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maxonik [38]
3 years ago
13

A freight train initially moving at +10m/s decelerates at a rate of -0.1m/s^2. How far does it move in 30 seconds?

Physics
2 answers:
Kaylis [27]3 years ago
8 0

Answer:

255 [m].

Explanation:

1) the formula is (d - requred distance, V₀ - initial velocity, t - elapsed time, a - deceleration):

d=V_0t+\frac{at^2}{2};

2) according to the formula above the required distance is:

d=10*30-0.5*0.1*900=300-45=255 [m].

siniylev [52]3 years ago
7 0
  • Initial velocity=u=10m/s
  • Acceleration=-0.1m/s^2=a
  • Time=t=30s

  • Distance=s

Apply second equation of kinematics

\\ \tt\longmapsto s=ut+\dfrac{1}{2}at^2

\\ \tt\longmapsto s=10(30)+\dfrac{1}{2}(-0.1)(30)^2

\\ \tt\longmapsto s=300+(-0.1)(450)

\\ \tt\longmapsto s=300-45

\\ \tt\longmapsto s=255m

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A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.
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Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

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We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2

(c) Moment of inertia :

The net torque acting on it is, \tau=I\alpha, I is the moment of inertia

Also, \tau=Fr

So,

I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2

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3 years ago
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Answer:

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Explanation:

Mass of the dart = 0.02kg, the spring was compressed to 6cm

Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m

Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m

Work needed to compress the spring = 0.036J

b) the total energy stored in the spring = the work done to compress the spring = 0.036J

c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.

d) 1/2mv^2 = 0.036

mv^2 = 0.036*2

v^2 = 0.036*2 / 0.02 = 3.6

v = √3.6 = 1.897 approx 1.9m/s

e) kinetic energy of the dart = work done against gravity to get the body to height h

Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward

0.036 = 0.02 * 9.81 * h

0.036 / ( 0.02*9.81) = h

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