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MrRissso [65]
3 years ago
13

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

ds, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 25 m/s. The masses of the two objects are 3.0 and 8.0 kg. Determine the final speed of the two-object system after the collision for the case when the large-mass object is the one moving initially and the case when the small-mass object is the one moving initially.
Physics
1 answer:
Lina20 [59]3 years ago
6 0

Large object moving, small object at rest:

v = 18.18m/s

Small object moving, large object rest:

v= 6.81m/s

Explanation:

p = mv

m1•Vi1 + m2•Vi2 = m1•Vf1 + m2•Vf2

First Scenario (large object moving):

For small object,

p=mv

p= 3kg • 0 = 0

For large object,

p= mv

p= 8kg • 25m/s = 200kg•m/s

Combined,

p1+p2 (small object + large object) = 200m/s

After the collision, there is one mass,

new mass is 8kg + 3kg = 11kg

So now , p = mv again. In this case

200m/s = (11kg)•v

v = (200kg•m/s)/(11kg) = 18.18 m/s

Second Scenario (small object moving):

p1v1 = 3kg•25m/s = 75kg•m/s

p2v2 = 8kg•0 = 0

p1v1 + p2v2 = 75kg•m/s

New mass = 8kg+ 3kg = 11kg

p = mv

75kg•m/s = 11kg•v

v = (75kg•m/s)/11kg = 6.81m/s

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Colt1911 [192]

Answer:

Ff = 839.05 N

Explanation:

We can use the equation:

Ff = μ*N

where <em>N</em> can be obtained as follows:

∑ Fc = m*ac   ⇒   N - F = m*ac = m*ω²*R    ⇒  N = F + m*ω²*R

then if

F = 32 N

m = 133 Kg

R = 0.635 m

ω = 95 rev /min = (95 rev / min)(2π rad / 1 rev)(1 min / 60 s) = 9.9484 rad /s

we get

N = 32 N + (133 Kg)*(9.9484 rad /s)²*(0.635 m) = 8390.53 N

Finally

Ff = μ*N = 0.10*(8390.53 N) = 839.05 N

3 0
3 years ago
If two objects of the same size move through the air at different speeds, which encounters the greater air resistance?a. The fas
blagie [28]

Answer:

Option A is correct.

(The faster object encounters more resistance)

Explanation:

Option A is correct. (The faster object encounters more resistance)

Air resistance depends on various factors:

  • Speed of the object
  • Cross-sectional area of the object
  • Shape of the object

Formula:

F=\frac{1}{2}C_d\rho A v^{2}

As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.

6 0
3 years ago
What is an error in a lab mean?
timurjin [86]
You have done something incorrectly. such as your data. 
5 0
3 years ago
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
KengaRu [80]

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A  = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m

Therefore, the length of the wire is 134.62 m.

6 0
2 years ago
An object begins x=75.2 m and undergoes a displacement of -48.7 m. what is its final position?
xz_007 [3.2K]

Answer:

26.5 m

Explanation:

x_{o} = initial position of the object = 75.2 m

x  = final position of the object

d  = displacement of the object = - 48.7  

Displacement of the object is given as the difference of final and initial position of the object

d = x - x_{o}

Inserting the values

- 48.7 = x - 75.2

x = 26.5 m

8 0
2 years ago
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