All categories

3 years ago

Determine the displacement of a plane that is uniformly accelerated from 66 m/s to 88 m/s in 12 s

1 answer:

german3 years ago

7 0

Displacement is the area under the graph for velocity-time graph so if plane accelerated uniformly from 66m/s to 88m/s the area will be the area of trapezium where formula for area under trapezium is 1/2 (sum of parallel lines) x height so the parallel lines in this case are 66m/s and 88m/s and height is time taken which is 12 so displacement is 1/2 (66+88)(12)= 924m

if you want to know the formula then it is S=1/2 (V+U)(t) where S is displacement V is final velocity U is initial velocity and t is time taken

You might be interested in

What is another name for the introitus? psych 230?

Vera_Pavlovna [14]

Vaginal opening. areola is the part of the breast.

7 0

3 years ago

Which technology is most efficient in mapping large areas of the ocean floor?

qaws [65]

''Sonar'' is the answer.

7 0

3 years ago

What is the overall charge of 1.5 x 10^10 electrons?

Troyanec [42]

Answer: Charge = -2.4x10^-9 Coulombs

Explanation:

The charge of one electron is e = -1.6x10^-19 C

Then, the charge of 1.5 x 10^10 electrons is equal to 1.5 x 10^10 times the charge of one electron:

Here i will use the relation (a^b)*(a^c) = a^(b + c)

Charge = ( 1.5 x 10^10)*( -1.6x10^-19 C) = -2.4x10^(10 - 19) C

Charge = -2.4x10^-9 C

7 0

3 years ago

A device for training astronauts and jet fighter pilots is designed to rotate the trainee in a horizontal circle of radius 11.0

kvv77 [185]

The velocity of the trainee is 29 m/s or 0.42 rev/s

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration (m / s²)v = final velocity (m / s)</em>

<em>u = initial velocity (m / s)</em>

<em>t = time taken (s)</em>

<em>d = distance (m)</em>

Centripetal Acceleration of circular motion could be calculated using following formula:

$\large {\boxed {a_s = v^2 / R} }$

<em>a = centripetal acceleration ( m/s² )</em>

<em>v = velocity ( m/s )</em>

<em>R = radius of circle ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

<u>Unknown:</u>

Velocity of Rotation = v = ?

<u>Solution:</u>

$F = ma$

$F = m\frac{v^2}{R}$

$7.80w = m\frac{v^2}{R}$

$7.80mg = m\frac{v^2}{R}$

$7.80g = \frac{v^2}{R}$

$7.80 \times 9.8 = \frac{v^2}{11.0}$

$v^2 = 840.84$

$v \approx 29 ~m/s$

$\omega = \frac{v}{R}$ → in rad/s

$\omega = \frac{v}{2 \pi R}$ → in rev/s

$\omega = \frac{29}{2 \pi \times 11.0}$

$\omega \approx 0.42 ~ rev/s$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Uniform Circular Motion : brainly.com/question/2562955
Trajectory Motion : brainly.com/question/8656387

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

6 0

3 years ago

Read 2 more answers

A 25.0 kg object moving at +15.0 m/s strikes a 15.0 kg

lesya692 [45]

The final velocity of the 15 kg mass is 18.33 m/s.

<h3>Conservation of linear momentum</h3>

The final velocity of the 15 kg mass can be determined by applying the principles of conservation of linear momentum as follows;

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\\\$

Where;

m₁ is the mass of the first object = 25 kg
u₁ is the initial velocity of the first object = 15 m/s
m₂ is the mass of the second object = 15 kg
u₂ is the initial velocity of the second object = -10 m/s
v₁ is the final velocity of the first object = -2 m/s
v₂ is the final velocity of the second object

Thus, the final velocity of the 15 kg mass after the collision is 18.33 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

8 0

2 years ago