Given:
Mass of object = 37 kg
Slope = 24 degrees
Let's solve the following questions.
A. Normal force acting on him.
To find the normal force acting on Matt, apply the formula below:
[tex]mg\cos (\theta)-N=0[/tex]Where N is the force.
m is the mass = 37 kg
g is the gravitational acceleration = 9.8 m/s^2
Θ = 24 degrees.
Thus, we have:
[tex]\begin{gathered} 37\ast9.8\cos (24)-N=0 \\ \\ 37\ast9.8(0.9135)-N=0 \\ \\ 331.25-N=0 \end{gathered}[/tex]Add N to both sides:
[tex]\begin{gathered} 331.25-N+N=0+N \\ \\ 331.25=N \\ \\ N=331.25N \end{gathered}[/tex]Therefore, the normal force acting on Matt is 331.25 N
B. Acceleration down the hill.
The acceleration down the hill will be the opposite side(side opposite the angle).
To find the acceleration down the hill, apply the formula below:
[tex]a=g\sin \theta[/tex]Thus, we have:
[tex]\begin{gathered} a=9.8\sin 24 \\ \\ a=9.8(0.4067) \\ \\ a=3.99m/s^2 \end{gathered}[/tex]Therefore, the acceleration down the hill is 3.99 m/s
C. Given:
Coefficient of friction between Matt and the hill is = 0.31
Let's find the acceleration assuming there is friction.
To find the acceleration, we have the formula:
[tex]Fg=m\ast a_g[/tex]Where:
Fg is the force due to gravity
m is the mass of Matt
Thus, we have:
[tex]Fg=37\ast3.99=147.5N[/tex]Also, let's find the force due to fricton using the formula:
[tex]F_f=uN[/tex]Where:
u is the coeficient of friction = 0.31
N is the normal force
We have:
[tex]F_f=0.31\ast147.5=45.7N[/tex]Thus, we have the formula:
[tex]F_g-F_f=m\ast a[/tex]Let's solve for a:
[tex]\begin{gathered} 147.5-45.7=37\ast a \\ \\ 101.8=37\ast a \\ \\ a=\frac{101.8}{37} \\ \\ a=2.75m/s^2 \end{gathered}[/tex]Therefore, the acceleration assuming there is friction is 2.75 m/s^2
ANSWER:
[tex]\begin{gathered} A\text{. 331.25 N} \\ \\ B.3.99m/s^2 \\ \\ \text{ C. 2.75 m/s}^2 \end{gathered}[/tex]Which of the following is not true of all waves?Waves carry energy from one place to another.Waves need a mechanical medium through which to propogate.Waves can be used to carry infomation across distances.For a wave propagating in a given medium, wavelength is inversely proportional to frequency.
Given:
Some statements about the waves
To find:
The statement which is not true of all waves
Explanation:
The waves carry energy from one place to another. The wavelength of a wave is,
[tex]\lambda=\frac{v}{f}[/tex]Here, 'v' is the speed of the wave and 'f' is the frequency of the wave.
So, the wavelength is inc=versely proportional to frequency in a given medium.
The waves line electromagnetic waves do not need any mechanical medium to propagate, they can travel without any media.
Hence, the statement "Waves need a mechanical medium through which to propogate" is not true.
Examine the diagram of the electromagnetic spectrum below.
Diagram of electromagnetic spectrum. Gamma waves have wavelengths of approximately 10 to the power of negative 12 meters. x-rays have wavelengths of approximately 10 to the power of 10 meters. Microwaves have wavelengths of approximately 10 to the power of negative 2 meters. Radio rays have wavelengths of approximately 10 to the power of 3 meters.
If a wave has a wavelength of 0.1 nanometer, it must be a(n)
gamma ray
microwave
radio wave
x-ray
If a wave has a wavelength of 0.1 nanometers, it must be a gamma ray.
High-frequency (or shortest wavelength) electromagnetic radiation with a large amount of energy is known as gamma rays. They can go through most materials. They can only be stopped by something solid, such a big concrete block or a lead block.
Gamma rays have frequencies above 10 Hz and wavelengths below 100 pm. They represent the most powerful type of electromagnetic radiation above 100 keV.
