\frac{ k }{ m }} [/tex]
[tex]{\omega}^{ 2 }=\frac {k}{m}[/tex]
where k- is Hooke's constant
m- is mass attached to a spring
According to the formula the mass is inversely proportional to the square of the angular frequency angular which means that increasing the mass attached to the spring will decrease the angular frequency.
The statement is false. Increasing the mass attached to a spring will decrease, not increase, the angular frequency of its vibrations due to the inverse proportional relationship between mass and angular frequency in a mass-spring system.
Explanation:The statement, 'Increasing the mass attached to a spring will increase the angular frequency of its vibrations', is false. According to the principles of simple harmonic motion, the angular frequency of a mass-spring system fundamentally depends on the mass and the force constant (spring constant) of the system. However, the relationship is inversely proportional; meaning that increasing the mass will actually decrease the angular frequency. This is because the angular frequency (ω) is calculated as the square root of the ratio of the force constant (k) to the mass (m: ω = sqrt(k/m). Hence, increasing the mass causes a decrease in the angular frequency, which consequently increases the period of the motion - the time it takes for one complete cycle of vibration to pass a given point.
Learn more about Simple Harmonic Motion here:https://brainly.com/question/28208332
#SPJ11
) The sprocket assembly on a bicycle consists of a chain and two sprockets, one on the pedal and the other on the rear wheel. If the sprocket on the pedal is 6 inches in diameter, the sprocket on the rear wheel is 4 inches in diameter, and the rear wheel is 26 inches in diameter, how fast is the bicycle traveling in miles per hour when the cyclist is pedaling at the rate of 1.1 revolutions per second? Round your answer to the nearest tenth mph.
Answer:
7.8 mph
Explanation:
Rate of cycling = 1.1 rev/s
Rear wheel diameter = 26 inches
Diameter of sprocket on pedal = 6 inches
Diameter of sprocket on rear wheel = 4 inches
Circumference of rear wheel = [tex]\pi d=26\pi[/tex]
Speed would be
[tex]\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s[/tex]
Converting to mph
[tex]1\ inch/s=\frac{1}{63360}\times 3600\ mph[/tex]
[tex]134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph[/tex]
The Speed of the bicycle is 7.8 mph
To determine the bicycle speed, calculate the rear wheel's circumference, use the gear ratio to find the rear wheel's rotation rate, and then convert the distance traveled per second into miles per hour. The resulting speed is 17.3 mph when pedaling at 1.1 revolutions per second.
Explanation:The student's question involves calculating the speed of a bicycle based on the diameters of the sprockets on the pedal and rear wheel, the diameter of the rear wheel, and the pedaling rate in revolutions per second. To find the speed in miles per hour, we first calculate the circumference of the rear wheel, which gives us the distance traveled per revolution. We then use the gear ratio to find the effective revolutions of the rear wheel based on the pedaling speed. The total distance traveled per second is then converted to miles per hour.
The diameter of the rear wheel is 26 inches, so its circumference is π(26 inches) ≈ 81.68 inches. The gear ratio is the diameter of the pedal sprocket (6 inches) over the diameter of the rear wheel sprocket (4 inches), giving us a 1.5:1 gear ratio. For every pedal revolution, the rear wheel rotates 1.5 times. Pedaling at 1.1 revolutions per second, the rear wheel rotates at 1.5 * 1.1 = 1.65 revolutions per second. The distance traveled per second is then 81.68 inches * 1.65 ≈ 134.77 inches per second.
To convert inches per second to miles per hour, we use the conversion factors 12 inches per foot and 5280 feet per mile, and 3600 seconds per hour. Thus, the bike speed is (134.77 inches/second) * (1 foot/12 inches) * (1 mile/5280 feet) * (3600 seconds/hour) ≈ 17.26 miles per hour. Rounded to the nearest tenth, the bicycle is traveling at 17.3 mph.
An unknown mass is hung from a very light spring with a spring constant of 13.2 N/m. Neglect the mass of the spring. If the period of oscillation of the mass-spring system is 2.60 s, then the unknown mass is kg.
