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vlada-n [284]
3 years ago
14

A jetliner rolls down the runway with constant acceleration from rest, it reaches its take off speed of 250 km/h in 1 min. What

is its acceleration? Express in km/h^2, does this result make sense?
Physics
1 answer:
nlexa [21]3 years ago
3 0

Answer:

Acceleration, a=14970.05\ km/h^2

Explanation:

Given that,

Initially, the jetliner is at rest, u = 0

Final speed of the jetliner, v = 250 km/h

Time taken, t = 1 min = 0.0167 h

We need to find the acceleration of the jetliner. The mathematical expression for the acceleration is given by :

a=\dfrac{v-u}{t}

a=\dfrac{250}{0.0167}

a=14970.05\ km/h^2

So, the acceleration of the jetliner is 14970.05\ km/h^2. Hence, this is the required solution.

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  • Since the increase in the applied force caused the increase in the cart's acceleration, force and acceleration are directly proportional to each other, which is in accordance with Newton's second law.

When we state something about the results on the basis whether the observed data supports the original hypothesis, we say that we are concluding the results.

What is the relationship between force and acceleration based on Newton's 2nd law?

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2 years ago
At what wavelength does Io reflect the most sunlight? A. Blue
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B. Ultraviolet.

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Please help me out! This is a big question!
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Read 2 more answers
Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial veloc
svet-max [94.6K]

Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(c) t₃=1.5 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

Displacement S from t₁ to t₅

Velocity V from t₁ to t₅

Solution

According to kinematic equation of motion and given information conclude that v is given by

v=v_{i}+gt\\

Also get the equation of displacement

S=v_{i}t+(1/2)gt^{2}

These two formula are used to find velocity as well as displacement for time t₁ to t₅

For t₁=0.5 s

v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\  And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m

For t₂

v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\  And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m

For t₃

v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\  And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m

For t₄

v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\  And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m

For t₅

v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\  And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m

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4 years ago
sound sensor display reads 40 dB inside and outside it reads 80dB. how many times louder is it outside than inside?
LuckyWell [14K]

The difference is (80 dB - 40 dB) = 40 dB.

The sound is 40 dB louder outside.

Each 10 dB means 10 times more sound power.

40 dB louder means 10x10x10x10 times more sound power.

That's <em>10,000 times</em> more sound power outside than inside.

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