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zzz [600]
3 years ago
10

A 5000-lb wrecking ball hangs from a 50-ft cable of density 10 lb/ft attached to a crane. Calculate the work done if the crane l

ifts the ball from ground level to 50 ft in the air by drawing in the cable.
Physics
1 answer:
aniked [119]3 years ago
3 0

Answer:

total work is = 52450 J

Explanation:

given data

mass =  5000-lb

density = 10 lb/ft

height = 50 ft

solution

as we will treat here cable and ball are separate  

and

here work need to lift cable is

w = (10Δy )(9.8 y )  j

and

now summing all segment of cable

so passing limit Δy to 0

so total work need

= \int\limits^{10}_0 {98y} \, dy    

= [49 y^2]^{50}_0

= 2450J

so lifting 5000 lb wrcking 50 m  required additional 5000 + 2450

so total work is = 52450 J

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A 55.0-g aluminum block initially at 27.5 degree C absorbs 725 J of heat. What is the final temperature of the aluminum? Express
andrew11 [14]

Answer:

Final temperature of the aluminum = 41.8 °C

Explanation:

We have the equation for energy

      E = mcΔT

Here m = 55 g = 0.055 kg

ΔT = T - 27.5

Specific heat capacity of aluminum = 921.096 J/kg.K

E = 725 J

Substituting

     E = mcΔT

     725 = 0.055 x 921.096 x (T - 27.5)

     T - 27.5 = 14.31

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Final temperature of the aluminum = 41.8 °C

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3 years ago
Eq-45 what type of boat requires navigation lights?
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All the boats operating at night requires Navigation light.

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If a circuit has a power of 50 W with a current of 4.5 A, what is the resistance in the circuit?
lawyer [7]

Answer:

{ \rm{power, \: p = current \times p.d}} \\ { \rm{50 = 4.5 \times (current \times resistance)}} \\ { \rm{50 = 4.5 \times (4.5 \times r)}} \\ { \rm{resistance =  \frac{50}{ {4.5}^{2} } }} \\  \\ { \rm{resistance = 2.5 \:  ohms}}

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1 year ago
A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k
Veronika [31]

Answer:

14.98\ \text{kg m/s}

45.26^{\circ}

Explanation:

P_1 = Initial momentum of the pin = 13 kg m/s

P_i = Initial momentum of the ball = 18 kg m/s

P_2 = Momentum of the ball after hit

55^{\circ} = Angle ball makes with the horizontal after hitting the pin

\theta = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}

Momentum in the y direction

P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}

(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}

The pin's resultant velocity is 14.98\ \text{kg m/s}

P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}

The pin's resultant direction is 45.26^{\circ} below the horizontal or to the right.

4 0
2 years ago
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