An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.​

Answer:

Explanation:

1 Object C has an acceleration that is greater than the acceleration for D.

2 B

3 17M

4 The velocity is zero.

5 a straight line with negative slope

just took it

Answer:

heat

Explanation:

a.

The components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N

The force on a current carrying conductor in a magnetic field is given by F = iL × B where i = current = 9.00 A, L = 25.0 cmk = 0.25 mk (since the conductor is along the z-direction). B = magnetic field. Since B has component Bx = -0.242T, By= -0.985, and Bz = -0.336, B = -0.242i + (-0.985j) + (-0.336)k = -0.242i - 0.985j - 0.336)k.

So, F = iL × B

F = 9.00 A{(0.25 m)k × [-0.242Ti + (-0.985Tj) + (-0.336T)k]T}

F = 9.00 A{(0.25 m)k × (-0.242T)i + (0.25 m)k × (-0.985Tj) + (0.25 m)k × (-0.336T)k]}

F = 9.00 A{-0.0605mT)k × i + (-0.24625 mT)k × j + (-0.084 m)k × k]}

F = 9.00 A{-0.0605mT)j + (-0.24625 mT) × -i + (-0.084 mT) × 0]}

F = 9.00 A{-0.0605mT)j + (0.24625 mT)i + 0 mT]}

F = -0.5445 AmT)j + (2.21625 AmT)i + 0 AmT]}

F = -0.5445j + 2.21625i + 0 k

F = (2.2163i - 0.5445j + 0 k) N

So, the components of the force are Fx = 2.2163 N, Fy = -0.5445 N and Fz = 0 N

b.

The magnitude of the net force on the wire is 2.282 N

The net force F = √(Fx² + Fy² + Fz²)

F = √[(2.2163 N)² + (-0.5445 N)² + (0 N)²)

F = √[(4.912 N)² + 0.2964 N)² + (0 N)²)

F = √[5.2084 N)²

F = 2.2822 N

F ≅ 2.282 N

So, the magnitude of the net force on the wire is 2.282 N

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I think static is the correct answer

Answer:

Explanation:

a) The spring force will equal the weight.

b) If up is positive

kx - mg = 0

mg = kx              kx = 25 N

c) m = kx/g = 25/10 = 2.5 kg

The gravitational force acting between the two bodies is given by:

F=G

r

2

m

1

m

2

G=

m

1

m

2

Fr

2

The dimension of the force is [MLT

−2

]

=

[M][M]

[MLT

−2

][L

2

]

=M

−1

L

3

T

−2

Answer:

Remember, NORTH ^, EAST >, SOUTH v, WEST <

Explanation:

It doesn't have to be a super complex answer. All you have to do is look to the left (west) of the North American plate. What are the 2 plates that you see? The Pacific and the Juan de Fuca, yeah? To the South, there is the Cocos amongst a few others.

I am not going to share the answer for sure as I haven't completed the test yet but that's how I'm solving it. You should write the answer in your own words anyways. Hope this helps! Have a good day :)

Answer:

The Juan de Fuca Plate and the Pacific Plate both border the west side of the North American Plate.

Explanation:

Edmentum

Answer:

Explanation:

conservation of momentum during the collision

0.035(214) + 0.15(0) = 0.185v

v = 40.486 m/s

The kinetic energy after impact will convert to gravity potential energy

(ignoring air resistance)

mgh = ½mv²

     h = v²/2g

     h = 40.486² / (2(9.8))

     h = 83.6303...

     h = 84 m

Answer:

Light can travel in three ways from a source to another location: (1) directly from the source through empty space; (2) through various media; (3) after being reflected from a mirror.

Explanation:

The magnitude of the load L on the crane before the collapse is 3211.81 N

To determine the magnitude of the load on the crane (L), we will need to make use of the equilibrium conditions of the torque.

It is always an ideal process to list out all the parameters given as this will let you understand how you can determine the answer to the question from the given parameters.

From the given information;

From the question;

the angle at which the crane is positioned can be determined by taking the tangent of the angle θ. i.e.