One of the most energetic types of light created in the universe's hotter regions are gamma rays. They are also created by radioactive material in space during supernova explosions.
Ionizing radiation, of which gamma rays are a kind, is quite harmful. Ionizing radiation is high-energy radiation that charges particles by removing electrons from their atoms.
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A5 kg box is at the top of a 2.7 m tall frictionless incline as shown in the diagram. It slides to the bottom of the incline, reaching a speed of 7.3m / s . What is the box's kinetic energy at the bottom of the incline? Whats the boxs ptential energy at the top of the incline?
ANSWER:
Kinetic energy: 133.2 J
Potential energy: 132.3 J
STEP-BY-STEP EXPLANATION:
Gven:
Mass (m) = 5 kg
Height (h) = 2.7 m
Speed (v) = 7.3 m/s
We calculate the kinetic energy and the potential energy using the respective formula in each case, as follows:
[tex]\begin{gathered} E_k=\frac{1}{2}\cdot m\cdot v^2=\frac{1}{2}(5)(7.3)^2=133.2\text{ J} \\ \\ E_p=m\cdot g\cdot h=(5)(9.8)(2.7)=132.3\text{ J} \end{gathered}[/tex]Therefore, the kinetic energy is equal to 133.2 joules and the potential energy is equal to 132.3 joules.
You are moving a dresser that has a mass of 36 kg; its acceleration is 0.5 m/s2. What is the force being applied?72 N18 N4.8 N35.5 N
ANSWER
18 N
EXPLANATION
Given:
• The dresser's mass, m = 36 kg
,• The dresser's acceleration, a = 0.5 m/s²
Find:
• The applied force, F
By Newton's second law of motion, the net force acting on an object is equal to the product between the object's mass and its acceleration,
[tex]F=ma[/tex]Assuming that there is no friction between the dresser and the floor,
[tex]F=36kg\cdot0.5m/s^2=18N[/tex]Hence, the force being applied is 18 N.
At the top of a raised platform, a ball is thrown vertically upward with a speed of 60 m/s. The ball rises up, then falls to the ground below the platform with a downward speed of 90 m/s.(a) For how much time was the ball in free-fall?(b) How far is the platform from the ground?
Before we begin to answer the question we need to notice that since the gravitaional acceleration will act on the motion this is a uniform accelerated motion. This means that we can use the following equations:
[tex]g=\frac{v_f-v_0}{t}[/tex][tex]y=y_0+v_0t+\frac{1}{2}gt^2[/tex][tex]v^2_f-v^2_0=2g(y-y_0)[/tex]Before we begin we establish that upward is the positve direction, this means that g=-10 m/s^2 and that the velocity of the ball when it reaches the ground below the platform is -90 m/s. We also establis that the origin of motion is at the platform
a)
To determine the time the ball is free-falling we need to know the maximum height the ball reaches before it begins falling.
-Motion from the platform to the maximum height.
In this case the initial velocity is 60 m/s while the final velocity is 0 m/s; furthermore the initial height is zero since our origin is on the platform.
The height it reaches the ball before it starts to fall can be obtained by the third equation:
[tex]\begin{gathered} 0^2-60^2=2(-10)(y-0) \\ -3600=-20y \\ y=-\frac{3600}{20} \\ y=180 \end{gathered}[/tex]Therefore the maximum height is 180 meters.
Now the time it takes the ball to reaches this height is given by the first formula:
[tex]\begin{gathered} -10=\frac{0-60}{t} \\ t=\frac{-60}{-10} \\ t=6 \end{gathered}[/tex]Hence it takes the ball six seconds to reach its maximum height.
Now, we know that in an accelerated uniform motion the time it takes to reach the maximum height is the same as the tame it takes to reach the initial height, therefore it takes 6 seconds for the ball to travel from the maximum height to the platform again.
