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butalik [34]
3 years ago
12

A golf ball is struck with a velocity of 80 ft/s as shown. Determine the speed at which it strikes the ground at b and the time

of flight from a to b
Physics
2 answers:
ziro4ka [17]3 years ago
8 0

Explanation:

The initial velocity at which the ball is struck is:  

v

A

=

80

m

/

s

 

The angle of inclination of the surface is:  

α

=

10

∘

The angle at which ball is projected is:  

β

=

45

∘

The total angle of inclination of projectile from horizontal is

θ

=

α

+

β

olga_2 [115]3 years ago
5 0

Answer:

Speed at which it strikes the ground= 67.4ft/s

Time of flight= 3.57secs

Explanation:

(Va)x= 80cos(10+45)°= 44.886

(Va)y= 80sin(10+45)°= 65.532

S=So + Vot + 1/2at^2

d sin 10°= 0 + 65.532t + 1/2 (-32.2)t^2

0.17d= 0 + 65.532t + (-16.1)t^2

Solving the equation gives

d=166ft

t= 3.57seconds

(Vb)x= (Va)x= 45.886

(Vb)y= 65.532 - 32.2(3.57)

(Vb)y = -49.357

Vb= sqrt(45.886^2+(-49.357^2))

Vb= 67.4ft/s

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You pull with a force of 77 N on a piece of luggage of mass 23 kg, but it does
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Answer:

The force of static friction acting on the luggage is, Fₓ = 180.32 N

Explanation:

Given data,

The mass of the luggage, m = 23 kg

You pulled the luggage with a force of, F = 77 N

The coefficient of static friction of luggage and floor, μₓ = 0.8

The formula for static frictional force is,

                                      Fₓ = μₓ · η

Where,

                                  η - normal force acting on the luggage 'mg'

Substituting the values in the above equation,

                                   Fₓ = 0.8 x 23 x 9.8

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Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N

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What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a
viktelen [127]

Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

 COP = 8.49

b

  COP_1 = 9.49  

Explanation:

From the question we are told that

     The lower operation temperature of refrigerator is  T_1 =  -8.00^oC =  265 \  K

     The upper operation temperature of the refrigerator is   T_2 =  23.2 ^oC =  296.2 \  K

Generally the refrigerators coefficient of performance is mathematically represented as

        COP =  \frac{T_1}{T_2 - T_1  }

=>     COP =  \frac{265}{296.2 - 265  }

=>     COP = 8.49

Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

            COP_1 =  \frac{T_2}{ T_2 - T_1}  

=>         COP_1 =  \frac{296.2}{ 296.2 - 265 }  

=>         COP_1 = 9.49  

8 0
3 years ago
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