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:

An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.

Explanation:

Answer:

An effect begins to alter movement, and the direction of moves in the circular path is known as centripetal force. Its measurable unit is Newton or Kilogram meter per square of the second. The product of mass and square of velocity divided by the radius of path travel by the body provide s the term centripetal force.

Explanation:

Answer:

The angle of incidence = The angle of reflection

Explanation:

For instance, the formula is <i = <r which means if the surface is smooth then the reflacted anglw will be equal according to the normal (dotted 90°) line.

Upon being reflected backwards the angle of incidence is simply the same, except from the other side

The energy in the system is given by the initial potential energy at the point 1.

The linear velocity at point 3, is approximately 33.59 m/s.

Reasons:

The parameters are;

Height at point 1, h₁ = 83 m

Radius of the ring = 8 cm

Mass of the ring, M = 8 kg

Height at point 2, h₂ = 32 m

At point 2, we have;

Change in potential energy = Kinetic energy

Which gives;

(83 - 32) × 9.81 × 8 = 0.5 × 8 × v² + 0.5 × 8 × 0.08² × (v/0.08)²

Which gives;

v ≈ 22.37 m/s

At point 3, the rotational kinetic energy remains constant while the

translational kinetic energy increases as follows;

K.E. at point 3 = Initial kinetic energy + Change in potential energy

Which gives;

K.E. at point 3 = 0.5 × 8 × v₃³ ≈ 0.5×8×22.37² + 32×9.81×8

[tex]v_3^2 = \dfrac{0.5 \times 8 \times 22.37^2 + 32 \times 9.81 \times 8}{0.5 \times 8} = 1128.15[/tex]

v₃ ≈ √(1128.15) ≈ 33.59

The linear velocity at point 3, v₃ ≈ 33.59 m/s

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The probable question parameters as obtained from a similar question online are;

Height at point 1, h₁ = 83 m

Radius of the ring = 8 cm

Mass of the ring, M = 8 kg

Height at point 2, h₂ = 32 m

The sun emits electromagnetic waves with a power of  

4.0 ∗ 10  (26)  W.

Answer:

According Lisa, both the ball and the balloon have the same forward velocity of Vx.

(D) is correct

I think force can accelerate

Explanation:

i think force of an object

Answer:

Explanation:

20 n is an unknown amount

If that is supposed to be 20 N(ewtons)

then W = Fd = 20(15) = 300 J

Answer: it will be 300 newton meters

Explanation:

Answer:

v down exponenet 1 brainlest

Explanation:

Answer:

v0 [vee nought] is the initial velocity when time=0

Answer:

Explanation:

s = s₀ + v₀t + ½at²

50 = 0 + 0(4) + ½a(4²)

50 = 8a

a = 50/8 = 6.25 m/s²

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]

The steepness of the slope is therefore;

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]

Where;

[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m

[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m

[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]

[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]

[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]

The maximum steepness of the slope where the truck can be parked is 54.55 %.

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Answer:

a thing or event that evokes a specific functional reaction in an organ or tissue.

or

a thing that rouses activity or energy in someone or something; a spur or incentive.

We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"

Answer:

From the question we are told

This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.

A) Resistivity is given as,

[tex]p = \frac{RA}{l}[/tex]

where,

[tex]R = \frac{V}{I}[/tex]

Therefore,

[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]

B) Conductivity is given as,

[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]

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Answer:

Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].

Explanation:

Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].

If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].

Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.

Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)

By Newton's second law of motion, the acceleration of this vehicle would be:

[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].

In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:

[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].

Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.

Explanation:

the speed of sound is affected by temperature and humidity

Answer:

Explanation:

If we assume the rod and sphere are of uniform construction so that their individual centers of mass are at their geometric centers, and that the rod end is attached to the surface of the sphere.

Balance moments about the rod free end of the assembly with its parts

(M + m)Cx = M(L/2) + m(L + R)

Cx = (M(L/2) + m(L + R)) / (M + m)

Answer:

in general context yes it is closest to 1.04

Explanation:

theres no right or wrong way to scientifically prove this though.

Overall in scale its closest to 1.04 hope that helped

Answer:

192

Explanation:

Hi there!

For this problem, we must use the conservation of angular momentum. This is an example of an inelastic "collision", so:

I₁w₁ + I₂w₂ = (I₁ + I₂)wf

We know that the moment of inertia of a disk is 1/2mR², so we can calculate the moments of inertia for both disks:

Disk 1: 1/2(2)(0.40²) = .16 kgm²/s

Disk 2: 1/2(2)(0.20²) = .04 kgm²/s

Plug in the values. Let counterclockwise be positive.

.16(-50) + .04(50) = (.16 + .04)wf

Solve:

wf = -30 rev/s

Answer:

By applying the definition of torques ( [tex]\vec \tau = \vec r \times \vec F[/tex] ) and them remembering a few tricks.

Namely: if you wrap your RIGHT hand fingers around something and stick your thumb out, the direction your finger wraps gives you the verse of rotation and the thumb the orientation of the torque. Bottom force (4N) will give a counterclockwise rotation, torque is pointing up; top force (3N) will give a clockwise rotation and its torque its pointing down (read up and down as if the sheet the image is printed on is on your table).

In terms of magnitude the trick is easy: You want to multiply the intensity of the force (3N and 4N) by the distance between the point and the line the force it is applied to (that is, you don't care about the length of r itself, but the distance at a right angle, which is 0.9 and 0.8m respectively.

