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Brums [2.3K]
2 years ago
14

A bulldozer attempts to drag a log weighing 500 N along the rough horizontal ground. The cable attached to the log makes an angl

e of 30° above the ground. The coefficient of static friction between the log and the ground is 0.50, and the coefficient of kinetic friction is 0.35. What minimum tension is required in the cable in order for the log to begin to slide?
Physics
1 answer:
Gemiola [76]2 years ago
6 0

Answer:

T= 224.01 N

Explanation:

in imminent motion we have to :

  • The frictional force reaches its maximum value
  • The system is in balance of forces

Data

W=  500 N :  weight of the log

μs = 0.5

μk = 0.35

α = 30°above the ground :  angle of the cable attached to the log

Newton's first law to the log:

∑F =0 Formula (1)

∑F : algebraic sum of the forces in Newton (N)

Forces acting on the log

T: cable tension for impending movement

N: normal force

W : weight

f: frictional force , f= μsN

We apply the formula (1)

∑Fx=0

Tx-f = 0

Tcosα-μsN=0

Tcos30°-0.5N=0 Equation (1)

∑Fy=0

N+Ty-W=0

N+Tsin30°-500=0

N= 500-Tsin30°  Equation (2)

We replace the value of N of the Equation  (2) in the equation (1)

Tcos30°-0.5(500-Tsin30°) = 0

Tcos30°+0.5Tsin30° = 0.5*500

T( cos30°+0.5*sin30°) = 250

(1.116) T = 250

T= 250/1.116

T= 224.01 N

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An engine does 15.0 kj of work while exhausting 37.0 kj to the cold reservoir. What is the efficiency of the engine?
Nataliya [291]

Answer:

28.8 %

Explanation:

The efficiency of an engine is given by:

\eta = \frac{W}{Q_{in}}

where

W is the useful work done

Q_in is the heat in input

For this machine, we have

W = 15.0 kJ is the work done

37.0 kJ is the heat exchausted to the cold reservoir, so the total amount of heat in input is

Q_{in} = 15.0 kJ + 37.0 kJ=52.0 kJ

And so the efficiency is

\eta=\frac{15.0 kJ}{52.0 kJ}=0.288

which corresponds to 28.8 %.

5 0
2 years ago
Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid
Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

8 0
2 years ago
Why is it important for a muscle to be attached to a fixed origin at one end and a moving insertion at the other? Discuss how th
vlabodo [156]
Muscles function only by contracting. This makes it necessary for one end of the muscle to be fixed and the other mobile.
Take the bicep for example.
Its origin is at the shoulder and its two heads connect to the bones of the forearm, the radius and ulna.
Now, had the muscle not been fixed at one end, and contracted, it would pull both our shoulder and forearm together resulting in an ineffective movement. The desired motion is to lift the forearm (proximal and distal movement) which can only be achieved if the bicep is fixed at the shoulder and allowed to move at the forearm.
6 0
2 years ago
The acceleration of an object as a function of time is given by a(t) = (1.00 m/s2)t2. If displacement of the object between time
jolli1 [7]

not enough information is given to determine the velocity of the object at time to=0.00s

3 0
2 years ago
A small crack occurs at the base of a 15.0-m-high dam. The effective area through which water leaves is 2.30 × 10-3 m2. (a) Igno
vova2212 [387]

Answer

given,                                              

height of the dam = 15 m            

effective area of water = 2.3 x 10⁻³ m²

Using energy conservation              

    m g h = \dfrac{1}{2}mv^2

    v= \sqrt{2gh}                  

    v= \sqrt{2\times 9.8 \times 15}

    v= \sqrt{294}              

           v = 17.15 m/s            

 discharge of water

      Q = A V                            

      Q = 2.3 x 10⁻³ x 17.15    

      Q = 0.039 m³/s

3 0
2 years ago
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