An elevator in a hotel starts at ground level. It rapidly moves up to the third floor, where it stops to unload passengers. The elevator then descends to the basement at a slower speed than before.

Identify whether or not the graph correctly represents the story. Explain your answer in 1-2 sentences.

The height of the cliff is 24.2 m.

Let:

Using the Second Equation of Motion:

                               [tex]h = ut + 1/2 at^{2} \\h = 0 + 1/2 (10)(2.2)^{2} \\h = 0 + 5(2.2)^{2} \\h = 5(4.84)\\h = 24.2m[/tex]

The Second Equation of Motion establishes a relationship between displacement and time. It can be used to calculate the displacement and time, in different conditions.

The cliff was 24.2m high above the water surface, into which the cliff diver dived.

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Mechanical advantage is a measure of the ratio of the output force produced to the input force applied in a system, mainly in the case of simple machines, such as levers and pulleys. It is a dimensionless quantity.

From the definition, the mechanical advantage of a simple machine is inversely proportional to the input force applied on the system. So, an increase in this force decreases the mechanical advantage of a simple machine.

For example, the mechanical advantage of a lever can be decreased by increasing the force on load arm or by reducing the force on the effort arm.

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Measures the diameter of a shotgun's barrel in terms of the quantity of lead balls needed to make the bore weigh one pound (12 gauge is the diameter of a lead ball weighing one-twelfth of a pound).

A shotgun has many projectile properties while a rifle only comes with one. Shotguns are helpful at close ranges when rifles are useful at a distance. Shotguns only have a front sight, whereas rifles have both front and rear sights.

Rifles are typically a little noisier than shotguns. Analyzing test results reveals that shotguns and rifles have decibel levels that are comparable. The typical decibel range

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ANSWER

B) 8.00 J

EXPLANATION

Given:

• The force applied to move the block, F = 4.00N

• The distance the block is moved, d = 2.00 m

Unknown:

• The work done on the block, W

If there is no friction between the block and the table, the only force acting upon the block in the direction of movement is the applied force F. Since the force is applied in the direction of movement, then the angle between the force and the displacement is 0, so the work done on the block is,

Hence, the work done on the block is 8.00 J.

Answer:

Resultant force of the given forces: approximately [tex]11.2\; {\rm N}[/tex] at approximately [tex]53^{\circ}[/tex] west of south.

Explanation:

The resultant force of these forces will be:

Refer to the diagram attached. The resultant [tex]5\; {\rm N}[/tex] force to the south and the [tex]10\; {\rm N}[/tex] force to the west are perpendicular to each other, forming the two legs of a right triangle. The hypotenuse of this right triangle will be the net effect of these two forces.

Apply Pythagorean's Theorem on this triangle to find the magnitude of this net effect:

[tex]\begin{aligned}(\text{length of hypotenuse}) &= \sqrt{10^{2} + 5^{2}} \approx 11.2\end{aligned}[/tex].

Hence, the magnitude of this net effect will be approximately [tex]11.2\; {\rm N}[/tex].

Let [tex]\theta[/tex] denote the angle between this resultant force and west. In this right triangle:

[tex]\begin{aligned} \tan(\theta) &= \frac{(\text{opposite})}{(\text{adjacent})} \\ &= \frac{5\; {\rm N}}{10\; {\rm N}} \\ &= \frac{1}{2}\end{aligned}[/tex].

[tex]\begin{aligned} \theta &= \arctan\left(\frac{1}{2}\right) \approx 27^{\circ}\end{aligned}[/tex].

Hence, the angle between this resultant force and south will be approximately [tex](90^{\circ} - 27^{\circ}) = 63^{\circ}[/tex]. This resultant force will be at approximately [tex]63^{\circ}[/tex] west of south.

An elliptical galaxy is a galaxy with an elliptical outline and a smooth distribution of brightness (no spiral arms).

In the field of astronomical physics, we can describe elliptical galaxies as stretched galaxies that have an elliptical outline. This type of galaxy is usually made through old stars and has an even, smooth distribution of light.

Although the elliptical galaxies are not a major part of the universe they make up 10–15% in the Virgo Supercluster.

