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2 years ago

Pube Goldberg machine is a complicated contraption designed to do a very simple task, like the one shown above

Physics

1 answer:

Elanso [62]2 years ago

6 0

Answer:

I think is 2

Explanation:

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Meclanical Energy

miss Akunina [59]

Answer:

a jet flying through the air making a paper airplane

7 0

3 years ago

w strong are you? Could you turn a moving car with just your own strength? No way! If a normal driver inputs about 50 N of force

malfutka [58]

The mechanical advantage of a simple machine is the measure of its amplified force gain.

The mechanical advantage is defined as the force amplified by a machine to the force required to generate such output.

$Mathematically\ mechanical\ advantage\ MA=\frac{F_{o}} {F_{i}}$

$F_{o} \ and\ F_{i}$ are the amplified force and applied force. We may also consider them as output and input force.

In the given question, the force given to the steering wheel is 50 N.

The output force produced by the steering wheel is 3750 N.

Hence the mechanical advantage will be-

$MA=\frac{F_{o}} {F_{i}}$

$=\frac{3750\ N}{50\ N}$

$=75$ [ans]

3 0

3 years ago

Look in the comments. I d a r e y o u

Alika [10]

Answer:

Where is the comments??

Explanation:

3 0

3 years ago

Read 2 more answers

In science, Bob learns that the energy of a wave is directly proportional to the square of the waves amplitude. If the energy of

KIM [24]

The energy of a wave is directly proportional to the square of the waves amplitude. Therefore, E = A² where A is the amplitude. This therefore means when the amplitude of a wave is doubled the energy will be quadrupled, when the amplitude is tripled the energy increases by a nine fold and so on.
Thus, in this case if the energy is 4J, then the amplitude will be √4 = 2 .

4 0

3 years ago

Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th

lapo4ka [179]

Answer:

$T_f=5.0116^{\circ}C$

Explanation:

Given:

mass of water, $m_w=0.25\ kg$

initial temperature of water, $T_i_w=20^{\circ}C$

initial temperature of pan, $T_i_p=173^{\circ}C$

mass of pan, $m_p=0.6\ kg$

mass of water evapourated, $m_v=0.03\ kg$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

specific heat of aluminium pan, $c_a=900\ J.kg^{-1}.K^{-1}$

latent heat of vapourization, $L=2256000\ J.kg^{-1}$

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

$m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})$

$0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)$

$T_f=5.0116^{\circ}C$

5 0

3 years ago