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.
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Answer:
physical change uwu owo hope this helps you uwu silly goose
Kinetic
energy is the energy that is possessed by an object that is moving. It is
calculated by one-half the product of the mass and the square of the velocity
of the object. We calculate as follows:<span>
For the truck,
KE = mv^2 / 2
KE = 20000 kg (110 km/h (1 h/3600 s)(1000 m / 1 km))^2 / 2
KE = 611111.11 J
For the astronaut,
KE = 80.0 kg (27500 km/h</span>(1 h/3600 s)(1000 m / 1 km)<span>)^2/2
KE = 611111.11 J
The kinetic energy possessed by the two bodies are the same.
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Answer:
A = -14.87 i ^ + 8.42 j ^ + 0 k ^
B = -25.41 i ^ -12.0 j ^ + 0 k ^
Explanation:
For this exercise let's use trigonometry by decomposing to vectors
vector A
module 17.1 with an angle of 150.5 counterclockwise.
Sin 150.5 =
/ A
cos 150.5 = Ax / A
A_{y} = A sin 150.5 = 17.1 sin 150.5
Aₓ = A cos 1505 = 172 cos 150.5
A_{y} = 8,420
Aₓ = -14.870
the vector is
A = -14.87 i ^ + 8.42 j ^ + 0 k ^
Vector B
= 28.1 sin 205.3
Bₓ = 28.1 cos 205.3
B_{y} = -12.009
Bₓ = -25.405
the vector is
B = -25.41 i ^ -12.0 j ^ + 0 k ^
Answer:
Explanation:
For image formation in objective lens
object distance u = 14 +1 = 15 mm
focal length f = 14 mm .
image distance v = ?
lens formula

Putting the values

v = 210 mm .
B )
magnification = v / u
= 210 / 15
= 14
size of image = 14 x 1.1 mm
= 15.4 mm
= 15 mm approx
C )
For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .
21 mm is the answer .
D )
overall magnification =

D = 25 cm , f_e = focal length of eye piece
= 14 x 250 / 21
= 166.67
= 170 ( in two significant figures )