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Pepsi [2]
3 years ago
15

A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0° with the horizontal. The coeff

icient of kinetic friction between slide and child is 0.120. (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of 0.559 m/s, what is her speed at the bottom?
Physics
1 answer:
natulia [17]3 years ago
6 0

Answer:

a

H  =212.6 \  J

b

v  =  7.647  \  m/s

Explanation:

From the question we are told that

   The child's weight is  W_c  =  287 \ N

    The length of the sliding surface of the playground is  L =  7.20 \  m

    The coefficient of friction is  \mu =  0.120

      The angle is \theta = 31.0 ^o

      The initial  speed is  u =  0.559 \  m/s

Generally the normal force acting on the child is mathematically represented as

=>    N  =  mg  *  cos \theta

Note  m *  g  =  W_c

Generally the frictional force between the slide and the child is    

         F_f  =  \mu *  mg  *  cos \theta

Generally the resultant force acting on the child due to her weight and the frictional  force is mathematically represented as

      F =m* g sin(\theta) - F_f

Here  F is the resultant force and it is represented as  F =  ma

=>   ma =   m* g sin(31.0)  - \mu *  mg  *  cos (31.0)

=>   a =  g sin(31.0)-  \mu *  g  *  cos (31.0)

=>  a =    9.8 *  sin(31.0) - 0.120 *  9.8  *  cos (31.0)

=>a =  4.039 \ m/s^2

So

   F_f  =  0.120  * 287  *  cos (31.0)

=> F_f  = 29.52 \  N

Generally the heat energy generated by the frictional  force which equivalent tot the workdone by the frictional force  is mathematically represented as

     H  =  F_f  * L

=>  H  = 29.52 *  7.2

=>  H  =212.6 \  J

Generally from kinematic equation we have that

    v^2  =  u^2  +  2as

=>  v^2  =  0.559^2  +  2 * 4.039 * 7.2

=>  v  =  \sqrt{0.559^2  +  2 * 4.039 * 7.2}

=>  v  =  7.647  \  m/s

   

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Answer:

4.16m/s²

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Fm is the moving force

\mu is  the coefficient of kinetic friction between the child and the slide

m is the mass

g is the acceleration due to gravity

a is the acceleration of the child

Substitute the given values and get the acceleration as shown;

35(9.8)sin27.5 - 0.415(35)(cos27.5) = 35a

158.38-12.88 = 35a

145.49 = 35a

a = 145.49/35

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A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After th
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Answer:

(a) The velocity of the bullet and B after the first impact is 4.4554 m/s.

(b) The velocity of the carrier is 0.40872 m/s.

Explanation:

(a) To solve the question, we  apply the principle of conservation of linear momentum as follows.

we note that the distance between B and C is 0.5 m

Then we  have

Sum of initial momentum = Sum of final momentum

0.03 kg × 450 m/s = (0.03 kg + 3 kg) × v₂

Therefore v₂ = (13.5 kg·m/s)÷(3.03 kg) = 4.4554 m/s

The velocity of the bullet and B after the first impact = 4.4554 m/s

(b) The velocity of the carrier is given as follows

Therefore from the conservation of linear momentum we also have

(m₁ + m₂)×v₂  = (m₁ + m₂ + m₃)×v₃

Where:

m₃ = Mass of the carrier = 30 kg

Therefore

(3.03 kg)×(4.4554 m/s) = (3.03 kg+30 kg) × v₃

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shusha [124]

Answer:

\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m

Explanation:

<u>Sum of Vectors in the Plane</u>

Given two vectors

\displaystyle \vec{v_1}\ ,\ \vec{v_2}

They can be expressed in their rectangular components as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_2}=

The sum of both vectors can be done by adding individually its components

\displaystyle \vec{v_1}+\vec{v_2}=

If the vectors are given as a magnitude and an angle (M\ ,\ \theta ), each component can be found as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_2}=

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\displaystyle \vec{v_1}=

\displaystyle \vec{v_1}=

The second vector has a magnitude of 2.71 m and an angle of -60°, so

\displaystyle \vec{v_2}=

\displaystyle \vec{v_2}=

The sum of the vectors is

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