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Katena32 [7]
3 years ago
7

A 6.2 kg object moving in the +x direction at 5.3 m/s collides head-on with an 7.8 kg object moving in the −x direction at 2.5 m

/s.
Find the final velocity of each mass for each of the following situations. (Take the positive direction to be +x.)
(a) The bodies stick together.
1 m/s (6.2 kg mass)
2 m/s (7.8 kg mass)
(b) The collision is elastic.
3 m/s (6.2 kg mass)
4 m/s (7.8 kg mass)
(c) The 6.2 kg body is at rest after the collision.
5 m/s (6.2 kg mass)
6 m/s (7.8 kg mass)
(d) The 7.8 kg body is at rest after the collision.
7 m/s (6.2 kg mass)
8 m/s (7.8 kg mass)
(e) The 6.2 kg body has a velocity of 4.0 m/s in the -x direction after the collision.
9 m/s (6.2 kg mass)
10 m/s(7.8 kg mass)
Physics
1 answer:
Crank3 years ago
8 0

Answer:

10ms kg 88.2

Explanation:

Toook test

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A uniform electric field of magnitude 4.6 ✕ 104 N/C is perpendicular to a square sheet with sides 3.0 m long. What is the electr
rodikova [14]

Answer:

41.4* 10^4 N.m^2/C

Explanation:

given:

E= 4.6 * 10^4 N/C

electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field

then electric flux = ∫ E*n dA

                            = ∫ 4.6 * 10^4 * 3*3

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3 0
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A piece of transparent material that is used to focus light and form a image
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A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0
3 years ago
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