Now we need to determine the time it takes the ball to fall from the height of the platform to the bottom; in this case the initial velocity is -60 m/s; this comes from the fact that the final velocity is the same at same heights in this kind of motion. Then, using the first formula we know that the time it takes is:
[tex]\begin{gathered} -10=\frac{-90-(-60)}{t} \\ t=\frac{-30}{-10} \\ t=3 \end{gathered}[/tex]Hence, it takes 3 seconds for the ball to travel from the platform to the ground.
Finally we add the three times to determine the total time of the free fall; therefore the time the ball is free falling is 15 seconds.
b)
To determine the height of the platform we can use the fact that the time it takes the ball to fall from this height to the ground is six seconds, then using the second equation we have that:
[tex]\begin{gathered} y=0-60(3)+\frac{1}{2}(-10)(3)^2 \\ y=-225 \end{gathered}[/tex]This means that the ball travels 225 meters downward from the height of the platform to the ground, therefore the platform is 225 meters above the ground.
As the object travels along the track, what is the maximum height that it reaches above point E?
Given:
Mass of object = 2 kg
At point E, the gravitational potential energy is 0.
Let's find the maximum height the object will reach above point E.
[tex]PE=mgh[/tex]Where:
m is the mass
g is acceleration due to gravity
h is the height.
The maximum height the object will reach above point E will be the height at Point C if the restriction is to the given points.
Point C is 20 m above the ground.
Therefore, the maximum height that it reaches above point E is 20 m.
ANSWER:
20 m.
Obtain the potential on the x-axis at x = 0 for the following point charge distributions on the x-axis: 200 μ C at x = 20 cm, -3 00 μ C at x = 30 cm and - 400 μ C at x = 40 cm.
Potential on x axis will be 9 * [tex]10^{6}[/tex] N [tex]m^{}[/tex]/[tex]C^{}[/tex] in all three cases
Electric potential is defined as the amount of work needed to move a unit charge from a reference point to a specific point against the electric field.
v1 = k ( q1 / r1)
= 9 * [tex]10^{9}[/tex] * ( 200 * [tex]10^{-6}[/tex] / 20 * [tex]10^{-2}[/tex] )
= 90 * [tex]10^{5}[/tex] = 9 * [tex]10^{6}[/tex] N [tex]m^{}[/tex]/[tex]C^{}[/tex]
v2 = k ( q2 / r2 )
= 9 * [tex]10^{9}[/tex] * (-300 * [tex]10^{-6}[/tex] / 30 * [tex]10^{-2}[/tex] )
= 9 * [tex]10^{6}[/tex] N [tex]m^{}[/tex]/[tex]C^{}[/tex]
v3 = k ( q3 / r3 )
= 9 * [tex]10^{9}[/tex] * 400 * [tex]10^{-6}[/tex] / 40 * [tex]10^{-2}[/tex]
= 9 * [tex]10^{6}[/tex] N [tex]m^{}[/tex]/[tex]C^{}[/tex]
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Block 1 (2 kg) is sliding to the right on a level surface at aspeed of 3 m/s. Block 2 (5 kg) is initially at rest, and block 1collides with it. After the collision, block 2 is moving to theright with a speed of 1.5 m/s. Calculate the magnitude anddirection of the velocity of block 1 after the collision.
Given:
The mass of block 1, m₁=2 kg
The mass of the block 2, m₂=5 kg
The initial velocity of the block 1, u₁=3 m/s
The initial velocity of the block 2, u₂=0 m/s
The velocity of the block 2 after the collision, v₂=1.5 m/s
To find:
The magnitude and direction of the velocity of block 2 after the collision.
Explanation:
From the law of conservation of momentum, the total momentum of blocks before the collision must be equal to the total momentum of the blocks after the collision.
Thus,
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]Where v₁ is the velocity of block 1 after the collision.
On rearranging the above equation,
[tex]\begin{gathered} m_1v_1=m_1u_1+m_2u_2-m_2v_2 \\ \implies v_1=\frac{m_1u_1+m_2u_2-m_2v_2}{m_1} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} v_1=\frac{2\times3+5\times0-5\times1.5}{2} \\ =-0.75\text{ m/s} \end{gathered}[/tex]The negative sign indicates that block 1 will be sliding to the left after the collision.
Final answer:
Thus the magnitude of the velocity of block 1 after the collision is 0.75 m/s.
