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Tju [1.3M]
3 years ago
6

Find the acceleration of a train whose speed increases from 9 m/s to 22 m/s in 160s

Physics
1 answer:
Rainbow [258]3 years ago
6 0
Acceleration = (v2-v1) / (t2-t1) = (22-9) / 160 = 13/160 m/s
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A toy of mass 0.190-kg is undergoing SHM on the end of a horizontal spring with force constant k = 350 N/m . When the toy is a d
vagabundo [1.1K]

Answer

a)0.0495 J

b)0.01681 m

c)0.7218 m/s

Explanation:

Given

Mass of the.toy M = 0.190 kg

force constant k = 350 N/m

Displacement from equilibrium x = 0.0140 m

Speed v = 0.400 m/s

a)What is the toy's total energy at any point of its motion?

The total energy at any point of it's motion can be calculated by adding together both the potential and kinetic energy of the toy, since it's posses potential energy when at rest and kinetic energy at motion

Total energy E = kinetic energy + potential energy

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.190)(0.4)² + ¹/₂ (350)(0.0140)²

E = 0.0495 J

Hence,the total energy is 0.0495 J

b) the amplitude of the motion can be calculated using below formula

Let amplitude = A

E = ¹/₂KA²

if we make Amplitude A the subject of the formula we have

A=√(2E/k)

But we have calculated our E up there, our K was given in question then if we substitute we have

A= √(2×0.0495)/350

Ans: 0.01681 m

Hence, our Amplitude is 0.01681 m

c) the the toy's maximum speed during its motion can be calculated using the expression below

Let maximum speed = vmax

E = (1/2)M * vmax^2

If we make vmax the subject of the formula we have

vmax =√(2E/m)

vmax= √(2×0.0495)/0.190

vmax=0.7218 m/s

Hence our vmax is 0.7218 m/s

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3 years ago
What is the peak emf generated (in V) by rotating a 1000 turn, 42.0 cm diameter coil in the Earth's 5.00 ✕ 10−5 T magnetic field
ELEN [110]

Answer:

5.77 Volt

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e = N x dФ / dt

e = N x A x dB/dt

e = (1000 x 3.14 x 0.21 x 0.21 x 5 x 10^-5) / (1.2 x 10^-3)

e = 5.77 Volt

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