Amplitude is the maximum displacement of the mass from the _______.
end point equilibrium point top point vibrational point
A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of +14.8 cm at time, t = 0.00 s. The displacement from equilibrium of the mass at time, t = 1.25 s is cm.
A ball of mass 1.55 kg is hung from a spring which stretches a distance of 0.3800 m. The spring constant of the spring (a scalar quantity) is N/m
A mass m on a spring with a spring constant k is in simple harmonic motion with a period T. If the same mass is hung from a spring with a spring constant of 3k, the period of the motion will be _______.
increased by a factor of the square root of 3 decreased by a factor of the square root of 3 decreased by a factor of 3 increased by a factor of 3
Answer:
Explanation:
Time period of oscillation of spring mass system
T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]
m = [tex]\frac{T^2\times k}{4\pi^2}[/tex]
= [tex]\frac{(2.6)^2\times 13.2}{4\pi^2}[/tex]
= 2.26 kg
Amplitude is the maximum displacement of the mass from the _equilibrium point .
maximum displacement from equilibrium of +14.8 cm at time, t = 0.00 s
x(t) = 14.8cosωt
ω = 2πn = 2π x 9.73 = 61.1 rad /s
x(t) = 14.8cos61.1 t
when t = 1.25
x(t) = 14.8cos61.1 x 1.25
= 14.8cos61.1 x 1.25
= 14.8cos76.38
= 8.22 cm
A ball of mass 1.55 kg is hung from a spring which stretches a distance of 0.3800 m
spring constant
k = mg / x
= 1.55 x 9.8 / .38
= 40 N / m
T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]
so if mass is hung from a spring with a spring constant of 3k
T decreases by a factor of the square root of 3
For the first question, the unknown mass is approximately 0.71 kg. For the second question, the term 'equilibrium point' completes the statement. For the third question, the displacement at time 1.25 seconds is -14.2 cm. For the fourth question, the spring constant k is approximately 40.0 N/m. For the fifth question, the period of the motion will be decreased by a factor of the square root of 3.
Explanation:The unknown mass hung from a spring can be calculated using the formula for the period of a mass-spring system, T = 2π√(m / k), where T is the period, m is the mass, and k is the spring constant. Rearranging for m gives us m = (T²k) / (4π²). Given that T = 2.60 seconds and k = 13.2 N/m, we find m = (2.60^2 * 13.2) / (4*π^2), which gives a mass of approximately 0.71 kg.
The amplitude is the maximum displacement of the mass from the equilibrium point.
The displacement of a harmonically oscillating mass can be found using the equation: x(t) = A*cos(2πft + φ), where A is the amplitude, f is the frequency, and φ is the phase angle. Given the information we have a displacement at 1.25 seconds of -14.2 cm.
The spring constant can be calculated using the equation k = mg / x, where m is the mass, g is gravity (9.81 m/s²), and x is the distance stretched. Given that the mass is 1.55 kg and the distance stretched is 0.3800 m, we find that the spring constant k is approximately 40.0 N/m.
If the same mass is hung from a spring with a spring constant of 3k, the period of the motion will be decreased by a factor of the square root of 3.
Learn more about Harmonic Motion here:https://brainly.com/question/32363393
#SPJ3
An aircraft is loaded 110 pounds over maximum certificated gross weight. If fuel (gasoline) is drained to bring the aircraft weight within limits, how much fuel should be drained?
Answer:
242 lts de fuel
Explanation:
We know density of fuel (is variable) but we can say that is close to 1 Kg/lts.
If we have 110 pounds over maximun certificated gross weight, we need to get rid of that 110 pounds or 110 * 2.2 = 242 Kgs
Now by rule of three
If 1 lt weight 1 kg.
x 242 kgs
x = 242 lts de fuel
To bring the aircraft within the maximum gross weight limits, approximately 18.33 gallons of fuel need to be drained, calculated by dividing the excess weight (110 pounds) by the weight of fuel per gallon (6 pounds/gallon).
To calculate how much fuel needs to be drained to bring an aircraft weight within the maximum certificated gross weight limits, we need to know two pieces of information: the weight by which the aircraft is over its maximum gross weight and the weight of the fuel that needs to be drained.
