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OverLord2011 [107]
3 years ago
13

200. newton·meters of work is put into a machine over a distance of 20. meters. The machine does 150. newton·meters of work as i

t lifts a load 10. meters high. What is the mechanical advantage of the machine?
Physics
1 answer:
dezoksy [38]3 years ago
3 0

Answer:

Answer:

Mechanical advantage of the machine is 1.5

Explanation:

The formula for mechanical advantage is:

MA = OutputForce/InputForce

To calculate this, we need the force of input and the force of output.

The formula to calculate the force given the work (W) and the distance (d) is:

F = W/d

Calculating the input force Fi:

Fi = 200Nm/20m = 10 N

Calculating the output force Fo:

Fo = 150Nm/10m = 15 N

Thus, the mechanical advantage MA is:

MA = 15N / 10N = 1.5

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In naming covalent compound (binary) based in IUPAC naming, we have 4 rules to be followed:

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2. The second element is named as if they are treated like an anion but put in mind that these are no ions in a covalent compound but we put -ide on the second element as if it is an anion.

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2 years ago
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What must occur for work to be done on an object?
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The answer is D) the object must move. The definition of work is movement (in any form).

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8 0
3 years ago
Astronauts on the first trip to Mars take along a pendulumthat has a period on earth of 1.50 {\rm s}. The period on Mars turns o
svetlana [45]

Answer:

3.7 m/s^2

Explanation:

The period of a simple pendulum is given by:

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum and g is the free-fall acceleration on the planet.

Calling L the length of the pendulum, we know that:

T_e = 2 \pi \sqrt{\frac{L}{g_e}}=1.50 s is the period of the pendulum on Earth, and g_e = 9.8 m/s^2 is the free-fall acceleration on Earth

T_m = 2 \pi \sqrt{\frac{L}{g_m}}=2.45 s is the period of the pendulum on Mars, and g_m = ? is the free-fall acceleration on Mars

Dividing the two expressions we get

\frac{T_e}{T_m}=\sqrt{\frac{g_m}{g_e}}

And re-arranging it we can find the value of the free-fall acceleration on Mars:

g_m = g_e \frac{T_e^2}{T_m^2}=(9.8 m/s^2)\frac{(1.50 s)^2}{(2.45 s)^2}=3.7 m/s^2

4 0
3 years ago
Stephen throws a 0.46 kg football at a speed of 14 m/s and it goes through the goal posts 2.4 m above where he threw it. What is
Dovator [93]
  • Answer:
  • \\  \\ 0.46 \times 14 {2 }^{2}  = 90.16 \\

5 0
2 years ago
By what factor would you need to adjust the length of a pendulum to make the period of 1/2 of what it used to be?
Irina-Kira [14]

Answer:

1/4

Explanation:

The period of a simple pendulum is:

T = 2π √(L/g)

At half the period:

T/2

= π √(L/g)

= 2π √(L/(4g))

= 2π √((L/4)/g)

So the length would have to be shortened by a factor of 1/4.

3 0
3 years ago
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