if an aircraft is in sluf and we know the weight of the aircraft, w, the wing area, s, the density and velocity, what else must we know to determine the coefficient of lift of the aircraft?

Answers

Answer 1

To determine the coefficient of lift of an aircraft in sluf, we must also know the angle of attack.

What is aircraft?
A vehicle that can fly by gaining support from of the air is called an aircraft. It uses either static lift, dynamic lift from an airfoil, or, in a few rare instances, the downward thrust from jet engines to counteract the force of gravity. Aeroplanes, helicopters, airships (including blimps), gliders, paramotors, as well as hot air balloons are a few examples of common types of aircraft. Aviation refers to the human activity which surrounds aircraft. Aeronautics is the study of aviation, which includes creating and building aircraft. Unmanned aerial vehicles may well be remotely controlled as well as self-controlled by onboard computers, unlike crewed aircraft that have an onboard pilot. Different factors, including lift type, aircraft propellant, usage, and others, can be used to categorise aircraft.

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Related Questions

Which of the following are the freezing and boiling points of water on the Centigrade scale?Select one:a. 32° and 100°b. 0° and 100°c. 32° and 212°d. 0° and 212°

Answers

Answer:

b

Explanation:

For C scale  , water freezes at 0 and boils at 100 C

the sweet spot on a baseball bat is 55 cm from the axis of rotation during the swing of the bat. if the sweet spot on the bat moves 36 m/s, what is the angular velocity of the bat (in rad/s)?

Answers

The Angular velocity is 54.55 rad/s


What is angular velocity?
The rate at which an angle between two bodies changes is known as angular velocity. Something changes when an object rotates or revolves around an axis. This displacement is depicted in the image by the angle formed by a line on just one body and a line on the other. Technical terms like degrees, 360-degree revolutions, as well as revolutions per minute have been frequently used to describe angles or angular displacements (rpm). Angles and angular velocities are typically expressed in radians as well as radians per second, respectively, in mathematics and physics.

These conversion factors link these measurements together:
A radian equals 1/180 (or 0.0175) of a degree, and a radian per second = 1/30 (or 0.105) of a radian.
Angular velocity W= U/r
Given V= 30m/s ; r=0.55mW=3/0.55 = 54.55 rad/s

Hence the answer is 54.55 rad/s

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To which category does galaxy #2 belong? Why does it belong in this category?

Answers

Galaxy #2 belongs to  spiral. This is because it has a bulge at the centre, a disk as well as the curved arms. Contains both Young and Old stars - contains gas and dust - contains more red giants and red supergiant's than disk - some regular rotation about centre.

What Is Dark Matter?

Dark matter is a form of matter thought to account for approximately 85% of the matter in the world. Dark matter consists of particles that do not absorb, reflect or emit light, so they cannot be detected by observing electromagnetic radiation. Dark matter is material that cannot be directly seen. We know  dark matter exists because  it affects objects  we can directly observe.

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Please help me with this, it's a Kinematics Equation 2 problem!

Answers

The distance covered by the object is given as,

[tex]d=ut+\frac{1}{2}at^2[/tex]

Plug in the known values,

[tex]\begin{gathered} d=(0\text{ m/s)(9 s)+}\frac{1}{2}(2m/s^2)(9s)^2 \\ =0\text{ m+}81\text{ m} \\ =81\text{ m} \end{gathered}[/tex]

Thus, the distance covered by the object is 81 m.

why are the rocky inner planets different than the gaseous outer planets? heat from the growing sun allowed rock to mainly form close in gases come from further away from the center of the solar system the rocky inner planets got rocks from the sun later solar wind pushing material away had more effect on metals gas has more momentum and flies further away naturally

Answers

The correct answer is option A.

The rocky inner planets are different than the gaseous outer planets because heat from the growing sun allowed rock to mainly form close in.

In our solar system, the inner planets are rocky while the outer planets are in a gaseous state.

Rocky planets are also called terrestrial planets as their outer surface is solid and composed of rocks and metal.

They have large thick gaseous atmospheres which are abundant in gases such as nitrogen, carbon dioxide, and oxygen.

While the outer planets are giant balls of gases that orbit around the sun.

Unlike terrestrial planets, these gaseous planets do not have thick atmospheres and are 2x larger in size than the rocky inner planets.

