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Zinaida [17]
3 years ago
12

. A truck is traveling at a speed of 25.0 m/sec along a level road. A crate is resting on the bed of the truck, and the coeffici

ent of static friction between the crate and the truck bed is 0.650. Determine the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck.
Physics
1 answer:
OLga [1]3 years ago
5 0

Answer:

98.12 m

Explanation:

Initial speed(v) = 25.0 m/s

Coefficient of static friction (u) = 0.650

The minimum value of the coefficient of kinetic friction should be the coefficient of static friction.

Kinetic friction is given as

umg = ma (a is the required acceleration)

We then have

a = -ug

= -(0.65*9.8)

= -6.37 m/s^2

The shortest distance d is obtained by

d = (V^2 - v^2)/ 2a

V = 0 m/s (final velocity)

d = (0 - 25^2) / 2(-6.37)

d = -625/-6.37

d = 98.12 m

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A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the
Ierofanga [76]

Answer:

The discharge rate is Q = 0.0192 \  m^3 /s

Explanation:

From the question we are told that

   The  diameter is  d =  60 \ mm   =  0.06 \ m

    The  head is  h  =  6 \ m

     The  coefficient of contraction is  Cc  =  0.68

     The  coefficient of  velocity is  Cv  =  0.92

The radius is mathematically evaluated as

         r =  \frac{d}{2}

substituting values

        r =  \frac{ 0.06 }{2}

        r =  0.03 \ m

The  area is mathematically represented as

      A =  \pi r^2

substituting values

      A =  3.142 *  (0.03)^2

      A = 0.00283 \ m^2

 The  discharge rate is mathematically represented as

        Q =  Cv *Cc  *  A  *  \sqrt{ 2 * g *  h}

substituting values

       Q = 0.68 *  0.92*   0.00283  *  \sqrt{ 2 * 9.8 *  6}

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6 0
3 years ago
I will be so thankful if u answer correctly!!​
olga_2 [115]
<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>

In this case, a frictional force F_{r} is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma    -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F -  F_{r}

m = 50kg

a = 0   [At constant velocity, acceleration is zero]

<em>Substitute these values into equation (i) as follows;</em>

F -  F_{r} = m x a

F -  F_{r} = 50 x 0

F -  F_{r} = 0

F =  F_{r}            -------------------(ii)

<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>

In this case, the same frictional force F_{r} is opposing the movement of the box.

∑F = 1.5F -  F_{r}

m = 50kg

a =  0.1m/s²

<em>Substitute these values into equation (i) as follows;</em>

1.5F -  F_{r} = m x a

1.5F -  F_{r} = 50 x 0.1

1.5F -  F_{r} = 5            ---------------------(iii)

<em>Substitute </em>F_{r}<em> = F from equation (ii) into equation (iii) as follows;</em>

1.5F - F = 5            

0.5F = 5            

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

<em />

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