When the IMA is compared to AMA in the real world, _____.

2 answers:

5 0

<span>

I believe the answer is A.) the value of the IMA is always larger than the AMA
(Not 100% on this, not my strongest topic, so I would double check)</span>

Korvikt [17]3 years ago

4 0

AMA is actual mechanical advantage in where it would compare the force in as well as out in using actual measurements in the real life or with friction while in IMA which is the ideal mechanical advantage no friction then is needed. The answer then is a. the value of IMA is larger than AMA.

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A circuit consists of a 12 V battery connected across a single resistor. If the current in the circuit is

finlep [7]

Answer:

4 Ohms

Explanation:

Apply the formula:

Voltage = I (current) . Resistance

You can change it the way you want to use for your purpose.

In this case...

R = V/I

R = 12/3

R = 4 Ohms (Ohm is the unit of measurement of eletrical resistance)

7 0

2 years ago

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$

$-W_{s} \times x + W_{b} \times 1.5$ = 0

x = $\frac{W_{b} \times 1.5}{W_{s}}$

= $\frac{332 N \times 1.5}{505 N}$

= 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

8 0

3 years ago

A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.

Airida [17]

Answer:

a) $v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )$ , b) $\Delta K = 9.559\times 10^{-5}\,J$

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

$(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v$