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ake a 40-pf capacitor and charge it to a potential difference of 500 v. then disconnect it from the battery and connect its terminals to those of an uncharged 10-pf capacitor. what are: (a) the

Answers

If we rearrange the circuit so that the inductor and capacitor were connected in series,the impedance would be  Decreased and the resonant frequency would be unchanged.

What is capacitor ?

Frequency is the rate at which something occurs over a particular period of time or in a given sample.

Sol-An LC circuit is made up of an inductor (L) and a capacitor (C). At the resonance condition of the LC circuit, the inductive reactance XL. becomes equal to the capacitive reactance Xc.

It is defined as-

XL= 2πfL

Xc= 1/2πfC

So when reactances are equal we have-

XL=Xc

2πfL= 1/2πfC

f^2=1/4π^2LC

f=1/2π√LC

The resonance condition is the same for both the parallel and series LC circuit, so the resonance frequency will not change.

The impedance of a parallel LC circuit is:

Z(w)= jL w^2-w^2•/w

In a series LC circuit, when w

, the impedance becomes equal to 0.

Therefore the impedance of a series LC circuit will decrease and the resonant frequency will remain the same.

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Which Combination of graphs best describes free-fall motion? [Neglect air resistance.]​

Answers

The combination of the graphs that shows a free fall are A and C.

What is the description of free fall?

We know that motion has to do with the change of position with time. When an object is falling freely, the velocity of the object would remain constant because the velocity or the speed does not change.

Recall that when we throw objects down from a height, the objects would fall at the same rate irrespective of the masses of the objects. This can be understood to mean that the acceleration that acts on the objects is the same and does not change.

However, the distance that is covered by the object does change with tome and this can be seen from the graph as we have in the question. We can now see that the graph that has a flat speed time graph and a straight distance time graph can describe on object that is undergoing free fall.

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a 2 900-kg truck moving at 11.0 m/s strikes a car waiting at a traffic light, hooking bumpers. the two continue to move together at 7.00 m/s. what was the mass of the struck car?

Answers

The mass of the car that was stuck is  1657 Kg.

What is the mass of the struck car?

We know that we have to use the laws of the conservation of the linear momentum here. It is known that the momentum before collision must be equal to the total momentum that we have after collision.

In this case, we have;

m1u1 + m2u2 = m1v1 + m2v2

m1  = mass of the truck

m2 = mass of the car

u1 and v1 = the initial and final velocities of the truck

u2 and v2 = initial and final velocities of the car

Again we are told that the two continue to move together hence v1 + v2 = v

m1u1 + m2u2 = (m1 + m2) v

(2900 * 11.0) + 0 = (2900  + m2) 7

31900 = 20300 + 7m2

31900 - 20300 = 7m2

m2 = 31900 - 20300 /7

m2 = 1657 Kg

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a square loop of wire with initial side length 20 cm is placed in a magnetic field of strength 0.5 t. the field is parallel to the axis of the loop. somebody carefully grabs the edges of the loop and begins pulling them apart such that the area of the loop increases at a rate of 20 cm2/s. what is the magnitude of the induced emf in the loop?

Answers

The magnitude of the induced emf in the loop is 1mV.

The formula to be used is:

E = B (dA/dt)

B = magnetic of magnetic field parallel to the axis of loop = 0.5 T

dA/dt = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Now, plugging in the values we get:

E = B (dA/dt)

E = (0.5) (20 x 10⁻⁴ )

E = 10 x [tex]10^{-4}[/tex] volts

E = 1 mV

Therefore,  the magnitude of the induced emf in the loop is 1mV.

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the chemical makeup of a star's surface is obtained by ... a. examining rocks recovered from the moon. b. examining rocks here on earth. c. examining material collecting on solar panels here on earth.

Answers

option c is the correct answer

What is Spectroscopy ?

It is a common method used by astronomers to determine the composition of stars, planets, and other objects. Today, this process uses instruments with gratings that distribute the light from the object by wavelength. This spread of light is called the spectrum.

The science of spectroscopy is very difficult. From the spectral lines, astronomers can not only determine the element, but also the temperature and density of this element within the star. The spectral lines also tell us about each of the star's magnetic fields. The width of the line tells us how fast the matter is moving.

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an archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.15 m/s on a smooth, slippery surface. the 22.5-g arrow is shot with a speed of 42.0 m/s and passes through the target, which is stopped by the impact. what is the speed of the arrow after passing through the target?

