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12345 [234]
3 years ago
9

Why does the momentum before a collision equal the momentum after a collision?

Physics
2 answers:
Art [367]3 years ago
8 0

Answer: Correct option is C) because momentum is conserved

 

Explanation:

According to law of conservation of linear momentum if net external force on a system is zero then the total momentum of the system will be conserved .

<u>For a collision occurring between object 1 and object 2 in an isolated system</u>

If we consider the two colliding object as a single system then while collision the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

Thus momentum before collision equal the momentum after a collision because momentum is conserved .  

kotykmax [81]3 years ago
3 0
The answer is C. Due to conservation of momentum.
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When the mass of an object increases, the forcé of gravity<br>​
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Answer:

increace

Explanation:

they are both going up

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2 years ago
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A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.
Ilya [14]

Answer:

t = 0.192 \mu m

Explanation:

Path difference due to a transparent slab is given as

\Delta x = (\mu - 1) t

here we know that

\mu = 1.79

now total shift in the bright fringe is given as

Shift = \frac{D(\mu - 1)t}{d}

Also we know that the fringe width of maximum intensity is given as

\delta x = \frac{\lambda D}{d}

now we have

\frac{D}{d} = \frac{\delta x}{\lambda}

now the shift is given as

Shift = \frac{(\mu - 1) t \delta x}{\lambda}

given that the shift is

Shift = 0.37 \delta x

here we have

0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}

now plug in all values in it

0.37 = \frac{(1.79 - 1) t}{411 \times 10^{-9}}

t = 0.192 \times 10^{-6} m

t = 0.192 \mu m

3 0
3 years ago
Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
Given the information below, estimate the total distance travelled during these 6 seconds using a left endpoint approximation. t
Nutka1998 [239]

Answer:

184 feets

Explanation:

Given the data:

time (sec) __ velocity (ft/sec)

0 __________30

1 __________ 54

2 __________56

3 __________34

4 __________ 8

5 __________ 2

6 __________22

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(3,4) ___f(3) = 34

(4,5) ___f(4) = 8

(5,6) __ f(5) = 2

Hence, the Total distance traveled during the 6 second interval is:

Change ; dT = 1

1 * (30 + 54 + 56 + 34 + 8 + 2) = 184

4 0
2 years ago
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MrRissso [65]
Answer is x hopefully this helps your stupid head
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