And the direction of block 1 after the collision is to the left.
A spring with a spring constant of 30.0N/m Is compressed 5.00m. What is the force that the spring would apply
Force of the Spring = -(Spring Constant) x (Displacement)
so , Here we need to put the values
F = 30* 5 = 150N
Here, the Force applied is 150N. So, the correct ans is C
A passenger train and freight train start from the same time and travel in opposite directions. The passenger train traveled 3 times as fast as the freight train. In 5 hours, they were 360 miles apart. Find the rate of each train.
Given that the relative distance between 2 trains, d = 360 miles
Time, t is 5 hrs.
Let speed of freight train be v, then the speed of passenger train will be 3v.
For relative speed, the speed of trains will be added.
Hence,
[tex]\begin{gathered} 3v+v=\frac{360}{5} \\ v=\frac{360}{5\times4} \end{gathered}[/tex]Thus, the speed of freight train is 18 miles/hour.
And the speed of passenger train is 3v = 54 miles/hour.
A 50 kW pump is used to pump up water from a mine that is 50 m deep. Find the mass of water that can be lifted by the pump in 1.4 min.
Given data
*The given power of the pump is P = 50 kW = 50 × 10^3 W
*The given depth is h = 50 m
*The given time is t = 1.4 min = (1.4 × 60) = 84 s
The mass is calculated by the power formula as
[tex]\begin{gathered} P=\frac{W}{t} \\ P=\frac{mgh}{t} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} 50\times10^3=\frac{m\times9.8\times50}{84} \\ m=8571.42\text{ kg} \end{gathered}[/tex]Hence, the mass of the water is m = 8571.42 kg
If a 150 pound weight is on a frictionless surface, raised at an angle of 35 degrees, what is the tension in the rope that keeps it from sliding down? What is the force perpendicular to the surface?
This is the given situation.
Where m is the mass of the block and g is the acceleration due to gravity. It is given in the question, mg=150 pound=68.04 kg.
There are two components of weight. One along with x-direction and the other with negative y-direction.
x-component is
[tex]mg\sin \theta[/tex]y-component is
[tex]mg\cos \theta[/tex]Tension on the string is equal to the x-component of the weight. and the normal force,i.e. perpendicular force is equal and opposite to the y component of the weight. But tension is in opposite direction to the x-component of weight and perpendicular force is opposite to the y-component.
Therefore the tension is,
[tex]T=-mg\sin \theta=-68.04\times\sin 35^o=-39.03\text{ N}[/tex]And the normal force is,
[tex]N=mg\cos \theta=60.04\times\cos 35^o=55.74\text{ N}[/tex]Therefore the magnitude of the tension on the string is 39.03 N
And the normal force is 55.74 N
4kg of steam is at 100°C and he is removed until there is water at 39°C how much heat is removed
1024.8 KJ Heat is removed when 4kg of steam is at 100°C and he is removed until there is water at 39°C
Mass =4 kg
ΔT=100−39=61 ∘C
Q=m×C×ΔT
C= specific heat capacity of water =4200J/(kgK)
Q=4×4200×61
=1024800 Joule.
=1024.8KJ
Heat is the amount of energy that flows from one body to another on its own as a result of their different temperatures, as opposed to internal energy, which is the sum of all the molecules' energies within an item. Although it is an energy form, heat is energy in motion. Heat is not a system's property. However, a temperature difference causes the transfer of energy as heat to take place at the molecular level.
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A shop sign weighing 215 N hangs from the end of auniform 175-N beam as shown in (Figure 1).Find the tension in the supporting wire (at 35.0 degrees)
In order to find the tension in the wire, let's first decompose it in its vertical and horizontal components:
[tex]\begin{gathered} T_x=T\cdot\cos (35\degree) \\ T_y=T\cdot\sin (35\degree) \end{gathered}[/tex]Now, since the system is stable, the sum of vertical forces is equal to zero, so we have:
[tex]\begin{gathered} T_y-175-215=0 \\ T_y-390=0 \\ T_y=390 \\ T\cdot\sin (35\degree)=390 \\ T\cdot0.57358=390 \\ T=\frac{390}{0.57358} \\ T=679.94\text{ N} \end{gathered}[/tex]So the tension in the wire is equal to 679.94 N.
