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3 years ago

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A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict

ionless surface. The object is released from rest when the spring is compressed 0.19 m.(a) Find the force on the object.1 N
(b) Find its acceleration at that instant 2 m/s2

1 answer:

Genrish500 [490]3 years ago

6 0

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

$F=157\times 0.19\\\\F=29.83\ N$

(B) The force is simply given by :

F = ma

a is acceleration at that instant

$a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2$

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Answer:

decibels (dB)

Explanation:

The sound intensity level is a quantity derived from the sound intensity.

The intensity of a wave is defined as the power of the source of the wave divided by the area through which the power of the wave is spread, mathematically:

$I=\frac{P}{A}$

where

P is the power of the source

$A=4\pi r^2$ is the surface area over which the wave spreads (assuming that the wave propagates in all directions, it corresponds to the surface area of a sphere of radius $r$ , where r is the distance between the source of the wave and the observer)

For sound waves, the intensity is often expressed using another unit, called decibel (dB), defined as follows:

$\beta(dB)=10Log_{10}(\frac{I}{I_0})$

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$\beta$ is the sound intensity level in decibels

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2 years ago

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

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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

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Explanation:

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2 years ago

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The arrow was moving at 15.9 m/s.

Explanation:

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3 years ago

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pishuonlain [190]

Answer:

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We use the Gauss Law and the superposition law in order to solve this problem.

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3 years ago

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