In this scenario, the aircraft is 110 pounds over the maximum certificated gross weight. In general, aviation gasoline (avgas) weighs about 6 pounds per gallon. To solve this problem, we will divide the excess weight by the weight of the fuel per gallon to find out how many gallons need to be drained.
So, the calculation is as follows:
Excess weight: 110 pounds.
Weight of fuel per gallon: 6 pounds/gallon.
Number of gallons to be drained = Excess weight / Weight of fuel per gallon.
Number of gallons to be drained = 110 pounds / 6 pounds/gallon.
Number of gallons to be drained = 18.33 gallons (approximately).
Therefore, to bring the aircraft back within weight limits, approximately 18.33 gallons of fuel need to be drained.
Calculate the angular velocity of the earth in its orbit around the sun.
Answer:
[tex]0.0172rad/day=1.99x10^{-7}rad/second[/tex]
Explanation:
The definition of angular velocity is as follows:
[tex]\omega=2\pi f[/tex]
where [tex]\omega[/tex] is the angular velocity, and [tex]f[/tex] is the frequency.
Frequency can also be represented as:
[tex]f=\frac{1}{T}[/tex]
where [tex]T[/tex] is the period, (the time it takes to conclude a cycle)
with this, the angular velocity is:
[tex]\omega=\frac{2\pi}{T}[/tex]
The period T of rotation around the sun 365 days, thus, the angular velocity:
[tex]\omega=\frac{2\pi}{365days}=0.0172rad/day[/tex]
if we want the angular velocity in rad/second, we need to convert the 365 days to seconds:
Firt conveting to hous
[tex]365days(\frac{24hours}{1day} )=8760hours[/tex]
then to minutes
[tex]8760hours(\frac{60minutes}{1hour} )=525,600minutes[/tex]
and finally to seconds
[tex]525,600minutes(\frac{60seconds}{1minute})=31,536x10^3seconds[/tex]
thus, angular velocity in rad/second is:
[tex]\omega=\frac{2\pi}{31,536x10^3seconds}=1.99x10^{-7}rad/second[/tex]
At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in simple harmonic motion, with a period of 12.5 h. How long does it take for the water to fall a distance 0.250d from its highest level?Number__________Unit____________.
Answer:
2.08335 hours
Explanation:
T = Time period = 12.5 h
Angular frequency is given by
[tex]\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{12.5}\\\Rightarrow \omega=0.50265\ rad/h[/tex]
The distance moved from highest to lowest level is given by
[tex]d=2x_{m}\\\Rightarrow x_m=\frac{d}{2}\\\Rightarrow x_m=0.5d[/tex]
At the ocean surface [tex]x_m=0.25d[/tex]
[tex]x_0=x_mcos(\omega t+\phi)[/tex]
[tex]\phi[/tex] = Phase constant = 0 as clock is started at [tex]x_0=x_m[/tex]
[tex]0.25d=0.5dcos(0.50265t)\\\Rightarrow cos(0.50265t)=\frac{0.25}{0.5}\\\Rightarrow 0.50265t=cos^{-1}0.5\\\Rightarrow t=\frac{1.047}{0.50265}\\\Rightarrow t=2.08335\ h[/tex]
The time taken for the water to fall the distance is 2.08335 hours
This star map shows stars as we see them in our sky from Earth, centered around the constellation Canis Major. Larger dots represent brighter stars, and a few of the brightest stars are identified. From this view alone, what can you conclude about Sirius?
Answer:
It has the greatest apparent brightness of any star in this region of the sky.
Explanation:
We can conclude that Sirius is brighter than the other stars by the question saying "Larger dots represent brighter stars, and a few of the brightest stars are identified."
It can be viewed that it has the greatest apparent brightness of any star in this region of the sky.
What is Sirius?The brightest star in the sky at night is Sirius. It gets its name from the Greek word for 'glowing' or 'scorching.'
Sirius is known as the Dog Star because it is the brightest star in the Canis Major constellation. It's extremely bright because it's one of the stars closest to our sun. The name could be derived from ancient Egypt.
