All categories

3 years ago

Find the acceleration of a train whose speed increases from 7m/s to 17 m/s in 120s

1 answer:

7 0

Initial speed of the train = 7 m/s
Final speed of the train = 17 m/s
Change of speed of the train = (17 - 7) m/s
   = 10 m/s
Time taken for the change of the speed of the train = 120 s
Then
Acceleration of the train = Change of speed of the train/Time taken for the change of speed
   = 10/120 m/s^2
   = 1/12 m/s^2
   = 0.083 m/s^2
So the acceleration of the train is 1/12 meter per second square or 0.083 m/s^2. I hope this is the answer you were looking for.

You might be interested in

(a) In the baggage claim area of an airport, a particular baggage carousel is shaped like a section of a large cone, steadily ro

Tom [10]

Answer:

Part a)

$F_f = 107.8 N$

Part b)

$\mu = 0.415$

Explanation:

Part a)

Time period of one revolution is given as

T = 42 s

now the angular speed of the belt is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{42}$

$\omega = 0.15 rad/s$

Now at rest position the force along the surface of carousel must be constant

so we will have

$mgsin\theta + m\omega^2 r cos\theta = F_f$

$30(9.81)sin20.5 + 30(0.15^2)(7.46)cos20.5 = F_f$

$F_f = 107.8 N$

Part b)

New Time period of one revolution is given as

T = 30 s

now the angular speed of the belt is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{30}$

$\omega = 0.21 rad/s$

Now at rest position the force along the surface of carousel must be constant

so we will have

$mgsin\theta + m\omega^2 r cos\theta = F_f$

$30(9.81)sin20.5 + 30(0.21^2)(7.94)cos20.5 = F_f$

$F_f = 112.9 N$

Also we know that in perpendicular direction also force is balanced

$F_n + m\omega^2 r sin\theta = mgcos\theta$

$F_n = mgcos\theta - m\omega^2 r sin\theta$

$F_n = 30(9.81)cos20.5 - 30(0.21)^2(7.94)sin20.5$

$F_n = 272 N$

now for friction coefficient we will have

$F_f = \mu F_n$

$112.9 = \mu 272$

$\mu = 0.415$

8 0

3 years ago

Two masses are attracted by a gravitational force of 7 N.

damaskus [11]

Answer:

The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.

Explanation:

Data

F = 7 N

F = ? if the masses is quartered

Formula

$F = \frac{Km1m2}{r2}$

Process

Normal conditions F = Km₁m₂/r² = 7

When masses quartered F = K(m₁/4)(m₂/4)/r² = ?

F = K(m₁m₂/16)/r²

F = K(m₁m₂/16r² = 7/16 = 0.4375 N

3 0

3 years ago

A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west

Minchanka [31]

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

$\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}$

(see the attached graphic)

We have

ducks (relative to wind) = 7.0 m/s in some direction θ relative to the positive horizontal direction, or

$\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)$

wind (relative to Earth) = 5.0 m/s due East, or

$\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)$

ducks (relative to earth) = some speed v due South, or

$\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)$

Then by setting components equal, we have

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0$

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

We only care about the direction for this question, which we get from the first equation:

$\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}$

$\cos\theta=-\dfrac57$

$\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)$

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies both equations. We want

$\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v$

which means θ must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

8 0

3 years ago

Which pair correctly identifies the strongest and weakest fundamental forces,

Kay [80]

Answer: Option B.

Strong nuclear force Gravity.

Explanation:

Nuclear force is the strongest force while Gravity is the weakest force. Strong nuclear force is the strongest force because it's act and hold atoms of matter together. It also binds neutrons and protons together to form atomic nuclei and it's has strength of 1.

Gravity is considered the weak force out of the four fundamental forces because it has a very small strength of 10^40 than electromagnetic force.

7 0

3 years ago

A car is traveling at a constant velocity of sok

Ber [7]

If velocity is constants there is no acceleration therefore acceleration will equal zero.

3 0

4 years ago