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N76 [4]
3 years ago
5

A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica

lly up from the ground with an initial velocity of 0=40.0 m/s . At what height from the ground will the two objects first meet?

Physics
1 answer:
-BARSIC- [3]3 years ago
7 0

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

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3 years ago
Read 2 more answers
A pulley with a mechanical advantage of 5 will require you to pull _____ times the amount of rope.
Marat540 [252]

Answer:

It will require a force of 1/5, answer A.

Explanation:

In the attached image we can see an example of an array of pulleys that will lift a 100 kg-f load.

If we analyze pulleys A,B, and C as in the image we can check a force in the cable of 20kg-f.

In the pulley D we have three forces of 20 kg-f each and those forces plus the forces in the pulley B, sum a total of 100 kg-f (60+40). This matches the mechanical advantage (100/5) = 20 kg-f

6 0
3 years ago
A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex
Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

\text{Efficiency of the heat engine} = \eta = 26\% = 0.26

\text{Energy taken in by the heat engine during each cycle} = Q_h

\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J

\eta = 1 - \frac{Q_{c}}{Q_{h}}

0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}

\frac{8.55* 10^{3}}{Q_{h}} = 0.74

Q_h = \frac{8.55*10^3}{0.74}

Q_h = 11.554*10^3J

PART A)

Work done by the heat engine in each cycle = W

W = Q_h-Q_c

W = 11.554*10^3J-8.55*10^3J

W = 3004J

According to the value given we have that,

P = 4.0kW

P = 4000W

Power is defined as the variation of energy as a function of time therefore,

P = \frac{W}{t}

4000W = \frac{3004J}{t}

t = \frac{3004}{4000}

t = 0.75s

Therefore the interval for each cycle is 0.75s

5 0
3 years ago
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