This star map depicts the stars as seen from Earth, centered on the constellation Canis Major. The brighter stars are represented by larger dots, and a few of the brightest stars are identified.
It has the highest apparent brightness of any star in this region of the sky, as can be seen.
Thus, this can be concluded regarding Sirius.
For more details regarding Sirius, visit:
https://brainly.com/question/29388328
#SPJ6
Your question seems incomplete, the missing image is attached below:
A mover pushes a 30.0 kg crate across a wooden floor at a constant speed of 0.75 m/s. If the coefficient of static friction for wood-on-wood is 0.20, what is the normal force exerted by the floor on the crate?
Answer:
294.3 N
Explanation:
In this situation, we are told that the crate is not accelerating in the horizontal plane. But also it is not accelerating in the vertical plane. Meaning that the sum of all vertical forces add up to zero.
Fnet = ma
Weight + Normal force = mass * acceleration
-(30 kg * 9.81 m/s²) + Normal force = 30.0 kg * 0 m/s²
Normal force = 294.3 N
The normal force exerted by the floor on the crate is equal to the weight of the crate when it's at rest. It can be calculated by multiplying mass and acceleration due to gravity. For a crate with a mass of 30.0 kg, the normal force would be 294 N.
Explanation:The key to answering this question involves understanding how friction and forces work. In this scenario, the crate is not accelerating, so the forces acting on it are balanced, meaning the upward normal force equals the downward gravitational force. The normal force exerted by the floor on the crate is determined by the weight of the crate, which can be calculated using the formula for weight, W = mg (mass times gravity). Here, the mass m is 30.0 kg and the acceleration due to gravity g is 9.8 m/s².
So, we can substitute these values into the formula to find the weight: W = (30.0 kg) * (9.8 m/s²) = 294 N. This weight also equals the normal force since the crate is not moving vertically, so the normal force exerted by the floor on the crate is 294 N.
Learn more about Normal Force here:https://brainly.com/question/30436126
#SPJ3
The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope. The sleds move on an icy surface. The dogs exert a 240-N force on the rope attached to the front sled. The front rope pulls horizontally. Find the acceleration of the sleds.
Answer:
Acceleration, [tex]a=1.2\ m/s^2[/tex]
Explanation:
Given that,
The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope, m = 100 kg
Force exerted by the doges on the rope attached to the front sled, F = 240 N
To find,
The acceleration of the sleds.
Solution,
Let a is the acceleration of the sleds. The product of mass and acceleration is called force. Its expression is given by :
F = ma
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{240\ N}{2\times 100\ kg}[/tex] (m = 2m)
[tex]a=1.2\ m/s^2[/tex]
So, the acceleration of the sleds is [tex]1.2\ m/s^2[/tex].
A gas, behaving ideally, fills a fixed volume container at a pressure P₁ and at a temperature T₁. The temperature of the container is changed to T₂. Using Avogadro’s, Charles’, or Boyle’s law, develop an expression that would solve for the new pressure P₂.a. V₁/T₁=V₂/T₂b. P₁*V₁=P₂*V₂c. P₁/T₁=P₂/T₂d. P₁*T₁=P₂*T₂
Answer:
c. P₁/T₁=P₂/T₂
Explanation:
neither Avogadro’s, Charles’, or Boyle’s law formula can be used, since some parameters like volume is not given,
to find P₂, given P₁, T₁, and T₂ we will therefore use Gay-lussac's law.
gay lussacs law state that, provided volume is kept constant, pressure is directly proportional to temperature.
the volume volume is said to be filled, i.e its is kept constants when temperature is change
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 5.0 m/s. Once free of this area, it speeds up to 12 m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. What is the final speed?
