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3 years ago

How long will be required for an object to go from a speed of 22m/s to a speed of 27m/s if the acceleration is 5.93m/s^2 ?

Physics

1 answer:

mario62 [17]3 years ago

4 0

Answer:

Required time, t = 0.84 seconds

Explanation:

It is given that,

Initial speed of an object, u = 22 m/s

Final velocity of an object, v = 27 m/s

Acceleration, a = 5.93 m/s²

We have to find the time required for an object to go a speed of 22 m/s to a speed of 27 m/s. It can be solved by using first equation of motion as:

$v=u+at$

Where

t = time

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

$t=\dfrac{27\ m/s-22\ m/s}{5.93\ m/s^2}$

t = 0.84 seconds

Hence, the time required for an object is 0.84 seconds.

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Answer:

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Explanation:

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3 years ago

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3 years ago

Formula for the distance (d) is given by d = rate*time. For example if you are traveling at 60 mph for 3 hours the distance trav

babunello [35]

Explanation:

Distance covered by the particle is given by:

Distance (d) = rate (v) × time (t)

Speed of Mary, v₁ = 50 mph

Speed of Jim, v₂ = 60 mph

It is assumed that, Mary and Jim leave at the same time. After one hour, Jim is 10 miles ahead.

Distance travelled by Jim, d₁ = (60t + 10)

Distance travelled by Mary, d₂ = 50t

The distance between Mary and Jim is greater than or equal to 100 miles.

$60t+10-50t\ge100$

$10t\ge90$

$t\ge9\ h$

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3 years ago

An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the

alexdok [17]

Answer:

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Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

$X=V_{o}*t + \frac{a*t^{2}}{2}$ Replacing the known values:

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t = 67.7s

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3 years ago

Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test c

amid [387]

Answer:

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Explanation:

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Then, it means that the net force acting on the test charge has a magnitude of √2F.

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3 years ago