Option(a) the mass of cart 2 is twice that of the mass of cart 1 is the right answer.
The mass of cart 2 is twice that of the mass of cart 1 is correct about the mass of cart 2.
Let's demonstrate the issue using variables:
Let,
m1=mass of cart 1
m2=mass of cart 2
v1 = velocity of cart 1 before collision
v2 = velocity of cart 2 before collision
v' = velocity of the carts after collision
Using the conservation of momentum for perfectly inelastic collisions:
m1v1 + m2v2 = (m1 + m2)v'
v2 = 0 because it is stationary
v' = 1/3*v1
m1v1 = (m1+m2)(1/3)(v1)
m1 = 1/3*m1 + 1/3*m2
1/3*m2 = m1 - 1/3*m1
1/3*m2 = 2/3*m1
m2 = 2m1
From this we can conclude that the mass of cart 2 is twice that of the mass of cart 1.
To learn more about inelastic collision visit:
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Answer:
Explanation:
side of the square loop, a = 7 cm
distance of the nearest side from long wire, r = 2 cm = 0.02 m
di/dt = 9 A/s
Integrate on both the sides

i = 9t
(a) The magnetic field due to the current carrying wire at a distance r is given by


(b)
Magnetic flux,





(c)
R = 3 ohm

magnitude of voltage is
e = 1.89 x 10^-7 V
induced current, i = e / R = (1.89 x 10^-7) / 3
i = 6.3 x 10^-8 A
Answer:
M
Explanation:
To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after
As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved
⇒
-
is the moment of interia of earth before impact -
is the angular velocity of earth about an axis passing through the center of earth before impact
is moment of interia of earth and asteroid system
is the angular velocity of earth and asteroid system about the same axis
let 
since 

⇒ if time period is to increase by 25%, which is
times, the angular velocity decreases 25% which is
times
therefore

(moment of inertia of solid sphere)
where M is mass of earth
R is radius of earth

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)
where
is mass of asteroid
⇒ 

=
+ 

⇒

Answer:
the wavelength is 9.8 meters
Explanation:
We can use the relationship:
Velocity = wavelenght*frequency.
Initially we have:
wavelenght = 4.9m
velocity = 9.8m/s
then:
9.8m/s = 4.9m*f
f = 9.8m/s/4.9m = 2*1/s
now, if the velocity is doubled and the frequency remains the same, we have:
2*9.8m/s = wavelenght*2*1/s
wavelenght = (2*9.8m/s)*(1/2)s = 9.8 m