Answer:
26m/s
Explanation:
Assuming that the acceleration is constant, we can start by calculating the train speed when it's free of the congested area:
[tex]a = \frac{\deltav}{\deltat} = \frac{12 - 5}{8} = \frac{7}{8} = 0.875 m/s^2[/tex]
Then with the same acceleration we can find out the final speed:
[tex]v = v_0 + at = 12 + 0.875*16 = 26m/s[/tex]
2. Which of the following is an example of a passive immunization? A. Antibodies against rabies given to someone who was bitten by a potentially rabid dog B. Heat-killed flu antigen grown in chicken eggs C. Live viral antigen given on a sugar cube to protect against polio D. Both A and B. E. Both B and C.
Answer:
A. Antibodies against rabies given to someone who was bitten by a potentially rabid dog
Explanation:
passive immunization is the immunity or protection against certain infectious diseases by injecting antigens or antibodies to the blood stream or certain antibodies can be gotten through colostrum or breast milk.
for option A. Antibodies against rabies given to someone who was bitten by a potentially rabid dog. the immunity against rabies infection is administer to the person
For option B:is only telling us the type of infection but not stating the immunity administered
option c can be fairly correct, because its another method of immunization but its not passive immunization
Suppose you are asked to find the amount of time t, in seconds, it takes for the turntable to reach its final rotational speed. Which of the following equations could you use to directly solve for the numerical value of t?
A) θ=θ0+ω0t+(1/2)αt2
B) ω=ω0+αt
C) ω2=ω02+2α(θ−θ0)
D) More information is needed before t can be found.
Answer:
option (D)
Explanation:
Here initial rotation speed is given, final rotation speed is given and asking for time.
If we use
A) θ=θ0+ω0t+(1/2)αt2
For this equation, we don't have any information about the value of angular displacement and angular acceleration, so it is not useful.
B) ω=ω0+αt
For this equation, we don't have any information about angular acceleration, so it is not useful.
C) ω2=ω02+2α(θ−θ0)
In this equation, time is not included, so it is not useful.
D) So, more information is needed.
Thus, option (D) is true.
The equation to directly solve for time t in this context is ω = ω₀ + αt, as it includes all necessary variables except for t, making it easy to solve algebraically. Hence, option(b) is correct.
The correct equation to use in this problem is ω = ω₀ + αt.
By rearranging this equation, you can directly find the value of t using given angular velocities and acceleration. This is applicable in analyzing rotational motion in physics.
To solve for the time t, we can rearrange the equation algebraically:
[tex]\\t = \frac{\omega - \omega_0}{\alpha}\\[/tex]
This equation directly relates
the final angular velocity (ω), initial angular velocity (ω₀), and angular acceleration (α) to time t.Given the initial and final angular velocities and the angular acceleration, you can plug these values in to find t.
According to the principle of conservation of momentum, A. the amount of momentum of all the objects in the universe is constant. B. in an open system, the total amount of momentum of the objects is conserved. C. in a closed system, the speed of any two colliding objects will remain constant. D. in a closed system, an object's momentum before a collision will equal its momentum after the collision.
The principle of conservation of momentum states that in a closed system, an object's momentum before a collision will equal its momentum after the collision.
Explanation:The principle of conservation of momentum states that in a closed system, the total momentum of the objects before a collision will be equal to the total momentum of the objects after the collision. This means that in a closed system, an object's momentum before a collision will equal its momentum after the collision.
A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx3 , where c = 8.0 N/m3 . The particle’s speed at A, where xA = 1.0 m, is 6.0 m/s. What is its speed at B, where xB = −2.0 m?
Answer:4.58 m/s
Explanation:
Given
mass of Particle [tex]m=4 kg[/tex]
[tex]F=-cx^3[/tex]
[tex]a=\frac{F}{m}[/tex]
[tex]a=-\frac{cx^3}{m}[/tex]
[tex]a=-\frac{8x^3}{4}[/tex]
[tex]a=-2x^3[/tex]
[tex]v\frac{\mathrm{d} v}{\mathrm{d} x}=-2x^3[/tex]
[tex]vdv=-2x^3dx[/tex]
integrating
[tex]\int_{6}^{v_b}vdv=\int_{1}^{-2}-2x^3dx[/tex]
[tex]\frac{v_b^2-6^2}{2}=-\frac{1}{2}\left [ \left ( -2\right )^4-\left ( 1\right )^4\right ][/tex]
[tex]\frac{v_b^2-36}{2}=-0.5\times 15[/tex]
[tex]v_b^2=36-15[/tex]
[tex]v_b=\sqrt{21}[/tex]
[tex]v_b=4.58 m/s[/tex]
A particle of mass 4.0 kg is constrained to move along the x-axis under a single force F(x) = −cx³ The speed of the particle at point B is 5.3 m/s.
