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Tasya [4]
3 years ago
9

A horse pulls a cart along a road with a force of 550 lbs. If the horse does 2,674,100 ftlbs of work by the time it stops, how f

ar did the horse pull the cart?
Engineering
2 answers:
garik1379 [7]3 years ago
8 0

Answer:

approximately 222,841.6666666667 feet

BartSMP [9]3 years ago
5 0

Answer:

When a horse pull a cart the action is on?

A horse is harnessed to a cart. If the horse tries to pull the cart, the horse must exert a force on the cart. By Newton's third law the cart must then exert an equal and opposite force on the horse. Newton's second law tells us that acceleration is equal to the net force divided by the mass of the system.

Explanation:

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Answer:

C. Neither Technician A nor B

Explanation:

Just took the test

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What are flip flops and what do they look like
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For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration
Jobisdone [24]

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

4 0
3 years ago
Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a p
oee [108]

Answer:

a) 24 kg

b) 32 kg

Explanation:

The gauge pressure is of the gas is equal to the weight of the piston divided by its area:

p = P / A

p = m * g / (π/4 * d^2)

Rearranging

p * (π/4 * d^2) = m * g

m = p * (π/4 * d^2) / g

m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg

After the weight is added the gauge pressure is 2.8kPa

The mass of piston plus addded weight is

m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg

56 - 24 = 32 kg

The mass of the added weight is 32 kg.

5 0
3 years ago
The drag force, Fd, imposed by the surrounding air on a
Ad libitum [116K]

Answer:

a)  23.551 hp

b)  516.89 hp

Explanation:

<u>given:</u>

F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2

<u>required:</u>

the power in hp

<u>solution:</u>

(F_{d})_{a}  =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}.............(1)

by substituting in the equation (1)

         =353.27 lbf

(F_{d})_{b}  =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}..........(2)

by substituting in the equation (2)

         = 2769.29 lbf

power is defined by

             P=F.V

     P_{a}=353.27*36.67

           =12954.411 lbf.ft/s

           =12954.411*.001818

           =23.551 hp

      P_{a}=2769.29*102.67

           = 284323 lbf.ft/s

           = 284323*.001818

           = 516.89 hp

3 0
3 years ago
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