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andrew-mc [135]

3 years ago

9

Some workers do 15000 J of work on a 3500 N crate to push it up a ramp. The ramp is 4 m high. What is the efficiency of the ramp

?

1 answer:

Alja [10]3 years ago

8 0

Answer:

93.3%

Explanation:

Efficiency of ramp is given by

$n=\frac {W_o}{W_i}\times 100\%$

Where n is the efficiency, W is work and subscripts o and i represent output and input respectively.

The work input is what the workers do and that is given as 15 kJ

The work output

Since work is product of force and perpendicular distance and substituting force with 3500 N and distance with 4 m we get that the work output is 3500*4=14000 J= 14 kJ

Finally, substituting 14 kJ and 15 kJ for work output and input respectively we deduce that

$n=\frac {14 kJ}{15 kJ}\times 100\%=93.3333333333/%\approx 93.3/%$

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Explanation:

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3 years ago

In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21

Lapatulllka [165]

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3 years ago

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snow_lady [41]

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s

Explanation:

It is given that,

Mass of the car, $m_1=1.1\times 10^3\ kg$

Mass of the truck, $m_2=2.3\times 10^3\ kg$

Initial velocity of the car, $u_1=22\ m/s$

Initial velocity of the truck, u₂ = 0

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