A train is rounding a circular curve whose radius is 2.55 x 102 m. At one instant, the train has an angular acceleration of 1.48
1 answer:
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Answer:
a=0.212 m/s²
Explanation:
Given that
q= 10⁻⁹ C
m = 5 x 10⁻⁹ kg
Magnetic filed ,B= 0.003 T
Speed ,V= 500 m/s
θ= 45°
Lets take acceleration of the mass is a m/s²
The force on the charge due to magnetic filed B
F= q V B sinθ
Also F= m a ( from Newton's law)
By balancing these above two forces
m a= q V B sinθ
a=0.212 m/s²
Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π × ................2
put here value
period of oscillations = 2π ×
period of oscillations = 0.6953
so period of oscillations is 0.695 second
RESULT: Twalk= 3.64 hr
Explanation:
Given that,
Wavelength of the light,
(a) Slit width,
The angle that locates the first dark fringe is given by :
(b) Slit width,
The angle that locates the first dark fringe is given by :
Hence, this is the required solution.