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3 years ago

A train moves from rest to a speed of 25 m/s in 30.0 seconds . What is the acceleration

1 answer:

7 0

When a car moves from rest then the initial velocity,
$u = 0m {s}^{ - 1}$

The final velocity,

$v = 25m {s}^{ - 1}$
and time,
$t = 30s$
Using the equation,
$v = u + at$
We have,

$25 = 0 + a(30)$
$25 = 30a$
$\frac{25}{30} = a$
$\therefore \: a = \frac{5}{6} m {s}^{ - 2}$

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a skier is gliding along 3.0 m/s on horizontal, frictionless snow. he suddenly starts down a 10 degrees incline. his speed at th

Elza [17]

... find length
(way 1) determine acceleration using force
only force act on skier is mg vertically. spilt vector we get force along the incline = mgsin(10) and f= ma so
ma = mgsin(10) or a = gsin(10)
a (along the incline)= gsin(10) = 10sin(10) = 1.74
v^2 = u^2 + 2as
15^2 = 3^2 + 2(1.74)s
s = 62.06 m

(way 2) using conservation of energy
energy (KE+PE) on top = energy at bottom
0.5m3^2 + mgh = 0.5m15^2 +0
h (height of incline) = (112.5 - 4.5)/10 = 19.8 m
length of incline = h/sin(10) = 62.2 m ; trigonometry

... find time
s = (u+v)t/2
t = 2s/(u+v) = 2(62.2)/(3+15) = 6.91 s

4 0

3 years ago

4. Assume a multiple level queue system with a variable time quantum per queue, and that the incoming job needs 50ms to run to c

Troyanec [42]

Answer:

Explanation:

For the completion of incoming job it will take 50ms

First queue takes 5ms quantum time and the subsequent queue takes double of the previous question

So,

First queue T_1 = 5ms

Second queue T_2 = 2 × T_1 = 2 × 5 = 10ms

Third queue T_3 = 2 × T_2 = 2 × 10 = 20ms

Fourth queue T_4 = 2 × T_3 = 2 × 20 = 40ms

Fifth queue T_5= 2 × T_4 = 2 × 40 = 80ms.

Now, the job will be done after the fifth queue.

So, after the first queue, the job is not completed, so, we have first interruption

After the second queue, the job is not completed, so, we have second interruption

After the third queue, the job is not completed, so we have third interruption.

After the fourth queue, the job is not yet completed, so we have the fourth interruption

And in the fifth queue the job is completed, so we don't have any interruption here.

So, the job will be interrupted 4 times and it will finished on the fifth queue.

6 0

3 years ago

If a cell phone is dropped from a very tall building, how far has the phone fallen after 2.7 seconds, neglecting air resistance?

Zielflug [23.3K]

The free fall of the phone is an uniformly accelerated motion toward the ground, with constant acceleration equal to
$g=9.81 m/s^2$

So, assuming the downward direction as positive direction of the motion, since the phone starts from rest the distance covered by the phone after a time t is given by
$y(t) = \frac{1}{2}gt^2$
And if we substitute t=2.7 s, we find the distance covered:
$y(t)= \frac{1}{2}(9.81 m/s^2)(2.7 s)^2=35.8 m$

6 0

3 years ago

Give examples of factors or criteria often used to determine higher wages for a particular job

marshall27 [118]

-- the applicant's previous experience at similar jobs;
-- the color of the applicant's hair;
-- the applicant's grammar and vocabulary;
-- where the applicant went to school;
-- the shirt the applicant wears to the job interview;
-- the applicant's favorite football team;
-- the applicant's self-confidence;

5 0

3 years ago

Read 2 more answers

The product of m1m2 in the formula for gravitational force is always

Nataly [62]

Answer:

The correct answer is the second option

Explanation:

The formula in question (for gravitational force) is provided below

F = G.m₁.m₂/r²

Where

F = force of gravity

G = gravity constant (6.674 × 10⁻¹¹ m.kg⁻¹.s⁻²)

m₁ and m₂ = are the masses of the objects

r² = distance between the centers of the two objects

From the above, <u>it can be deduced that m₁ and m₂ are masses and hence can only be positive. Positives multiplied together only gives positive, thus the second option is correct.</u>

8 0

4 years ago