[tex]\mathbf{tan \ \theta = \dfrac{h}{d-s}}[/tex]

[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{h}{d-s} \Big )}[/tex]

[tex]\mathbf{\theta = tan^{-1} \Big ( \dfrac{2.070 }{5.580 - 0.522} \Big )}[/tex]

[tex]\mathbf{\theta =22.26^0}[/tex]

Consider the equilibrium conditions of the torques with respect to the magnitude of the load at point P.

∴

[tex]\mathbf{Tsin \theta (d-s) - W_L (d-x) -(Mg) (\dfrac{d}{2}) = 0}[/tex]

By making the magnitude of the load [tex]\mathbf{W_L}[/tex] the subject of the formula, we have:

[tex]\mathbf{W_L = \dfrac{Tsin \theta (d-x) -(Mg) (\dfrac{d}{2})}{ (d-s) } }[/tex]

[tex]\mathbf{W_L = \dfrac{(11650 )sin (22.26) (5.580-1.350) -(88.50\times 9.81) (\dfrac{5.580}{2})}{ (5.580-0.522) } }[/tex]

[tex]\mathbf{W_L = 3211.81 \ N }[/tex]

Therefore, we can conclude that the magnitude of the load is 3211.81 N

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Answer:

jsbdhdndmlsusgsbkaksudgnslsosufhbf ffb

Answer:

circuit

Explanation:

Answer:

Explanation:

Without friction, a roller coaster continuously converts potential energy to kinetic energy and back again. Total energy will be constant.

Let m be the mass of the car and ground level is the origin.

on the 5.5 m hill, total energy is

E = PE + KE

E = mgh + ½mv²  

E = m(9.8)(5.5) + ½m(9.3)² = 97m J

a) The maximum height will occur when the total energy is all potential energy.

E = mgh

h = E/mg

h = 97m/m(9.8) = 9.9 m  

As this value is greater than the height of the third hill at 5.5 + 4.0 = 9.5 m The car will cross the last hill with some remaining velocity in kinetic energy.

b) As 9.5 m is greater than 9.3 m, the 9.5 m hill will have more of the total energy of the system as potential energy, This mean there is less kinetic energy and therefore less velocity (and speed) on top of the 9.5 m hill.

c) KE = E - PE

KE = 97m - m(9.8)(9.5 - 1.0)

KE = 97m = 83.3m

KE = 13.7m = ½mv²

v² = √(2(13.7)

v = 5.2345...

v = 5.2 m/s

Hi there!

We can begin by using the work-energy theorem in regards to an oscillating spring system.

Total Mechanical Energy = Kinetic Energy + Potential Energy

For a spring:

[tex]\text{Total ME} = \frac{1}{2}kA^2\\\\\text{KE} = \frac{1}{2}mv^2\\\\PE = \frac{1}{2}kx^2[/tex]

A = amplitude (m)

k = Spring constant (N/m)

x = displacement from equilibrium (m)

m = mass (kg)

We aren't given the mass, so we can solve for kinetic energy by rearranging the equation:

ME = KE + PE

ME - PE = KE

Thus:

[tex]KE = \frac{1}{2}kA^2 - \frac{1}{2}kx^2\\\\[/tex]

Plug in the given values:

[tex]KE = \frac{1}{2}(20)(0.3^2) - \frac{1}{2}(20)(0.3^2) = \boxed{0 \text{ J}}[/tex]

We can also justify this because when the mass is at the amplitude, the acceleration is at its maximum, but its instantaneous velocity is 0 m/s.

Thus, the object would have no kinetic energy since KE = 1/2mv².

Answer:

.192 kg x m^2

Explanation:

I= mass of a times radius of a squared + mass of b times radius of b squared +...

I= .6 kg x .4m^2 + .6 kg x .4m^2

= .192 kg x m^2

Hope this helps :)

Answer:

The objects outside the reference frame aren't moving. It appears this way since the vehicle you are inside is moving, but unless the objects are people, animals, or other vehicles, the objects aren't moving.