As the elliptical galaxy is made from broken stars, their size may vary from being a dwarf elliptical to a supergiant.

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A car is driving with a constant velocity around a circular track. The track’s radius is 156.1 m, and the car returns to its starting position in

49.3 s the car’s centripetal acceleration is 0.327 m/sec².

Rate of change of velocity with respect to time is called acceleration.

when the magnitude of velocity is constant and direction is changing

than the acceleration is centripetal acceleration.  

Given that is question radius of track  is 156.1 m and  car returns to its starting position in 49.3 s.

Velocity of car = Circumference of Circle / total time

                        = 352.308/ 49.30

                 v     = 7.14 m/sec

Centripetal acceleration = v²/r

                                         = 51.06/156.1

                                         = 0.327 m/sec²

The car’s centripetal acceleration is  0.327 m/sec².

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Mechanical energy  type of energy is the sum of kinetic and potential energy in an object that is used to do work.

Potential energy plus kinetic energy are combined to form mechanical energy. It refers to the power underlying an object's location and motion. A compressed spring, for instance, has mechanical energy in the form of potential energy, while a moving vehicle has mechanical energy in the form of kinetic energy.

A sequence of blades installed on a rotor shaft in a turbine generator are pushed by a flowing fluid, including such air, water, steam, combustion gases, or water. A generator's rotor shaft rotates or spins due to the fluid's force on the blades. In turn, the generator converts the rotor's physical (kinetic) energy into electricity.

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The two stones will fall at the different time. But the difference between the time is just of one second.

The distance of the cliff is also considered. Here, the speed or velocity is asked. To calculate this we have to use the equation of motion.

There are three equations of motion.

We have;

The kinematic equation of motion that can be used is s = ut - (1/2)·gt²

For the first stone, we have, s₁ = ut₁ - (1/2)·gt₁²

For the second stone, we get; s₂ = ut₂ - (1/2)·gt₂²

g = The acceleration due to gravity ≈ 9.81 m/s²

s = The distance below the cliff top

The initial velocity of the stones, u = 0

To calculate the change of distance between the two stones,

we will subtract the two equation of motions are

s₁-s₂

With all the values kept together in the equation of motion, we can easily find the change of distance between the two stones.

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Given data

*The given mass of the car is m = 1610 kg

*The given effective force constant is k = 5.75 × 10^4 N/m

(a)

The formula for the period of oscillation of a 1610 kg car is given as

Substitute the known values in the above expression as

Hence, the time period of oscillation of a 1610 kg car is T = 1.05 s

(b)

As from the given data, the amplitude of the oscillation of the car decreases by a factor of 5.00. Then, the expression for the amplitude of the oscillation, and the damping constant (b) is calculated as

Substitute the known values in the above expression as

Hence, the damping constant is b = 9871.2 kg/s

If two covalently bonded atoms move closer than a distance of the bond length, the potential energy of the atoms increases

Stronger bonds have lower potential energy than weaker ones. The weaker connection has a higher potential energy. Ionic bonds are stronger than covalent ones, hence covalent bonds have higher potential energy.

The tremendous repulsive interactions between the positively charged nuclei of the bound atoms cause the energy of the system to dramatically rise if the atoms move closer to one another than the ideal bond distance.

If there is an attractive force between the two, the potential energy will rise with more separation, but if there is a repulsive force, the potential energy will fall with greater separation.

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The problem describes a situation where a bullet is launched at a certain angle respect to the horizontal to hit in the target, given the launching conditions and the distance to the target. The formulas to compute the magnitudes involved in the projectile motion are

x=[tex]v0[/tex] cos0 t

Y=[tex]V0[/tex] sin0 t

xm=[tex]v0[/tex]² sin 20/g

[tex]Xm[/tex] is the maximum horizontal distance the object can reach when it returns to the ground (or launching height).

let's check if X is computed correctly .