Answers

24.46 m/s is the speed of the arrow after passing through the target.

p=mv

Conservation of momentum: m1v1=m2v2

The target has a mometum of -(.3kg)(2.15m/s)

the arrow has a mometum of (.0225kg)(42.0m/s)

After impact the mometum of the target is 0 Ns since its velocity is zero.

After impact the velocity of the arrow can be determined with conservation of momentum calculations.

Δp(arrow)=-Δp(target)

m(arrow)v(arrow)i+m(target)v(target)i=m(arrow)v(arrow)f+m(target)v(target)f

(.0225kg)(42.0m/s)-(.3kg)(2.15m/s)=(.0225kg)vf(arrow)

the final velocity of the arrow should be equal to((.0225kg*42.0m/s)-(.3kg*2.15m/s))/.0225kg

v(arrow) final = 24.46 m/s

What is momentum?Momentum is a quantity  used to describe the state of motion of an object with  non-zero mass. Therefore momentum can be applied to any moving object. If

       m  is the mass of a body and v is the velocity  of that body, the impulse momentum  can be expressed as  p=m.v

Since velocity is considered  a vector, momentum is also considered a vector. This means that the impulse has a significant magnitude and direction—an impulse indicated by a vector or  arrow. For example, if a ball of a certain mass moves with a certain speed, it has momentum. When the ball hits the wall, it stops and thus carries its momentum over the wall. Therefore momentum is always conserved.

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a window-mounted air conditioner removes 3.5 kj from the inside of a home using 1.75 kj work input. how much energy is released outside and what is its coefficient of performance?

Answers

The amount of heat released outside is 3.5 kJ and the coefficient of performance is 2.

What is efficiency of a machine?

The efficiency of the a machine or any device is the measure of effectiveness of the device. Efficiency describes how a machine converts input energy to output energy without wasting much of the input energy.

Mathematically, efficiency of a machine is given as;

E = output energy / input energy  x 100%

The coefficient of performance  (COP) of a heat pump, refrigerator or air conditioning system is a ratio of useful heating or cooling provided to work required.

The energy released outside = 3.5 kJ

COP = 3.5 kJ / 1.75 kJ

COP = 2

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mass of bullet 0.25kg, velocity of bullet 230kg, mass of gun 9kg. What is the recoil speed of the gun

Answers

The recoil speed of the gun is 6.39m/s


What is recoil speed of the gun ?
After firing the bullet, the bullet moves forward with a large momentum and the gun moves backward with an equal momentum. Let m and M be the masses of bullet and gun, respectively. If v and V are the velocities of the bullet and gun, respectively after the firing then:-

0= mv+ MV

Therefore V=-mv/M


The negative sign indicates that the gun moves in a direction to that of
the bullet.
m1v1 = m2v2

m1 = mass of bullet

v1 = velocity of bullet

m2 = mass of gun

v2 = velocity of gun

m1v1=m2v2

0.25*230=9*v2

v2=0.25*230/9 = 6.39m/s


The recoil speed of the gun is 6.39m/s
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two satellites orbit the earth in stable orbits. satellite a is three times the mass of satellite b. satellite a orbits with a speed vvv at a distance rrr from the center of the earth. satellite b travels at a speed that is greater than vvv . at what distance from the center of the earth does the satellite b orbit?

Answers

Satellite B's radius will be less than Satellite A's radius.

m = Satellite Mass

v = Satellite Velocity

r = Orbital radius of the satellite

Relating the gravitational and centripetal forces

mv^2/ r= GmM/r^2

v^2/r=GM/r^2

v^2=GM/r

v = sqrtGM/r

As can be shown, the mass of the satellite has no bearing on the relationship between velocity and radius.

This implies that the radius must shrink for v to increase

Here, vb>va

Therefore, satellite b's radius will be smaller than satellite a's radius.

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A person of weight 600n at the bottom of a mountain climbs to the top. the gravitational field strength changes from 10.00n/kg at the bottom to 9.97n/kg at the top.his mass is unchanged as he climbs. what are his mass and his weight at the top of the mountain

Answers

At the summit of the mountain, the mass is 60.18 kg, the weight is 600 N, and the mass is unaltered.