Answer:

a

Explanation:

sana po makatulong <3♡♡

The rolling disk's initial angular momentum is mR√[2(gR + v²)]/2

Using the law of conservation of energy, the initial mechanical energy E of the disk equals its final mechanical energy E' as it climbs the step.

So, E = E'

1/2Iω + 1/2mv² + mgh = 1/2Iω' + 1/2mv'² + mgh'

where I = rotational inertia of disk = 1/2mR² where m = mass of disk and R = radius of disk, ω = initial angular speed of disk, v = initial velocity of disk, h = initial height of disk = 0 m, ω' = final angular speed of disk = 0 rad/s (assumung it stops at the top of the step), v' = final velocity of disk = 0 m/s (assumung it stops at the top of the step), and h' = final height of disk = R/2.

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

1/2Iω² + 1/2mv² + mgh = 1/2Iω'² + 1/2mv'² + mgh'

1/2(1/2mR² )ω² + 1/2mv² + mg(0) = 1/2I(0)² + 1/2m(0)² + mgR/2

mR²ω²/4 + 1/2mv² + 0 = 0 + 0 + mgR/2

mR²ω²/4 + 1/2mv² = mgR/2

R²ω²/4 = gR/2 + 1/2v²

R²ω²/4 = (gR + v²)/2

ω² = 2(gR + v²)/R²

ω² = √[2(gR + v²)/R²]

ω = √[2(gR + v²)]/R

Since angular momentum L = Iω, the rolling disk's initial angular momentum is

L = 1/2mR² ×√[2(gR + v²)]/R

L = mR√[2(gR + v²)]/2

the rolling disk's initial angular momentum is mR√[2(gR + v²)]/2

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Answer:

200 kilocalories

Explanation:

Answer:it’s C

Explanation:

by using graph of velocity vs. time for two objects, Item 4 and Item 5 statement makes an accurate comparison of the motions for objects. thus option C is correct.

velocity is the  rate of change of the position of the object with respect to  reference and it is complicated but velocity is basically speeding a particular object in a specific direction.

Velocity is a vector quantity which means both magnitude (speed) and direction are combinedly define define velocity. The SI unit of velocity is meter per second (ms-1) and the magnitude or the direction of velocity of a body changes leads to acceleration.

Speed and velocity are the two closest term but the major difference between speed and velocity is that speed gives us an idea that the object with the faster rate of movement r where as  velocity  speed up as well as  tells us the direction of the body

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Hi there!

Part A:

The only quantities explicitly given to us are:

3. mass (m)

4. Maximum velocity (vmax)

5. Amplitude (A)

8. Position x from equilibrium

Part B:

To solve, we must begin by calculating the force constant, 'k'.

We can use the following relationship:

[tex]v = \sqrt{\frac{k}{m}(A^2-x^2)[/tex]

We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:

[tex]\frac{v^2}{(A^2-x^2)} = \frac{k}{m}[/tex]

[tex]\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44[/tex]

Now, we can find the velocity at 4.7cm (0.047m) using the equation:

[tex]v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s[/tex]

Plug this value into the equation for kinetic energy:

[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}[/tex]

Part C:

The potential energy of a spring is given as:

[tex]U = \frac{1}{2}kx^2[/tex]

Find 'k' using the derived quantity above:

[tex]\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m[/tex]

Now, calculate potential energy:

[tex]U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}[/tex]

Answer:

Explanation:

Δx means a change in the magnitude of the x variable, often used in reference to a number line on the horizontal axis of a plot.

Answer:

the answer is weight=10N

Answer:

[tex]\boxed {\boxed {\sf 9.8 \ Newtons}}[/tex]

Explanation:

Weight is also called the force of gravity. This force acts on all objects at all times, pulling them down toward the center of the Earth.

It is calculated by multiplying the mass by the acceleration due to gravity.

[tex]F_g=mg[/tex]

The mass of the object is 1 kilogram. This scenario is occurring on Earth, so the acceleration due to gravity is 9.8 meters per second squared.