The set of data is correct. We can now proceed to the required distance. We know the bullet is shot directly to the target, so the angle is 0, keeping the same speed. The new maximum distance is 150 m, so we can compute the time needed to complete the flight

since

Let's find the time from the formula

X=[tex]V0[/tex] cos0 t

t=[tex]Xm[/tex]/[tex]V0[/tex] cos0

t=150/100=1.5

Y=(9.8)(0.66)²÷2

Y=-2.134

The bullet will hit 2.134 meters below the target

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a. The speed of the satellite is  244 m/s

b. The period of the satellite is 1.72 × 10⁷ s

c. The radial acceleration of the satellite is 8913 rad/s

The speed of the satellite is given by v = √(GM/R) where

So, substituting the values of the variables into the equation, we have that

v = √(GM/R)

v = √(6.67 × 10⁻¹¹ Nm²/kg² × 5.972 × 10²⁴ kg /6380 km + 300 km)

v = √(6.67 × 10⁻¹¹ Nm²/kg² × 5.972 × 10²⁴ kg /6680 km)

v = √(39.83324 × 10¹³ Nm²/kg ÷ 6.680 × 10⁶ m)

v = √(5.963 × 10⁻³ × 10⁷ Nm²/kg)

v = √(5.963 × 10⁴ Nm/kg)

v = 2.44 × 10² m/s

V = 244 m/s

So, the speed of the satellite is  244 m/s

The period of the orbit is given by T = 2πR/v where

So, substituting the values of the variables into the equation, we have

T = 2πR/v

T = 2π × 6680 km/244 m/s

T = 2π × 6680 × 10³ m/244 × 10⁵ m/s

T = 13660π × 10³ m/244 × 10⁵ m/s

T = 41971.68 × 10³ m/244 × 10⁵ m/s

T = 17201.5 × 10³ s

T = 1.72015 × 10⁷ s

T ≅  1.72 × 10⁷ s

So, the period of the satellite is 1.72 × 10⁷ s

The radial acceleration of the satellite is given by a = v²/R where

Substituting the values of the variables into the equation, we have

a = v²/R

a = (244 m/s)²/6680 km

a = 59536 m²/s²/6680 × 10³ m

a = 8.913 × 10³ rad/s

a = 8913 rad/s

So, the radial acceleration of the satellite is 8913 rad/s

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The direction of the velocity of the clod of dirt relative to the truck just before it hits is in the same direction as that of the truck.

The direction of the velocity of the clod of dirt relative to the ground just before it hits is in the opposite direction as that of the ground.

The relative velocity of an object in motion is the velocity of that object when viewed from a reference point.

The direction of the velocity of the clod of dirt before it hits the ground is in the same direction as that of the truck.

However, compared to the ground, the direction of the relative velocity is in the opposite direction.

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The larger the velocity changes measured for the planet, the larger will be the mass of the star and smaller will be the orbit's radius.

v = √ ( G M / r )

v = Orbital velocity

G = Gravitational constant

M = Mass of the central body

r = Radius of orbit

v ∝ √ M

v ∝ 1 / √ r

From the above equation is is clear that velocity of the planet increases with the increase of star's mass and increases with the decrease of planet's orbital radius.

Therefore, the larger the velocity changes measured for the star, the larger will be the mass of the star and smaller will be the orbit's radius.

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The speed with which they threw the bundle is  14 m/s.

We know that the speed of the bundle can be obtained from the use of the equations of kinematics. Now we can use the equation of the upward motion under gravity here.

Using;

v^2 = u^2 - 2gh

v = final velocity

u = initial velocity

g = acceleration due to gravity

h = height

Now;

v = 0 m/s at the maximum height

u^2 = 2gh

We now have;

u = √2gh

u = √ 2 * 9.8 * 10

u = 14 m/s

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When a ladder with 20 ft. long is pulled along the floor away with a constant rate 1.2 ft/s, then the top of the ladder is moving toward the floor with velocity - 0.9 ft/s and there is no acceleration.

Let:

x = the distance between the bottom of the ladder from the wall

y =  the distance between the top of the ladder from the floor

When x = 12 ft, then using the Pythagorean Theorem:

x² + y² = 20²    (Equation 1)

12² + y² = 20²

y² = 20² - 12² = 16²

y = 16 ft.