What is Mass?Weight is the force of gravity acting on an item, whereas mass is the amount of matter that makes up the object. Weight varies from place to place yet mass is constant everywhere you are in the cosmos. Kilograms are used to measure mass.

Let the mass on earth be m.

Given, W= 600N, g= 9.97 N/kg(top of mountain)

As we know:

The formula for Weight = mass x acceleration due to gravity at place W=m*g

Now, calculate as follows:

m = W/g= 600*9.97mass = 60.18 kgW= m*g= 60.18 * 9.97= 600NWeight = 600N

Therefore, at the summit of the mountain, the mass is 60.18 kg, the weight is 600 N, and the mass is unaltered.

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Consider a 3-atom molecule a-b-a for which b has a total of only two valence electrons. If you assume that b is a group ii metal, what is the bond angle for a-b-a? please answer in degrees to the nearest whole number.

Answers

The bond angle of the molecule to the nearest degree is 120 degrees.

What is the VSEPR theory?

According to the valence shell electron pair repulsion theory (VSEPR). We know that the bond angle that an atom has with its neighbors which also call the dihedral angle of the compound is affected by the number of atoms that surround the central atom in the molecule.

Let us recall that when we have two atoms that surround the central atom then we are going to have a three atom compound which has a trigonal planar geometry and has a bond angle of 120 degrees.

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A rocket is launched from the surface of the earth with a speed of 9.0x103 m/s. What is the maximum altitude reached by the rocket? (MEarth=5.98x1024 kg, REarth=6.37x106 m)

Answers

From the Law of conservation of energy, we know that the sum of the kinetic and potential energy of the rocket is the same at the surface of the Earth and at the maximum altitude. Nevertheless, the kinetic energy of the rocket when it is at the maximum altitude is 0:

[tex]\begin{gathered} K_1+U_1=K_2+U_2 \\ K_2=0 \\ \Rightarrow K_1+U_1=U_2 \end{gathered}[/tex]

The kinetic energy is given by:

[tex]K=\frac{1}{2}mv^2[/tex]

On the other hand, the gravitational potential energy for big changes in altitude (comparable to the radius of the Earth) is given by the expression:

[tex]U=-\frac{GMm}{r}[/tex]

Where M is the mass of the Earth, m is the mass of the rocket, r is the distance from the center of the Earth to the rocket and G is the gravitational constant:

[tex]G=6.67\times10^{-11}N\cdot\frac{m^2}{\operatorname{kg}}[/tex]

At the beggining of the movement, the value of r corresponds to the radius of the Earth:

[tex]U_1=-\frac{GMm}{R_E}[/tex]

At the end of the movement, the value of r corresponds to the radius of the Earth plus the maximum altitude h:

[tex]U_2=-\frac{GMm}{R_E+h_{}}[/tex]

Substitute the expressions for U_1, K_1 and U_2 and simplify the equation by eliminating the factor m:

[tex]\begin{gathered} \frac{1}{2}mv^2-\frac{GMm}{R_E}=-\frac{GMm}{R_E+h} \\ \Rightarrow\frac{1}{2}v^2-\frac{GM}{R_E}=-\frac{GM}{R_E+h} \end{gathered}[/tex]

Isolate the term GM/(R_E+h):

[tex]\Rightarrow\frac{GM}{R_E+h}=\frac{GM}{R_E_{}}-\frac{1}{2}v^2[/tex]

Divide both sides by the factor GM:

[tex]\Rightarrow\frac{1}{R_E+h}=\frac{1}{R_E}-\frac{v^2}{2GM}[/tex]

Take the reciprocal to both sides of the equation:

[tex]\Rightarrow R_E+h=\frac{1}{\frac{1}{R_E}-\frac{v^2}{2GM}}[/tex]

Isolate h:

[tex]h=\frac{1}{\frac{1}{R_E}-\frac{v^2}{2GM}}-R_E[/tex]

Substitute the values of each variable: R_E=6.37x10^6m, M=5.98x10^24kg, G=6.67x10^-11 N*m^2/kg^2, and v=9.0x10^-3 m/s:

[tex]\begin{gathered} h=\frac{1}{\frac{1}{6.37\times10^6m}-\frac{(9.0\times10^3\cdot\frac{m}{s})^2}{2(6.67\times10^{-11}N\cdot\frac{m^2}{kg^2})(5.98\times10^{24}kg)}}-6.37\times10^6m \\ =18.03\times10^6m-6.37\times10^6m \\ =11.7\times10^6m \end{gathered}[/tex]