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Jennifer uses many forms of energy to get ready for school in the morning. Which table best matches each form of energy to Jennifer's actions?
Answer:
c I think
Explanation:
pretty sure cuz of common sense
Answer:
A form of energy you use when cheering for your favorite team. answer choices. sound energy. electrical energy. thermal energy. electromagnetic energy.
Explanation:
Hoped it helped :))))
A student measures the voltage and current between two points in an electrical circuit. If the voltage is 110 V and the current is 0.75 A, what is the resistance, according to Ohm's law?Α. 147 ΩΒ. 109 ΩC. 0.007 ΩD. 82.50 Ω
In order to calculate the resistance, we can use the formula below (Ohm's law):
[tex]R=\frac{V}{I}[/tex]If the voltage is 110 V and the current is 0.75 ohms, the resistance will be:
[tex]\begin{gathered} R=\frac{110}{0.75}\\ \\ R=147\text{ ohms} \end{gathered}[/tex]Therefore the correct option is A.
Calculate the acceleration of the elevator for each 5 second interval
Given,
The weight of the student, W=500 N
Thus the mass of the student is given by,
[tex]m=\frac{W}{g}[/tex]Where g is the acceleration due to gravity,
On substituting the known values,
[tex]\begin{gathered} m=\frac{500}{9.8} \\ =51.02\text{ kg} \end{gathered}[/tex]For the first 5 seconds, the net force acting on the student is 0 N. Thus scale reads only his weight. As the net force is zero the acceleration of the elevator is also zero.
In the next 5 intervals, the net force acting on the student is F=200 N, as seen from the diagram.
Thus the acceleration of the elevator is given by the equation,
[tex]F=ma[/tex]Where a is the acceleration of the elevator.
On substituting the known values,
[tex]\begin{gathered} 200=51.02\times a \\ \Rightarrow a=\frac{200}{51.02} \\ =3.92m/s^2 \end{gathered}[/tex]Thus the acceleration in this interval is 3.92 m/s²
During the interval, 10s-15s, the net force acting on the student is zero as seen from the graph. Thus the acceleration of the elevator is also zero.
During the interval, 15 s-20 s, the net force on the student is F=-200 N as seen from the diagram.
Thus the acceleration is,
[tex]\begin{gathered} F=ma \\ \Rightarrow a=\frac{F}{m} \\ a=\frac{-200}{51.02} \\ =-3.92m/s^2 \end{gathered}[/tex]Thus the accelerating in this interval is -3.92 m/s²That is the elevator is accelerating downwards.
I need help with one of my physics question
Outline alpha, betta and gamma radiation in terms of depth of tissue penetration, ionizing effect and speed of radiation.
Penetration and Ionizing effect gamma, betta, and alpha radiation.
The speed of radiation is alpha, betta, and gamma radiation.
A gamma ray, also known as gamma radiation, is a penetrating shape of electromagnetic radiation arising from the radioactive decay of atomic nuclei. It consists of the shortest wavelength electromagnetic waves, typically shorter than the ones of X-rays.
Gamma rays have a lot of penetrating strength that several inches of dense material like lead, or maybe some feet of concrete may be required to forestall them. Gamma rays can skip absolutely through the human body; as they pass through, they are able to purpose ionizations that harm tissue.
These are a number of the most deadly radiation known. If a person happened to be close to a gamma-ray producing item, that they had been fried in an instantaneous. honestly, a gamma-ray burst should affect existence's DNA, inflicting genetic harm lengthy after the burst is over.
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Determine whether the statement is true or false.2 ∈ {x | x ∈ N and x is even}Is the statement true or false?❑ True❑ False
ANSWER
True
EXPLANATION
The set is the set of all the natural even numbers. The natural numbers are {1, 2, 3, 4, ...} and the natural even numbers are {2, 4, 6, 8, ...}. Therefore, it is true that 2 belongs to this set.