Take the derivative of equation 1 with respect to t

2x . dx/dt + 2y. dy/dt = 0

Substitute x = 12, y = 16, dx/dt = 1.2 ft/s

2 . 12 . (1.2) + 2 . 16 . dy/dt = 0

dy/dt = - 0.9 ft/s

Hence, the velocity is - 0.9 ft/s

Since the floor is moving with a constant rate, the acceleration is zero.

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The answer is true about the cabins in commercial airliners that require pressurization.

Airplanes are pressurized because the air is very thin at the high altitude where they fly. The passenger jet has a cruising altitude of about 30,000 - 40,000 feet. At this altitude or height, humans can't breathe very well and our body gets less amount of oxygen. Most aircraft cabins are pressurized to an altitude about 8,000 feet. This is called cabin altitude. Aircraft pilots have access to the control's mode of a cabin pressure control system and if needed it can command the cabin to depressurize.

So we can conclude that cabins in commercial airliners require pressurization because of the greater pressure of the surrounding environment.

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Answer:

the velocity of a ball at highest point is 8.5m/s

Explanation:

: The Freefall Problem

Xue,is standing on the roof of a building. Emily is standing below and tosses a ball straight upwards to Xue. It travels up past

Xue, comes back down and Xue reaches out and catches it. Xue catches the ball 6.0 m above Emily’s hands. The ball was

travelling at 12.0 m/s upwards the moment it left Emily’s hand. We would like to know how much time this trip takes.

1. Represent. Complete part A below. Indicate the y-origin for position measurements and draw a sign convention where

upwards is positive. Label the important events.

2. Represent. Complete part C below. Make sure the two graphs line-up vertically. Draw a single dotted vertical line

through the graphs indicating the moment when the ball is at its highest.

A: Pictorial Representation

Sketch, coordinate system, label givens, conversions, describe events

Event 1:

Event 2:

Event 3:

C: Physics Representation

Motion diagram, motion graphs, key events

The total length of the path traveled by an object is the distance. The change in position, from one event to another is the

displacement. Distance is a scalar quantity and displacement is a vector quantity.

3. Reason. Explain why this in this example it is relatively easy to find the displacement, but harder to find the distance.

4. Reason. The BIG 5 equations are valid for any interval of motion where the acceleration is uniform. Does the ball

accelerate uniformly during events 1 and 3? Explain D: Mathematical Representation

Describe steps, complete equations, substitutions with final statement isThere are multiple ways of solving…one

will require the use of the quadratic formula.

The other way will take additional steps.

x = -b+-under root b^2-4ac/2a

The net force on an object if a  force of 30 N to the right is applied to an object. An opposite force of 20 N to the left is applied to the same object is 10 N to the right. And the right option is D.

The net force is the vector sum of all the forces that act upon an object.

To calculate the net force applied to the object, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence the right option is D. 10 N to the right.

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The material that finally worked in developing the light bulb is

Carbonized cardboard/ filament.

The formation of the light bulb was an important contribution to physics and mankind. However, various experiments were conducted in different labs to produce a light bulb.

The scientists of the olden times were having trouble in finding a material that could burn for a long and not have such a short lifespan.

It was the lab in which Thomas Edison worked that first used carbonized carbon filament in the light bulb. Their research proved to be a breakthrough because carbon filaments have a longer lifespan and burned effectively when electricity was provided.

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The springs inside a bathroom scale will be more compressed.

What is acceleration due to gravity?

In physics, gravitational acceleration is the acceleration of an object in free fall in a vacuum (hence no resistance). This is a steady increase in velocity caused solely by the gravitational pull. All objects accelerate at the same speed in a vacuum, regardless of their mass or composition . The measurement and analysis of these velocities is known as gravimetry.

At a fixed point on the surface, the magnitude of the Earth's gravity is the combined effect of gravity and the centrifugal force due to the rotation of the Earth. Free-fall acceleration at various points on the Earth's surface ranges from 9.764 to 9.834 m/s2 (32.03 to 32.26 ft /s2) depending on altitude, latitude and longitude . The old standard is precisely defined as 9.80665 m/s2 (32.1740 ft/s2). Locations that deviate significantly from this value are called gravity anomalies. Other effects such as lift and air resistance are not taken into account.

When the elevator accelerates and shows a higher weight reading. As the lift accelerates and descends, the springs in the bathroom scale are compressed less and the displayed weight value is reduced.