Therefore, the maximum altitude reached by a rocket with an initial speed of 9.0x10^3m is:

[tex]11.7\times10^6m[/tex]

a. What is pressure? Write down its Sl unit​

Answers

Answer:

Pascal (Pa)

Explanation:

the si for pressure is the pascal pa equal to one newton per sqaure meter (N/m^2, or kgxm-^1xs-^2)

[tex] \boxed {\huge \: \sf \red{~~~Answer~~~} \: }[/tex]

Pressure is defined as the physical force exerted on an object.

SI unit of pressure is Pascal (pa).

[tex] \\ \\ [/tex]

______________________

a tank is 10 m long, 4 m wide, 4 m high, and contains kerosene with density 820 kg/m3 to a depth of 3.5 m. (use 9.8 m/s2 for the acceleration due to gravity.) (a) find the hydrostatic pressure on the bottom of the tank.

Answers

The hydrostatic pressure on the bottom of the tank is 197.083 kN.

Given,

Length of tank = 10 m

Width of tank = 8 m

height of tank = 4 m

density of kerosene = 820 kg/m³

depth of oil = 3.5 m

A. Hydrostatic pressure at the underside of the tank:

P = ρ g h

P = 820 x 9.81 x 3.5

P = 28154.7 Pa

B. Hydrostatic force at the underside of the tank:

F = P A

⇒A  = 10 x 8 = 80 meter²

⇒F = 28154.7 x 80

    = 2252.38 kN

C. Hydrostatic force at one edge:

Given,

Width = 8 m

depth = z

A = 4 y

dA = 4 dy

P = ρgh

P = ρgz = ρg(y-0.5)

F = P dA

F = ρg(y-0.5)(4dA)

F = 24.5 ρg

Hydrostatic force at one end = 24.5 x 820 x 9.81 = 197.083 kN.

The pressure exerted by a liquid thanks to gravity at any point in equilibrium. Hydrostatic pressure is proportional to the depth measured from the surface because the weight of the liquid increases when a downward force is applied. Hydrostatic pressure is the force produced by still or still water. this is often the constant force that water pressure exerts on underground walls.

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a driver who does not wear a seat belt continues to move forward with a speed of 18.0 m/sm/s (due to inertia) until something solid like the steering wheel is encountered. the driver then comes to rest in a much shorter distance - perhaps only a few centimeters. find the net force acting on a 65.0 kgkg driver who is decelerated from 18.0 m/sm/s to rest in 5.00 cmcm .

Answers

The net force acting on: 2.113 x 10⁵ N.

First, we need to calculate the deceleration of the driver by using 3rd equation of motion:  2as = Vf² - Vi²

where,

a = deceleration = ?

s = distance = 5 cm = 0.05 m

Vf = Final Velocity = 0 m/s

Vi = Initial Velocity = 18 m/s

Therefore,

2a(0.05 m) = (0 m/s)² - (18 m/s)²

a = (- 324 m²/s²)/0.1 m

a = - 3240 m/s²

where the negative sign represents a deceleration

From Newton's Second Law of Motion:

F = ma

F = (65 kg)(-3240 m/s²)

F = - 2.106 x 10⁵ N

So, the closest answer is:

F = 2.113 x 10⁵ N

What is force ?

In physics, a force is an effect that can change the motion of an object. A force can cause an object with mass to change speed, ie. accelerate A force can also be intuitively described  as a push or a pull. A force has both magnitude and direction, so it is a vector quantity.

When a body tends to change or change its state under the influence of an external cause, it is called a force. A body can also change its shape, size and direction when a force is applied. For example, kicking a ball, pulling and pushing a door.

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Jack and Jill takes 14 minutes to walk 340 m [West], then 129 m [South]. Calculate
a. their total displacement
b. their average speed
c. their average velocity

Answers

Answer:

a. their total displacement = 364 meters

b. their average speed      = 33.5 meters/minute

c. their average velocity    =  26 meters/minute SW

Explanation:

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A freight train moves due north with a speed of 8 m/s. The mass of the train is 6 x 106 kg. How fast would a 1500 kg automobile have to be moving due north to have the same momentum as the train

Answers

The freight train has a momentum of mv = (4.5 × 105 kg)(1.4 m/s) = 6.3 ×105 kg-m/s.