At this point, assuming "upwards" (relative to the plane of the sheet that is) torques positive, the 3N force gives you a torque of [tex]- 3N \times 0.9m = - 2.7N\cdot m[/tex] and the 4N force provides [tex]+4N\times 0.8 m = +3.2 N\cdot m[/tex]

Answer:

4600kg

Explanation:

Density=mass÷volume

920=m/5

m=920×5=4600kg

The elevation at the top of the hill is 1,653.85 m.

The given parameters;

The elevation at the top of the hill is calculated as follows;

[tex]P.E = mg\Delta h\\\\P.E = mg(h_2 -h_1)\\\\h_2 -h_1 = \frac{P.E}{mg} \\\\h_2 = \frac{P.E}{mg} + h_1\\\\h_2 = \frac{9.2 \times 10^5 }{72 \times 9.8} \ + \ 350 \ m\\\\h_2 = 1,653.85 \ m[/tex]

Thus, the elevation at the top of the hill is 1,653.85 m.

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a. Assuming all energy involved is conserved, at the lowest point of the swing (which includes the moment the student grabs the rope), the student only has kinetic energy,

K = 1/2 m (3.5 m/s)²

and at the highest point of the swing, the student only has potential energy

P = mgh

The energies at the bottom and top of the swing must be equal, so

1/2 m (3.5 m/s)² = mgh

h = (3.5 m/s)² / (2g)

h = 0.625 m ≈ 63 cm

b. In part (a), we found the relationship

h = v²/(2g)

If we cut the speed v in half, we replace v in the equation above with v/2 :

h = (v/2)²/(2g)

and simplifying this gives

h = (v²/4)/(2g) = 1/4 • v²/(2g)

The factor of 1/4 tells you that reducing the speed by a factor of 1/2 reduces the height by a factor of 1/4. So he can swing as high as

1/4 (3.5 m/s)²/(2g) = 0.15625 m ≈ 16 cm

Answer:

c.

Explanation:

If the two plates that meet at a convergent plate boundary both are of oceanic crust, the older, denser plate will subduct beneath the less dense plate. The older plate subducts into a trench, resulting in earthquakes. Melting of mantle material creates volcanoes at the subduction zone.

When two oceanic plates converge, the denser plate will end up sinking below the less dense plate, leading to the formation of an oceanic subduction zone. Old, dense crust tends to be subducted back into the earth. An example of a subduction zone formed from a convergent boundary is the Chile-Peru trench….

Answer:

a divergent plate boundary

Answer:

V2 = (V1 - u) / (1 - V1 u / c^2)

V1 = speed of ship in observer frame = .99 c    to right

u = speed of frame 2 = -.99 c   to left relative to observer

V2 = speed of V1 relative to V2

V2 = (.99 - (-.99 ) / (1 - .99 (-.99)) c

V2 = 1.98 / (1 + .99^2) c = .99995 c

This question involves the concepts of the law of conservation of momentum, velocity, and mass.

The two quantities, the students should multiply before and after the collision are "A. mass and velocity".

According to the law of conservation of momentum, In an isolated system, the total momentum of the system before the collision is always equal to the total momentum of the system after the collision.

To prove the law of conservation of momentum, consider two balls of masses ‘m₁’ and ‘m₂’, moving with velocities ‘u₁’ and ‘u₂’, respectively, such that u₁ is greater than u₂. After some time, these balls collide with each other and their velocities become ‘v₁’ and ‘v₂’, respectively.

This situation is illustrated in the attached picture.

So, according to the law of conservation of momentum:

Total Momentum Before Collision = Total Momentum After Collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

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Sorry this is a year late, but here it is for those of you who are stuck on the same thing.

NOTE: Please Check and Confirm That You Are On The Same Assignment with The Same Questions and Number of Questions. Thank You and Good Luck!

1. Mass & velocity

2. The total momentum after the collision is the same as the total momentum before the collision.

3. 0.54 kg⋅m/s

4. The system has external forces, such as friction and air resistance, acting on it.

5. 3.0 m/s

Answer:

do it got a picture

on the edge

Explanation:

This problem is describing a pump from which water is discharged at 1 MPa and 165 °C and is asking for the specific volume and internal energy at those conditions, thus, we can use the steam tables for resolving this requirement.

First of all, we need to remember that water can be a saturated liquid, vapour or liquid-vapour mixture, and this is determined for the temperature and pressure it is at.

In this case, we find that at 165 °C the saturation pressure is about 0.6178 MPa; this means we are referring to a saturated liquid so that both the specific volume and internal energy can be simply read from the steam tables as vf and uf as follows:

[tex]v=0.001127\frac{m^3}{kg}\\\\u=761.67 \frac{kJ}{kg}[/tex]

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Explanation:

The torque exerted on the wrench is

[tex]\tau = Fd\sin{\theta} = Fd[/tex]

since the force is applied perpendicular to the wrench, in which sin90 = 1. The torque then is

[tex]\tau = (185\:\text{N})(0.11\:\text{m}) = 20.35\:\text{N-m}[/tex]

9514 1404 393

Answer:

  about 5.89 seconds

Explanation:

The penny will hit the ground at the same time it would if it were simply dropped. The equation for the vertical motion is ...

  h(t) = -4.9t^2 +170 . . . . . where 170 is the initial height in meters

h(t) = 0 when ...

  4.9t^2 = 170

  t = √(170/4.9) ≈ 5.89

The penny will hit the ground in about 5.89 seconds.