The potential energy at any point x is given by:
F(x) = -cx³
U(x) = -∫F(x) dx
U(x)= ∫cx³dx
U(x) = c/4 × x⁴ + C
The total mechanical energy E remains constant:
E = K + U
E_A = K(A) + U(A)
E_A = (1/2)mv(A)²+ c÷4 × x(A)⁴ + C
At point B, the potential energy U(B) is:
U(B) = c/4 × x(B)⁴ + C
The total mechanical energy at point B is:
E(B) = K(B) + U(B)
Since the total mechanical energy is conserved, we have:
E(A) = E(B)
(1/2)mv(A)² + c÷4 × x(A)⁴ + C = (1/2)mv(B)² + c÷4 × x(B)⁴ + C
(1/2)(4.0)(6.0)² + ((8.0) ÷ 4) × (1.0)⁴ = (1/2)(4.0)v(B)² + ((8.0) ÷ 4) × (-2.0)⁴
72.0 + 2.0= 2.0 × v(B)² + 16.0
56.0 J = 2.0× v(B)²
v(B)² = 28.0
v(B) = ±5.3 m/s
Since speed cannot be negative, the speed of the particle at point B is 5.3 m/s.
To know more about the speed:
https://brainly.com/question/30462853
#SPJ6
A man lifts a 25.9 kg bucket from a well and does 5.92 kJ of work.The acceleration of gravity is 9.8 m/s^2 . How deep is the well? Assume that the speed of the bucket remains constant as it is lifted. Answer in units of m.
The depth of the well can be calculated using the formula for gravitational potential energy, taking the work done to lift the bucket as equal to the gravitational potential energy gained by the bucket. This gives us the equation: Depth of the well (h) = Work done (W) / (mass (m) * acceleration due to gravity (g)), which we solve to find the depth.
Explanation:The physics concept involved in this question is called gravitational potential energy, which refers to the energy an object possesses because of its higher position in a gravity field.
In such cases, the energy can be calculated using the formula for work, which is force times distance. When you're lifting an object, the force you're countering is the force of gravity on the object, which is its mass multiplied by the acceleration due to gravity (9.8 m/s^2). This amount of work is equivalent to the gravitational potential energy gained by the bucket when it is lifted.
So, the formula we'll use is: Work = mgh (mass * gravity * height)
We know the work done (5.92 kJ or 5920 J), the mass (25.9 kg) and the acceleration due to gravity (9.8 m/s^2). We want to find height (h), which in this context, means the depth of the well.
Rearranging the formula and solving for h gives us: h = Work / (m*g). By substituting the known values, h = 5920 J / (25.9 kg * 9.8 m/s^2).
By solving this, we get the depth of the well.
Learn more about Gravitational Potential Energy here:https://brainly.com/question/23134321
#SPJ11
In order from lowest urgency to highest, which sequence properly ranks the product categories issued by the National Weather Service? Note that not all product categories are issued for all hazard types.
A. Watch, Advisory, Outlook, Warning
B. Outlook, Watch, Advisory, Warning
C. Outlook, Advisory, Watch, Warning
D. Advisory, Outlook, Watch, Warning
Answer:
In order from lowest urgency to highest, the sequence which properly ranks the product categories issued by the National Weather Service are as follows:
B- Outlook, Watch, Advisory, Warning.
Explanation:
An outlook for a hazardous weather describes the potential hazardous weather of concern in day 1 through 7. There are total two segments of the outlook, one for the marine zones and the second for the land based zones.
A watch is issued when there is the possibility of hazardous weather within 48 hours, it does not guarantee that a hazardous weather is going to come, it just reminds the possibilities of any such weather to come.