An automobile with a mass of 1800 kg would need to go at a speed of 6.3  × 105 kg-m/s ÷ 1800 kg = 350 m/s in order to have a momentum of the same magnitude. (Roughly 800 mph)

A moving body's amount of motion is referred to as momentum. It has units of kg m/s and is theoretically represented as p = m * v.

The momentum equation connects the total force exerted on a fluid element to its acceleration or rate of change of momentum and is a formulation of Newton's Second Law. You're certainly familiar with the formula F = ma, which links applied force and acceleration in solid mechanics analysis.

Momentum is the result of a particle's mass and velocity. Being a vector quantity, momentum possesses both magnitude and direction. According to Isaac Newton's second equation of motion, the force applied on a particle is equal to the time rate of change of momentum.

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Determine normal boiling point of chloroform if its heat of vaporization is 31. 4 KJ/mol and it has a vapor pressure of 190 mmHg at 25 °C

Answers

Answer:

The answer is 61.45 °C

Explanation:

Clausius-Clapeyron equation:

P2= Normal pressure

T2= Normal boiling point

ln(p2/p1)=−(ΔHv/R)*(1/T2 − 1/T1)

ln(760/190) = –(31400/8.314)*[(1/T2) – (1/298)]

T2 = 334.6 K = 61.45 °C

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Slim Jim, continually maintaining his svelte body, lifts a 70 kg barbell 1.4m above the grounda) How much energy did the barbell have when it was on the ground?b) How much energy does it have after being lifted 1.4m? What kind of energy does it have afterbeing lifted? Where did it come from?c) How much work did Jim do to the lift the object?dIf he lifted it in 1.5s how much power did he use?

Answers

Given,

The mass of the barbell, m=70 kg

The height to which Slim Jim lifts the barbell, h=1.4 m

a)

When the barbell was on the ground, it will have zero kinetic energy as it has no velocity. And if assume the height of ground as zero meters, then its potential energy is also zero.

Thus when the barbell was on the ground, its energy was zero joules.

b)

The energy of the barbell when it is at a height of h is given by,

[tex]E=\text{mgh}[/tex]

Where g is the acceleration due to gravity.

On substituting the known values,

[tex]\begin{gathered} E=70\times9.8\times1.4 \\ =960.4\text{ J} \end{gathered}[/tex]

Thus the energy that the barbell has after being lifted 1.4 m is 960.4 J

This energy is the energy stored in the barbell due to its position. Thus the energy stored in the barbell is the potential energy.

Slim Jim has to do some work to lift the barbell to the given height. This work done will be stored in the barbell in the form of potential energy. That is, the energy of the barbell is supplied to it from Slim Jim through the work.

c)

All the work done by Jim will be stored in the barbell in the form of potential energy. Thus, the work done by Jim is equal to the potential energy of the barbel.

Therefore, the work done by Jim is 960.4 J

d)

Given,

The time interval, t=1.5 s

The power is given by,

[tex]P=\frac{W}{t}[/tex]

Where W is the work done by Jim.

On substituting the known values,

[tex]\begin{gathered} P=\frac{960.4}{1.5} \\ =640.27\text{ W} \end{gathered}[/tex]

Thus the power used by Jim is 640.27 W

now assume that the crate is sliding along the floor with constant velocity. what is the magnitude fp of the force you need to exert on the crate to make it continue to slide along the floor with constant velocity?

Answers

The magnitude[tex]F_{P}[/tex]  of the force you need to exert on the crate to make it continue to slide along the floor with constant velocity is  437.5N.

How friction force can be calculated?

The Normal Force (in a horizontal plane, it is comparable to the mass by gravity) and the Kinetic Coefficient of Friction must be multiplied in order to obtain the Friction Force, which must be used in order to solve this problem.

Given that the sentence does not specify the value of the coefficient of friction, I will ascribe a value of 0.5 between the crate and the ground. If a different value has been specified for this coefficient, the below equation will simply need to be changed to reflect it.

[tex]F_{P}[/tex] = µK N

µK = Friction coefficient

Normal Force is N.

Replacing,

[tex]F_{P}[/tex] = 0.5 × 875

[tex]F_{P}[/tex] = 437.5N.

Consequently, the crate is subject to a friction force of 437.5N.