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Answers:

Distance = 40 meters

Displacement = 0 meters

============================================

Reason:

The distance traveled is simply the sum of the two distances 20 and 20 to get 20+20 = 40. It's what anyone would expect when they ask "how far did the person travel in total?"

The displacement is the change in position from start to end. In this case, the student ends up back where they started, meaning no change has happened. If the teacher closed their eyes and opened them after the student gets back, then for all the teacher knows, the student hasn't moved at all.

Put another way, displacement tells us how far away the student has ended up compared to the start position. But since they went back to the start, no change has occurred.

Given data

*The given mass of the car is m = 1475 kg

*The maximum tension is T = 2551 N

The formula for the maximum possible acceleration of the truck is given by Newton's second law as

Substitute the known values in the above expression as

Hence, the maximum possible acceleration of the truck is a_max = 1.72 m/s^2

The elasticity and Poisson's ratio for this metal is 0.232.

What is ratio?

The realation between two numbers which shows how much bigger one quantity is than another.

Sol-

As per the given question

P=190KN

d=16 mm

Lo=50mm

X=50.1349-50=0.1349mm

Y=15.99-16=-0.01mm

The formula-

E=ó/€

Ó=P/A

A=r/4 d^2 =π/4(16)^2=201.062 mm

ó={190(1000)}201.062=944.982 Mpa

E=944.982/0.002698=350.253 GPa

€y=-0.000625

v=0.232(answer)

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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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Answer:

25000Nm⁻² using rounded g figure

Explanation:

The box is in cm. It needs to be converted into SI metres.

1m = 100cm

1cm = ( 1 / 100 )m

2cm = ( 1 / 50 )m

A = ( 1 / 2500 )m²

Assume that g = 10ms⁻¹. Use your own definition.

F = 10N

P = F / A

P = 10N( 2500 )m⁻²

25000Nm⁻²

Answer:

Explanation:

Given:

m = 2 kg

a = 2 cm = 0.02 m

g =9.8 m/s²

_______________

P - ?

Area of the cube:

S = a² = 0.02² = 0,0004 m²

Gravity:

F = m·g = 2·9.8 = 19.6 N

Pressure:

p = F / S =  19.6 / 0,0004 = 49 000 N/m²

The acceleration of the plane is:  -5.59 m/s²

The formula for uniformly varied rectilinear motion (UVRM) and procedure that we will use to solve this exercise is:

a = (vf - vi) /t

Where:

Information about the problem:

Applying the acceleration formula we have:

a = (vf - vi) /t

a = (0 m/s - 43.0 m/s) /7.69 s

a = (-43.0 m/s) /7.69 s

a =  -5.59 m/s²

It is a physical quantity that indicates the variation of velocity as a function of time, it is expressed in units of distance per time squared e.g.: m/sec2 ; km/h2

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A 10 kg box is pulled across a level floor, where the coefficient of kinetic friction is 0.35. The horizontal force is required for an acceleration of 2.0 m/s2 will be 54.3 N

Newton’s second law states that the acceleration of an object depends upon two variables – the net force acting on the object and the mass of the object.  The acceleration of the body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body.

equation of motion will be

F(net) =  mass * acceleration

F - fr = m * a

where

F = force by which the box is moving

fr = force due to friction

m = mass of the object

a = acceleration by the object is moving

F = fr + m * a

  = (mu * N) + ma                                          where N = mg

  = (mu * mg) + ma

  = (0.35 * 10 * 9.8) + 10 * 2

  = 54.3 N

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The acceleration of the object will be 13.33 m/s^2.

According to Newton's second law of motion,

The magnitude of Force is equal to the product of the mass of the body and the acceleration it is given ( or achieves ).

The units of force, mass and acceleration are Newton, kg and m/s^2 respectively.

In numerical terms,

Force = mass × acceleration

i.e. F = m × a

Given,

mass of the body = 15 kg

Force applied to the body = 200 N.

Putting these values in the equation,

200 = 15 × a

=> a = 200/15

=> a = 13.33 m/s^2.

Therefore, the acceleration of the given body will be 13.33 m/s^2.

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