Advisory comes third in the urgency ranking, we can explain this through an example: a winter weather advisory may be issued for amount of freezing rain or when there are chance of 2 to 4 inches of snow. And, in winter weather, an warning may be issued when there is one fourth inch or more of ice accumulation.
The National Weather Service orders their product categories based on urgency: Outlook, Watch, Advisory, and finally Warning. An outlook is prospective, a watch means conditions are favorable, an advisory means the event is likely, and a warning means the event is expected within 24 hours.
Explanation:The National Weather Service issues their product categories in a sequence based on the impending urgency of the weather condition. The correct order from lowest to highest urgency is: Outlook, Watch, Advisory, and then Warning.
An Outlook is used when hazardous weather is possible over the next week. A Watch is used when conditions are favorable for a hazard to occur; it typically means there is a possibility of the occurrence within the next 48 hours. An Advisory is used when a hazardous event is occurring, imminent or likely, but it is considered less severe than a warning.
A Warning is issued when a hazardous weather event is occurring, imminent or has a very high probability of occurring; it typically means the event is expected within the next 24 hours. Therefore, the correct answer is B. Outlook, Watch, Advisory, Warning.
Learn more about National Weather Service Product Categories here:https://brainly.com/question/14035979
#SPJ12
A 7.29 kg bowling ball moves at 1.73 m/s. How fast must a 2.28 g Ping-Pong ball move so that the two balls have the same kinetic energy? Answer in units of m/s.
The speed of the ping-pong ball must be 97.8 m/s
Explanation:
The kinetic energy of an object is the energy possessed by an object due to its motion. It is calculated as
[tex]E_k = \frac{1}{2}mv^2[/tex]
where
m is the mass of the object
v is its speed
Here we have a bowling ball that has
Mass: m = 7.29 kg
Speed: v = 1.73 m/s
So its kinetic energy is
[tex]E_k = \frac{1}{2}(7.29)(1.73)^2=10.9 J[/tex]
Now we want the ping-pong ball to have the same kinetic energy, so
[tex]E_k = 10.9 J[/tex]
its mass is
[tex]m=2.28 g = 2.28\cdot 10^{-3} kg[/tex]
So, we can find the speed it should have by re-arranging the equation:
[tex]v=\sqrt{\frac{2E_k}{m}}=\sqrt{\frac{2(10.9)}{2.28\cdot 10^{-3}}}=97.8 m/s[/tex]
Learn more about kinetic energy:
brainly.com/question/6536722
#LearnwithBrainly
Technician A says that magnetic lines of force can be seen by placing iron filings on a piece of paper and then holding them over a magnet. Technician B says that the effects of magnetic lines of force can be seen using a compass. Which technician is correct?
Answer:
Both A and B are correct
Explanation:
Magnetic line of force can be seen by placing iron fillings on a piece of paper and then hold a magnet over them, the iron fillings align themselves in the magnetic field of the magnetic along the line of forces thus showing the loops of the magnetic line of forces.Also, we know that unlike poles attract where as like poles repel each other. This fact can be proved when we hang a bar magnet it aligns itself in the North-South direction because of the the magnetic field due to the alignment of the geographical North-South poles and the poles of the bar magnet and the compass works the same way.Thus both the technicians are correct.
Answer:
Technician A is correct
Magnetic lines of force can be seen by placing iron filings on a piece of paper and then holding them over a magnet.
Explanation:
As we know that the magnetic field lines are always originated out from the North pole of magnet and then terminate at south pole of the magnet
Here we know that when magnetic material aligns itself in the direction of external magnetic field
So here when we put the iron filings on the cardboard and then place a magnet near it then all the iron filing will align itself in the direction of the magnetic field
Now if we place a compass near the magnet then it will aligns itself in the tangential direction of the magnetic field at the same position
So here we can say that Technician A is correct
Which type of intermolecular attractive force operates between all molecules?