Substitute the values into the formula.

[tex]F_g= 1 \ kg *9.8 \ m/s^2[/tex]

Multiply.

[tex]F_g= 9.8 \ kg*m/s^2[/tex]

Convert the units. 1 kilogram meter per second squared is equal to 1 Newton, so our answer of 9.8 kilogram meters per second squared is equal to 9.8 Newtons.

[tex]F_g= 9.8 \ N[/tex]

A 1 kilogram mass at Earth's surface weighs 9.8 Newtons.

Answer:

Explanation:

As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.

Ff = Fcosθ

if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.

μN = Fcosθ

μ(mg - Fsinθ) = Fcosθ

μmg = Fcosθ + μFsinθ

100μ = F(cos30 + μsin30)

F = 100μ / (cos30 + ½μ)

Ff = 100μcos30 / (cos30 + ½μ)

A 100 N create is being pulled at a constant velocity by a rope a 30 degrees to the horizontal as depicted in the diagram given in question the force of friction Ff = 100μcos30 / (cos30 + ½μ).

A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

As the velocity is constant, Net force is zero. This means that the friction force must equal the applied force in the horizontal direction.

Ff = Fcosθ

if we had a coefficient of kinetic friction μ, we could quantify the friction force more precisely.

μN = Fcosθ

μ(mg - Fsinθ) = Fcosθ

μmg = Fcosθ + μFsinθ

100μ = F(cos30 + μsin30)

F = 100μ / (cos30 + ½μ)

Ff = 100μcos30 / (cos30 + ½μ)

the force of friction Ff, is 100μcos30 / (cos30 + ½μ).

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Answer:D: the velocity is zero

Explanation:

Answer:

440 m

Explanation:

S=(u+v) t / 2

S = (11+33) × 20/2

S= 44× 20/2

S=440 m

[tex]\omega = 21.8\:\text{rad/s}[/tex]

Explanation:

We know that there are [tex]2\pi[/tex] radians in one revolution and 60 seconds in one minute so we can easily convert the rev/min unit to rad/s using the following conversion factors:

[tex]208\:\dfrac{\text{rev}}{\text{min}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}}×\dfrac{1\:\text{min}}{60\:\text{s}}[/tex]

[tex]\;\;\;\;\;=21.8\:\text{rad/s}[/tex]

Answer:

[tex]1.74\times10^3 kg; 100m[/tex]

Explanation:

Step a: mass of the car. Let's grab the definition of kinetic energy: [tex]K= \frac12 mv^2[/tex]. We have K, we have v (which we should convert in meters per second, dividing by 3.6) to get:[tex]4.22\times10^5 = \frac12m(22.03)^2 \rightarrow m= 2\times4.22 / 495.22 \times 10^5 = 1.74 \times 10^3 kg[/tex]

Point a is done.

Now for the (b)reaking part. (I'm sorry, it's an horrible joke, but I couldn't resist)

In theory we have the mass, we have the force, so we could find the acceleration, find how long it takes to slow down, and then find the distance traveled. Too long. Let's do things more easily: when the car slows down to 56 km/h it will have a different kinetic energy. The difference in kinetic energy is the work done by the breaking force ofer the slowing distance.

[tex]K_f-K_i=W[/tex] A quick note on signs: if you look carefully the final kinetic energy will be less than the initial value, thus the work will be negative: it means it's correct, since the work is against the motion, slowing it down. Let's get calculating, first by converting 56 kmh in m/s (15,56 m/s), then finding the final kinetic energy:

[tex]K_f =\frac12 (1.74\times10^3) (15.56)^2 =2.11 \times 10^5 J[/tex]

The difference will be the work done by the force, or

[tex](2.11 - 4.22) \times 10^5 = \vec F\cdot \vec x=Fx[/tex] where  we are assuming that force and displacement have the same line of actions to simplify the dot product.

[tex]2.11\times 10^5 = 2100x = 1.00\times 10^2 m[/tex]