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victus00 [196]
3 years ago
12

A jetliner, traveling northward, is landing with a speed of 69 m/s. Once the jet touches down, it has 750 m of runway in which t

o reduce its speed to 6.1 m/s. Compute the average acceleration (magnitude and direction) of the plane during landing.
Physics
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

Acceleration, a=-3.14\ m/s^2

Explanation:

It is given that,

Initial speed of the jetliner, u = 69 m/s

Final speed of the jetliner, v = 6.1 m/s

Distance covered, s = 750 m

We need to find the average acceleration of the jetliner. It can be calculated using third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(6.1)^2-(69)^2}{2\times 750}          

a=-3.14\ m/s^2

So, the average acceleration of the plane during landing is -3.14\ m/s^2. Hence, this is the required solution.

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Answer:

Explanation:

Torque on a loop in a magnetic field

Maximum torque = M B

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M = area x current

= 3.14 x .10² x i

Maximum torque = M B

1 x 10⁻³ = 3.14 x .10² x i x .65 x 10⁻⁴

i = 490 A

Current = 490 A.

3 0
3 years ago
Please help me with this question. BRAINLIEST if answered.
IceJOKER [234]

... The top branch of the 3-branched parallel block ... the 9 and 6 in series ...
is equivalent to a single resistor of 15 ohms.

... The 3-branched parallel block boils down to (30, 10, and 15) in parallel.
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... The 5-ohm-equivalent block and the 20-ohm resistor form a
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8 0
3 years ago
A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
n200080 [17]

Answer:

a. t = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}  b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0}  +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0}  +/- \sqrt{v_{0} ^{2} - gD} }{g}

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

 0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁  = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

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timofeeve [1]

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For example, the (absolute) percent difference between 3 and 6 is

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which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take x from the left column and y from the right column.

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\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%

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3 0
2 years ago
Why doesn’t a machine that increases force break the law of conservation of energy?
USPshnik [31]

Answer:

A machine in which work input equals work output. energy can be used to do work, work can be used to transfer energy. The change in the kinetic energy of an object is equal to the net work done on the object.

hope this helps

8 0
3 years ago
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