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Bad White [126]
3 years ago
10

A plane accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long. Assume that the plane reaches

the required takeoff velocity at the end of the runway. What is the time tTO needed to take off? Express your answer in seconds using three significant figures.
Physics
1 answer:
tiny-mole [99]3 years ago
3 0

Answer:

26.8 seconds

Explanation:

To solve this problem we have to use 2 kinematics equations: *I can't use subscripts for some reason on here so I am going to use these variables:

v = final velocity

z = initial velocity

x = distance

t = time

a = acceleration

{v}^{2}  =  {z}^{2}  + 2ax

v = z + at

First let's find the final velocity the plane will have at the end of the runway using the first equation:

{v}^{2}  =  {0}^{2}  + 2(5)(1800)

v = 60 \sqrt{5}

Now we can plug this into the second equation to find t:

60 \sqrt{5}  = 0 + 5t

t = 12 \sqrt{5}

Then using 3 significant figures we round to 26.8 seconds

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A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleratio
8090 [49]

Question:

A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero

Answer:

24 m/s

Explanation:

Given:

x=(24t - 2.0t³)m

First find velocity function v(t):

v(t) = ẋ(t) = 24 - 2*3t²

v(t) = ẋ(t) = 24 - 6t²

Find the acceleration function a(t):

a(t) = Ẍ(t) = V(t) = -6*2t

a(t) = Ẍ(t) = V(t) = -12t

At acceleration = 0, take time as T in velocity function.

0 =v(T) = 24 - 6T²

Solve for T

T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2

Substitute -2 for t in acceleration function:

a(t) = a(T) = a(-2) = -12(-2) = 24 m/s

Acceleration = 24m/s

4 0
3 years ago
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle wi
Gnom [1K]

Answer:

(a) Acceleration of positron is 6.03 x 10¹³ m/s²

(b) Speed of positron after 8.70 x 10⁻⁹ s is 5.24 x 10⁵ m/s

Explanation:

Given :

Constant electric field, E = 343 N/C

Mass of positron, m = 9.1 x 10⁻³¹ kg

Charge of positron, q = +e = 1.6 x 10⁻¹⁹ C

(a) Coulomb force on the positron is determine by the relation :

F = q x E    ....(1)

But, force is also equals to product of mass and acceleration. So,

F = ma  .....(2)

Here a is acceleration.

From equation (1) and (2).

m x a = q x E

a=\frac{qE}{m}

Substitute the values of q, E and m in the above equation.

a=\frac{1.6\times10^{-19}\times 343}{9.1\times10^{-31} }

a = 6.03 x 10¹³ m/s²

(b) Initially, the positron is at rest, so its initial speed is zero.

The equation of motion for positron is :

v = u + at

Here v is final speed, u is initial speed and t is time.

Since, u is zero, so the equation becomes :

v = at

Substitute 8.70 x 10⁻⁹ s for t and 6.03 x 10¹³ m/s² for a in the above equation.

v = 6.03 x 10¹³ x 8.70 x 10⁻⁹ m/s

v = 5.24 x 10⁵ m/s

 

6 0
3 years ago
While traveling along a highway a driver slows from 24 m/s to 15 m/s in 12 seconds. What is the automobile’s acceleration?
Nimfa-mama [501]

Average Acceleration = (change in speed) / (time for the change)

Change in speed = (ending speed) - (starting speed)

                             =   15 m/s - 24 m/s  =  -9 m/s

Acceleration =  (-9 m/s) / (12 sec) =  - 0.75 m/s² .

6 0
3 years ago
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swat32
Nuclear fusion is the process by which multiple nuclei join together to form a heavier nucleus.

Nuclear fusion of light elements releases the energy that causes stars to shine and hydrogen bombs to explode.

Nuclear fusion of heavy elements (absorbing energy) occurs in the extremely high-energy conditions of supernova explosions.

Nuclear fusion in stars and supernovae is the primary process by which new natural elements are created.

It is this reaction that is harnessed in fusion power.

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6 0
3 years ago
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borishaifa [10]

Answer:

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