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Complete question is "Suppose you have to move a heavy crate of weight 875 {\rm N} by sliding it along a horizontal concrete floor. You push the crate to the right with a horizontal force of magnitude 300 {\rm N}, but friction prevents the crate from sliding. Now assume that the crate is sliding along the floor with constant velocity. What is the magnitude F_p of the force you need to exert on the crate to make it continue to slide along the floor with constant velocity?"

What are three sources of errors in a center of gravity lab experiment

Answers

The centre of gravity, in physics, is an imaginary point in a body of matter where for convenience in certain calculations,

What are three sources of errors in a center of gravity?

Sources of errors for the center of gravity of an irregularly shaped object: -environmental error: when the wind blows it may remove the irregularly shaped object from equilibrium. add: density variations, shape variations say, thickness], inability to measure precisely, inability to compute precisely. For example, as a result of a number of the center of gravity measurements, we may have the best estimate of the true value for the acceleration due to gravity, the center of gravity is the point between which the force of gravity move on a thing or system. In most mechanics problems the gravitational pasture is assumed to be constant. The center of gravity is then in entirely the same position as the center of mass.

So we can conclude that It is important to know the center of gravity because it predicts the behavior of a moving body when acted on by gravity.

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two boxes of the same mass are connected by a string of negligible mass and pulled across a level tabletop with negligible friction by another horizontal string. the force applied by the leading string is f. m m t f a how does the magnitude of the force t exerted by the string connecting the boxes compare to the magnitude f?

Answers

The importance of the pressure t exerted via the string connecting the bins evaluate to the value f is T< f.

A force is a power that can trade the movement of an object. A force can encourage an item with mass to exchange its tempo, i.e., to boost up. stress can also be described intuitively as a push or a pull. A strain has both importance and path, making it a vector quantity.

considering both masses together

The acceleration of the system = a

a = f/m+m = f/2m

Now, considering one mass towards left m

T = ma

T = (f/2m)

m = f/2

T = f/2

T<f

Therefore, T must have a magnitude less than f.

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A constant force is used to punch a box along the floor a distance of 3 meters by applying 300j what is the magnitude of force?

Answers

The magnitude of the force is equal to 100 newtons.

How do you prove that?

It's easy! You just need to divide the work by the displacement:

300 = F · 3

300 = 3F

300/3 = 3F/3

100 = F

F = 100

Here, F is the magnitude of force. The standard unit for force is newton (N). Recall that in physics, work is mathematically defined as W = F · x where W is work, F is force, and x is displacement. The standard unit for work is joule (J) and the standard unit for displacement is meter (m).

You are given that the work applied is equal to 300 joules and the displacement is equal to 3 meters. The force used to punch a box along the floor would then be 100 newtons.

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compare and contrast the energy of moving water and the energy of water that is stored.
I need to get it done

Answers

The main difference between the energy of moving water and the energy of stored water is that the energy of moving water has kinetic energy whereas the energy of water that is stored has potential energy.

What are kinetic and potential energy?

Potential energy is defined as the stored energy in any object due to its position while on the other hand, while Kinetic energy is defined as the energy an object has due to its motion. The energy stored in the water molecules which is used for the movement of these molecules is called water potential. Stored water has potential energy but the kinetic energy is present only in the moving molecules of water.

So we can conclude that moving water has kinetic energy while on the other hand, the energy of water that is stored has potential energy.

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7 The distance-time graph for a motorcyclist riding
off from rest is shown in Figure 1.2.18.
a Describe the motion.
b
Calculate how far the motorbike moves in
30 seconds.
c Calculate the speed.

Answers

Answer:

a) Uniform rectilinear motion

b) 600 m

c) 20 m/s

Explanation:

a) average velocity is described as the change of position divided by the time of travel, so v= ∆x/∆t. This is a linear function (y = mx+ q) which is the same structure of the equation needed to calculate average velocity (x2= x1 + vt). If you do the ratio between a distance and the corresponding time you'll see that the average velocity is the same, which is typical of this kind of motion.

b) At the the time of 30 seconds the corresponding distance on the y axis is 600 m.

c) Again, v=∆x/∆t and we can rewrite is as v=x2-x1/t. If you take the same data we've used before you can write v= 600m-0/ 30s= 20 m/s. The average velocity is 20 m/s.

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