A racing car of mass 1500 kg, is accelerating at 5.0 m/s2, is experiencing a lift force of 600 N [up}, due to its streamlined shape, and grounding effects of 1000 N [down], due to air dams and spoilers. Find the driving force needed to keep the car going given that μk = 1.0.
Given data:
* The acceleration of the car is,
[tex]a=5ms^{-2}[/tex]* The mass of the car is,
[tex]m=1500\text{ kg}[/tex]* The force acting on the car in the upward direction is,
[tex]F_1=600\text{ N}[/tex]* The force acting on the car in the downward direction is,
[tex]F_2=1000\text{ N}[/tex]* The coefficient of kinetic friction is,
[tex]\mu_k=1[/tex]Solution:
The weight of the car is,
[tex]\begin{gathered} w=mg \\ w=1500\times9.8 \\ w=14700\text{ N} \end{gathered}[/tex]The normal force acting on the car is,
[tex]\begin{gathered} F_N=w+F_2-F_1 \\ F_N=14700+1000-600 \\ F_N=15100\text{ N} \end{gathered}[/tex]The frictional force acting on the car is,
[tex]\begin{gathered} F_k=\mu_kF_N \\ F_k=1\times15100 \\ F_k=15100\text{ N} \end{gathered}[/tex]According to newton's second law, the force acting on the car is,
[tex]\begin{gathered} F=ma \\ F=1500\times5.0 \\ F=7500\text{ N} \end{gathered}[/tex]The net force acting on the car in terms of the applied force and frictional force is,
[tex]\begin{gathered} F=F_a-F_k \\ F_a=F+F_k \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} F_a=7500+15100 \\ F_a=22600\text{ N} \end{gathered}[/tex]Thus, the driving force required to maintain the motion of the car is 22600 N.
A long, thin wire offers more resistance to an electrical current than a thick, short wire would.TrueFalse
Resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.
Thus, the given statement is true.
A PVC pipe has a length of 53.594 centimeters.a. What are the frequencies of the first three harmonics when the pipe is open at both ends? b. What are the frequencies of the first three harmonics when the pipe is closed at one end and open at the other?
Given:
Length of pipe, L = 53.594 cm
Let's solve for the following:
• (a). What are the frequencies of the first three harmonics when the pipe is open at both ends?
To find the frequency of the first harmonics, apply the formula:
[tex]f_1=\frac{v}{2l}[/tex]Where:
v is the speed of sound = 343 m/s
l is the length of the pipe in meters.
Where:
100 cm = 1 m
53.594 cm = 0.53594 m
Hence, for the frequency, we have:
[tex]\begin{gathered} f_1=\frac{343}{2*0.53594} \\ \\ f_1=319.99\approx320\text{ Hz} \end{gathered}[/tex]For the frequency of the second harmonics, we have:
[tex]\begin{gathered} f_2=2f_1 \\ \\ f_2=2*320 \\ \\ f_2=640\text{ Hz} \end{gathered}[/tex]For the frequency of the third harmonics:
[tex]\begin{gathered} f_3=3f_1 \\ \\ f_3=3*320 \\ \\ f_3=960\text{ Hz} \end{gathered}[/tex]• Part B.
,• What are the frequencies of the first three harmonics when the pipe is closed at one end and open at the other?
For the first harmonics when the pipe is closed at one end, apply the one end open harmonics equation.
[tex]\begin{gathered} f_1=\frac{v}{4l} \\ \\ f_1=\frac{343}{4*0.53594} \\ \\ f_1=160\text{ Hz.} \end{gathered}[/tex]The next 2 harmonics will be the odd multiples of the first.
Hence, we have:
[tex]\begin{gathered} f_3=3f_1 \\ f_3=3*160 \\ f_3=480\text{ Hz.} \\ \\ \\ f_5=5f1 \\ f_5=5*160 \\ f_5=800\text{ Hz.} \end{gathered}[/tex]Therefore, the frequencies of of the first three harmonics when the pipe is closed at one end and open at the other are:
160 Hz, 480 Hz, 800 Hz.
ANSWER:
(A). 320Hz, 640 Hz, 960 Hz.
