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Anarel [89]
3 years ago
12

You push a 45 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. The coefficient of kinetic friction is 0.25. N

ow you double the force on the box. How long would it take for the velocity of the crate to double to 2.0 m/s
Physics
1 answer:
tigry1 [53]3 years ago
8 0

Answer:

       t = 0.408 s

Explanation:

In this exercise we will use Newton's second law to find the acceleration and the kinematic relationships to find the time.

We use Newton's second law, for the first condition where the acceleration is zero

X axis

           F - fr = 0

Y Axis  

          N- W = 0

The expression for friction force is

        fr = μ N

       fr = μ mg

We substitute

      F = μ m g

This is the value for the applied force, now we apply the second condition, in this case the system has an acceleration

X axis

          2F- fr = ma

         a = (2F -fr) / m

Axis y

         N -W = 0

The friction force has the expression

        fr = μ N

        fr = μ W

We substitute

       a = (2F - μ m g) / m

       a = (2 μ mg - μ mg) / m

       a = μ g (2-1) = μ g

We calculate

      a = 0.25  9.8   1

      a = 2.45 m / s²

With this acceleration value we use the kinematics ratio to find the time

       v = v₀ + a t

The initial speed is 1 m / s and the final speed is 2 m / s

       t = (v-v₀) / a

       t = (2-1) / 2.45

       t = 0.408 s

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The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov
victus00 [196]

The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

=0.10368kg m²

l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

#SPJ4

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Answer:

0.3 %

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A grating whose slits are 3.2x10^-4 cm apart produces a third-order fringe at a 25.°0 angle. What is the wavelength of light tha
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Answer:

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then, by Bragg's law:

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λ = d×sin(Ф)/n

λ = (3.2×10^-4 cm)×sin(25.0°)/3

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  ≈ 4.51×10^-7 m

Therefore, the light used has a wavelenght of 4.51×10^-7 m.

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