Answer:
Hydrogen bonding
Ionic bonding
Non- induced dipole forces
Ion dipole forces
Van-der wall forces
Explanation:
Intermolecular forces are the forces which operates at molecular level they may be attractive or repulsive in nature. But the given question ask only about attractive molecular forces. These are of basically five types:-
1. Hydrogen bonding- It is the force of attraction between hydrogen atom and an electro-negative atom.
2 .Ionic bonding- It is the force of attraction between cationic and anionic pair.
3.Non - Induced dipole forces- It is the dipole- dipole interaction between permanent electromagnetic materials.
4.Ion-dipole forces- These bonds are stronger than hydrogen bond and also stronger than dipole-dipole bond.
Van-der walls forces- It is the force of attraction between uncharged molecules.
The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume that the photons spread out spherically. The area of a sphere is 4????????2.
Answer:
The distance from the radio station is 0.28 light years away.
Solution:
As per the question:
Distance, d = 4 ly
Frequency of the radio station, f = 854 kHz = [tex]854\times 10^{3}\ Hz[/tex]
Power, P = 50 kW = [tex]50\times 10^{3}\ W[/tex]
[tex]I_{p} = 1\ photon/s/m^{2}[/tex]
Now,
From the relation:
P = nhf
where
n = no. of photons/second
h = Planck's constant
f = frequency
Now,
[tex]n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s[/tex]
Area of the sphere, A = [tex]4\pi r^{2}[/tex]
Now,
Suppose the distance from the radio station be 'r' from where the intensity of the photon is [tex]1\ photon/s/m^{2}[/tex]
[tex]I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}[/tex]
[tex]1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}[/tex]
[tex]r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m[/tex]
Now,
We know that:
1 ly = [tex]9.4607\times 10^{15}\ m[/tex]
Thus
[tex]r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly[/tex]
Two parallel conducting plates are connected to a constant voltage source. The magnitude of the electric field between the plates is 2,000 N/C. If the voltage is doubled and the distance between the plates is reduced to 1/5 the original distance, the magnitude of the new electric field is
A) 800 N/C
B) 1600 N/C
C) 2400 N/C
D) 5000 N/C
E) 20000 N/C
Answer:
E) E₂= 20000 N/C
Explanation:
The electric field between two parallel conductive plates is thus calculated
E= V/d
Where:
E: Electric field (N/C)
V: voltage (V)
d : distance between the plates (m)
Problem development
E₁= V₁/d₁ = 2000 N/C
E₂= V₂/d₂
[tex]E_{2} = \frac{2V_{1} }{\frac{d_{1} }{5} }[/tex]
[tex]E_{2} = 10(\frac{V_{1} }{d_{1} })[/tex]
E₂= 10* (2000)
E₂= 20000 N/C
Final answer:
When the voltage between two parallel conducting plates is doubled and the distance between them is reduced to 1/5 of the original distance, the magnitude of the new electric field is 20,000 N/C.
Explanation:
The question relates to the electric field between two parallel conducting plates connected to a voltage source. The original electric field's magnitude is given as 2,000 N/C, and we're asked to find the new electric field magnitude when the voltage is doubled and the distance between the plates is reduced to 1/5 of the original distance. The electric field (E) between two parallel plates is given by the equation E = V/d, where V is the potential difference (voltage) between the plates, and d is the distance between them. When the voltage is doubled, V becomes 2V, and when the distance is reduced to 1/5, d becomes d/5. Thus, the new electric field E' = (2V)/(d/5) = 10(V/d) = 10E. So, the new electric field strength is 10 * 2,000 N/C = 20,000 N/C, which corresponds to option E).
Which fossil fuel is produced as a by-product that occurs when bacteria decompose organic material under anaerobic conditions?
Bio-gas is the naturally produced fossil fuel, a by-product when bacteria decompose organic material under anaerobic conditions.
Explanation:Organic matter particularly waste material is broken down by bacteria through fermentation in an environmental condition without any presence of oxygen. This process of decomposition leads to formation of bio-gas with "carbon dioxide and methane" in a 2:3 ratio.
The above biological process is termed as bio-digestion or anaerobic digestion. Methane is flammable and thus bio-gas can be used as "energy source", a waste-to-energy transformation. The remaining decomposed matter is ideal as manure for plants due to its rich nutrient level.