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ddd [48]
2 years ago
5

A plane accelerates to a velocity of 240 m/s in 11 s by which time it had traveled 1,400 m down the runway what were it average

and initial velocities
Physics
1 answer:
Alex Ar [27]2 years ago
5 0

The average velocity is 127.27 m/s while the initial velocity is 14.55 m/s

According to Newton's law of motion:

s = [(v + u)/2]t

Where s is the distance moved = 1400 m, v is the final velocity = 240 m/s, u is the initial velocity, t is the time taken = 11 s, hence:

1400 = [(240 + u)/2)11

2800 = (240 + u)11

u + 240 = 254.55

u = 14.55 m/s

The average velocity = (v + u)/2 = (240 + 14.55)/2 = 127.27 m/s

Therefore the average velocity is 127.27 m/s while the initial velocity is 14.55 m/s

Find out more at: brainly.com/question/13639113

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The friends take a few minutes to discuss their ideas. Which of Amy's comments with respect to a charged particle moving in a ma
Marizza181 [45]

Answer:

The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false.

Explanation:

The equation for the magnetic force on a charge q moving at velocity v on a magnetic field B is given by the (vectorial) Lorentz Force Law F=qv\times B

From it we can clearly see that the <em>magnitude of the magnetic force </em>exerted on the particle is <em>proportional to the magnitude of the charge q and to the speed v of the particle</em>, and that it is also <em>perpendicular to the particle's velocity</em>. This means that at each instant it moves perpendicularly to the force, so <em>the work done by the magnetic force on the particle is zero</em>.

The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false not only for this but for any force, a force always perpendicular to a velocity will curve the trajectory.

7 0
2 years ago
For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km
muminat

Answer with Explanation:

We are given that

y=(18-2x^2) km

x=(4t-3)km

Differentiate x and y w.r.t t

\frac{dx}{dt}=4

\frac{dy}{dt}=-4x\frac{dx}{dt}

\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)

v_x=\frac{dx}{dt}=4

v_y=\frac{dy}{dt}=-16(4t-3)

Substitute t=1

v_x=4

v_y=-16(4-3)=-16

Magnitude of velocity=\mid v\mid=\sqrt{v^2_x+v^2_y}

\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s

Hence, the magnitude of the missile's velocity=16.49 m/s

a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0

a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64

Substitute t=1

a_x=0,a_y=-64

\mid a\mid=\sqrt{a^2_x+a^2_y}

\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2

Hence, the magnitude of acceleration when t=1 s=64m/s^2

3 0
2 years ago
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

\theta = \omega t\\\\\theta = (100 \frac{rev}{\min}  \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi  \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

6 0
3 years ago
In a given chemical reaction the energy of the products is less than the energy of the reactants. Which statement is true for th
Westkost [7]

Answer:

B. Energy is released in the reaction.

7 0
3 years ago
Two cars are driving on a straight section of the interstate in opposite directions with 70 mph. They are one mile apart.
Mashutka [201]
They are trailing the same speed as it states in the question they are heeding toward each other a 70 mph <span />
5 0
3 years ago
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