(B). 160 Hz, 480 Hz, 800 Hz.
I’m working on this study guide but I’m stock on the first equation
To find the energy needed to jump from one state to another we just need to subtract the initial state energy to the final state. In this case the final state will have -13.6 eV of energy and the initial state has -122.4 eV of eneregy, then we have:
[tex]-13.6-(-122.4)=108.8[/tex]Therefore, the energy needed is 108.8 eV
An elevator uses 100 000J of electrical energy to raise a load of 800 N through a height of 40m in a time of 20s. What fraction of the energy input was NOT transferred to the load? A) 32% B) 68% C) 80%D) 97%
W: work
F: force required to lift load; F = 800 N
d: distance lifted: d = 40 m
W = Fd = 800*40
W = 32000 J
If 100000J were exerted on a load for which only 32000J was necessary, the other 100000-32000 = 68000J was not transferred to the load.
68000/100000 = 0.68 = 68%
Question 4 What FITT Principle describes what kind of exercise you do? O Type O Frequency Time O Indoor
Answer:
Type
FITT is acronym that stands for Frequency, Intensity, Time, and Type.
Answer:
D) type
Explanation:
How may we need to be more intentional on viewing parenting roles differently in order to most benefit our children/students?
We can be more intentional on viewing parenting roles differently in order to most benefit our children/students through-
1. Being relational
2. Being consistent
3. Being Instructive
What is intentional Parenting?
Having a strategy and setting priorities for your time and energy is all that constitutes intentional parenting. Our daily decisions and the commitments you make are then influenced by these priorities.Being an intentional parent entails understanding that the time we spend with our children is valuable and finite, and that the choices we make about how to spend that time will have an impact for a lifetime.
Thankfully, successful parenting doesn't demand perfection in these areas of instruction and punishment, but it does call for us to be thoughtful. Therefore, we must keep the following in mind to be intentional parents:
1. Parenting with intention involves relationships.
As children grow older and their lives full with milestones and events, life becomes increasingly busy. Throughout the hectic times of school and extracurricular activities, look for methods to interact frequently. Look for chances to spend time with each of your kids alone. Find moments throughout the day to have fun, play, and converse, even if it is only for a little while.
2. Consistent (even relentless) attention is a hallmark of intentional parenting.
It calls for us to be persistent in our efforts to connect with our children, refusing to give up on potential future connections just because the current one falls short of our expectations. Parenting is not a chore for parents who do it intentionally. Every chance they have to affect their children is seen by them as a wonderful gift.
3. Parenting with intention is instructive.
Kids that grow up in intentional, relational families are frequently eager and willing to learn. Every day, search for opportunities to reinforce prior teachings or find instructive situations. Next, schedule specific, devoted times to concentrate on some larger goals.
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The position of an object is given by the formulasx = -3t² + 2t -4 and y = -2t³ + 6t² +1A) What is the speed at t = 1sB) What is the acceleration at t = 1s
We are given that the position of an object if given by the following equations:
[tex]\begin{gathered} x=-3t^2+2t-4 \\ y=-2t^3+6t^2+1 \end{gathered}[/tex]To determine the velocity we will determine the derivative of each of the functions. For "x" we have:
[tex]x=-3t^2+2t-4[/tex]Finding the derivative with respect to time we get:
[tex]\frac{dx}{dt}=\frac{d}{dt}(-3t^2+2t-4)[/tex]Now we distribute the derivative on the left side:
[tex]\frac{dx}{dt}=\frac{d}{dt}(-3t^2)+\frac{d}{dt}(2t)-\frac{d}{dt}(4)[/tex]For the first derivative we will use the rule:
[tex]\frac{d}{dt}(at^n)=ant^{n-1}[/tex]Applying the rule we get:
[tex]\frac{dx}{dt}=-6t^{}+\frac{d}{dt}(2t)-\frac{d}{dt}(4)[/tex]For the second derivative we use the rule:
[tex]\frac{d}{dt}(at)=a[/tex]Applying the rule we get:
[tex]\frac{dx}{dt}=-6t^{}+2-\frac{d}{dt}(4)[/tex]For the third derivative we use the rule:
[tex]\frac{d}{dt}(a)=0[/tex]Applying the rule we get:
[tex]\frac{dx}{dt}=-6t^{}+2[/tex]Now, since the velocity is the derivative with respect to time of the position and this is and we determine the derivative for the x-position what we have found is the velocity in the x-direction, therefore, we can write:
[tex]v_x=-6t^{}+2[/tex]Now we substitute the value of time, t = 1, we get:
[tex]\begin{gathered} v_x=-6(1)+2 \\ v_x=-6+2 \\ v_x=-4 \end{gathered}[/tex]Now we use derivate the function for "y":
[tex]\frac{dy}{dt}=\frac{d}{dt}(-2t^3+6t^2+1)[/tex]Using the same procedure as before we determine the derivative:
[tex]\frac{dy}{dt}=-6t^2+12t[/tex]This is the velocity in the y-direction:
[tex]v_y=-6t^2+12t[/tex]Now we substitute the value of t = 1:
[tex]\begin{gathered} v_y=-6(1)^2+12(1) \\ v_y=6 \end{gathered}[/tex]Now, the speed is the magnitude of the velocity, the magnitude is given by:
[tex]v=\sqrt[]{v^2_x+v^2_y}[/tex]Substituting the values we get:
[tex]v=\sqrt[]{(-4)^2+6^2}[/tex]Solving the operations:
[tex]v=7.21[/tex]Therefore, the speed is 7.21 m/s.
To determine the acceleration we will determine the derivative of the formulas for velocities:
[tex]v_x=-6t^{}+2[/tex]Now we derivate with respect to time:
[tex]\frac{dv_x}{dt}=a_x=-6[/tex]Now we use the function for the velocity in the y-direction:
[tex]\frac{dv_y}{dt}=a_y=-12t^{}+12[/tex]Now we substitute the value of t = 1:
[tex]\begin{gathered} a_y=-12(1)^{}+12 \\ a_y=0 \end{gathered}[/tex]Since the acceleration in the y-direction is zero, this means that the total acceleration is the acceleration in the x-direction, therefore, the magnitude of the acceleration is:
[tex]a=6[/tex]Classify a sample of matter as a pure substance or mixture based on the number of elements or compounds in the sample.
A pure substance has constant chemical composition and properties.
The element cannot be separated into simples substances.
A mixture is composed of 2 or more pure substances that retain their individual identities but cannot be separated by physical methods.
As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper applies an average force of 800. newtons on the boat for 0.30 seconds as he jumps.a. How much momentum does Jasper’s 80.-kilogram body have as it lands on the dock?b. What is Jasper's speed on the dock?
a) Take into account that the force can be written as follow:
[tex]F=\frac{\Delta p}{\Delta t}[/tex]where F is the force, Δp the change in the momentum and Δt the time interval in which the force is applied.
Moreover, consider that the change in the momentum can be written as follow:
[tex]\Delta p=m\Delta v[/tex]where m is the mass and Δv is the change in the velocity.
By replacing the previous expression into the formula for the force F, you can solve for Δv to determine the change in the velocity of Jamper due the force he applies to the boat:
[tex]\begin{gathered} F=\frac{m\Delta v}{\Delta t} \\ \Delta v=\frac{\Delta t\cdot F}{m} \end{gathered}[/tex]In this case, m=80.0kg, Δt = 0.30s and F = 800.0N.
Then, for the change in the speed you obtain:
[tex]\Delta v=\frac{0.30s\cdot800.0N}{80.0kg}=3.0\frac{m}{s}[/tex]It means that related to the boat (where you can consider that Jasper is at rest) the speed of Jasper is 3m/s. However, it is necessary to take into account the speed of the boat before Jasper jumps to the dock, which is 3.0m/s.
The speed of Jasper when he lands on the dock is then:
3.0m/s + 3.0m/s = 6.0m/s
The momentum is the product of mass and speed, then, the momentum of jasper is:
p = (80.0kg)(6.0m/s) = 480.0kg*m/s
b) Based on the previous calculations, you have that the Jasper's speed on the dock is 6.0m/s