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evablogger [386]
3 years ago
11

The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict

ion between the block and the plane is μk = 0.24. Determine the distance the block will slide before it stops.
Physics
2 answers:
nikklg [1K]3 years ago
3 0

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

Furkat [3]3 years ago
3 0

Answer:

The distance the block will slide before it stops is x= 3.4 m .

Explanation:

m1= 0,02 kg

V1= 400 m/s

m2= 2 kg

V2= ?

g= 9.8 m/s²

μ= 0.24

N = m2 * g = 19.6 N

Fr= μ * N

Fr= 4.704 N

due to the conservation of the amount of movement:

m1*V1 = m2*V2

V2= 4 m/s

Fr = m2*a

a= Fr/m2

a= -2.352 m/s²

matching momentum and amount of movement

Fr*t=m2*V2

t= 1.7 sec

x= V2*t - (a*t²)/2

x= 3.4 m

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Rudik [331]

Answer:

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Explanation:

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4 0
2 years ago
Does a rigid object in uniform rotation about a fixed axis satisfy the first and second conditions for equilibrium?
Vladimir [108]

Answer:

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A body under uniform rotation is experiencing a centripetal force all the time so F net ≠ 0

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F is along r

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7 0
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Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the
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Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

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V =\frac{Kq}{r}

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

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6 0
3 years ago
Please help
WITCHER [35]

Gravity is the force that pulls you down.
(This is kind of a duh! question ... How do we know
which way is "down" ?  We feel gravity, and we call
that the "down" direction.)

Magnetic force holds things to fridge doors.

Contact forces need to touch something in order to
exert their force.
Example:  Gravity is NOT a contact force.

I don't know about "rubbing things away".
This might be a description of friction, but if so,
it's not a good one.

Buoyant force is what keeps floating things floating.

Air resistance slows things down when they move in air.

4 0
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ioda

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The horizontal component of the velocity is 2.68\ \text{m/s}.

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