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zlopas [31]
2 years ago
6

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

complete. What is the magnitude of the centripetal acceleration during the turn?
a. 12 m/s2
b. 9.8 m/s2
c. 0 m/s2
d. 6.9 m/s
e. 8.1 m/s2
Physics
1 answer:
ankoles [38]2 years ago
5 0

Answer:

The magnitude of the centripetal acceleration during the turn is a=12.04\ m/s^2.

Explanation:

Given :

Speed to the airplane in circular path , v = 115 m/s.

Time taken by plane to turn , t= 15 s.

Also , the plane turns from east to south i.e. quarter of a circle .

Therefore, time taken to complete whole circle is , T=t\times 4=60\ s.

Now , Velocity ,

v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.

Also , we know :

Centripetal acceleration ,

a=\dfrac{v^2}{r}

Putting all values we get :

a=12.04\ m/s^2.

Hence , this is the required solution .

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Answer:139.32kgm/s

Explanation:

Mass(m)=0.162kg

Velocity(v)=860m/s

Momentum=mass x velocity

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3 0
1 year ago
Lorenzo is making a prediction. “I learned that nonmetals increase in reactivity when moving from left to right. So I predict th
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That prediction is not correct because Xenon is extremely stable; column 18 of the periodic table contains the noble gasses, which are stable because their outer-most energy levels are completely filled. Having the octet (8) of valence electrons means that the element no longer needs to lose or gain electrons to gain stability.

The column 17 elements are unstable because they only have one valence electron short of the stable octet configuration of the noble gasses.
6 0
2 years ago
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Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov
frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v_{1} be the velocity of ball A along positive x axis

v_{2} be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v_{1}

∴  v_{1} = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v_{2}

2 = u +  v_{2} → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 =  v_{2} - u → equation 2

By adding both equations 1 and 2 we get

v_{2} = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

                         = 8·85×m J

Total momentum = m×√(2² + 3·7²)

                             = m× 4·21 kgm/s

3 0
2 years ago
An astronaut takes what he measures to be a 10-min nap in a space station orbiting Earth at 8000 m/s. A signal is sent from the
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Answer:

longer than

Explanation:

given,

time of nap = 10 min

speed of orbiting earth = 8000 m/s

c is the speed of light

using the equation of time dilation

t' = \dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}}}

now inserting all the values

t' = \dfrac{10}{\sqrt{1-\dfrac{8000^2}{3\times 10^8)^2}}}

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t' = 10.001 s

on solving the above equation we will get a value greater than 10minutes.

hence, On earth time of nap measured will be longer than 10 min

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Ff = a*Q

where Q is weight of the toy and a is friction

if we express a we get
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