A 1200-kg station wagon is moving along a straight highway at Another car, with mass 1800 kg and speed has its center of mass 40
1 answer:
Answer:
Center of mass lies 24 m in front of center of mass of second wagon.
Explanation:
Suppose A 1200 kg station wagon is moving along a straight highway at 12.0 m/s. Another car with mass 1800 kg and speed 20.0 m/s.
Given that,
Mass of first wagon = 1200 kg
Mass of second wagon = 180 kg
Distance = 40 m
We need to calculate the position of the center of mass of the system
Using formula of center mass
Hence, Center of mass lies 24 m in front of center of mass of second wagon.
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Answer:
Total contraction on the Bar = 1.22786 mm
Explanation:
Given that:
Total Length for aluminum bar = 600 mm
Diameter for aluminum bar = 40 mm
Hole diameter = 30 mm
Hole length = 100 mm
elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²
compressive load P = 180 KN = 180 × 10³ N
Calculate the total contraction on the bar = ???
The relation used in calculating the contraction on the bar is:
The relation used in calculating the total contraction on the bar can be expressed as :
Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)
i.e
Total contraction on the Bar =
Let's find the area of cross section without the hole and with the hole
Area of cross section without the hole is :
Using A = πd²/4
A = π (40)²/4
A = 1256.64 mm²
Area of cross section with the hole is :
A = π (40²-30²)/4
A = 549.78 mm²
Total contraction on the Bar =
Total contraction on the Bar =
Total contraction on the Bar = 2.117( 0.398 + 0.182)
Total contraction on the Bar = 2.117*(0.58)
Total contraction on the Bar = 1.22786 mm
Answer:
t = 96.1 nm
Explanation:
For strong reflection through liquid layer we know that the path difference between two reflected light rays must be integral multiple of wavelength
now we know that the path difference of two reflected light from thin liquid layer is given as
here we know that
t = thickness of layer
N = 0 (for minimum thickness of layer)
now we have
Answer:
Explanation:
Let T be the tension .
Applying newton's second law on the downward movement of the bucket
mg - T = ma
On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum
TR = Iα
TR = Ia/ R
T = Ia/ R²
Replacing this value of T in the other equation
mg - T = ma
mg - Ia/ R² = ma
mg = Ia/ R² +ma
a ( I/ R² +m)= mg
a = mg / ( I/ R² +m)
mg - T = ma
mg - ma = T
mg - m x mg / ( I/ R² +m) = T
mg - m²g / ( I/ R² +m ) = T
mg - mg / ( 1 + I / m R² ) = T
b ) T = Ia/ R²
I = TR² / a
c ) Moment of inertia of hollow cylinder
I = 1/2 M ( R² - R² / 4 )
= 3/4 x 1/2 MR²
= 3/8 MR²
I / R² = 3/8 M
a = mg / ( I/ R² +m)
a = mg / ( 3/8 M + m )
T = Ia/ R²
= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²
=