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2 years ago

The IMA of a lever can be increased by decreasing the length between the applied effort and the pivot.

Physics

2 answers:

AVprozaik [17]2 years ago

4 0

Explanation:

The ideal mechanical advantage of a lever is defined as the ratio of applied resistance force to the applied effort force. It can be also defined as the ratio of effort arm to the load arm. It is given by :

$IMA=\dfrac{d_1}{d_2}$

$d_1$ = effort arm

$d_2$ = load arm

If the length between the applied effort and the pivot is decreased, the IMA of a lever will decreased. This is due to the direct relation between IMA and effort arm. So, the given statement is false.

mina [271]2 years ago

3 0

<span>We never really used the acronym "IMA", or ideal mechanical advantage, but I'm assuming you are trying to increase the leverage and ease the effort. If so, the answer is false. You want larger movement on the effort side, and smaller movement on the resistant side of the fulcrum.</span>

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bazaltina [42]

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2 years ago

A 40 kg person sits on top of a 400 kg rock. What is the person’s weight? 390 N

algol13

The weight of the person is given by:

W = mg

W = weight, m = mass, g = gravitational acceleration

Given values:

m = 40kg, g = 9.81m/s²

Plug in and solve for W:

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2 years ago

How much speed is needed for an object to travel 250 km in 1.5 hrs? *<br> Your answer

NemiM [27]

Answer: 166.67km/hr

Explanation:

Given the following :

Distance traveled = 250km

Time taken = 1.5 hours

Recall :

Speed = Distance traveled / time taken

Speed = 250 km / 1.5 hours

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3 years ago

Which of these is not an example of

prohojiy [21]

Answer:

A) The process that give us day and night.

Explanation:

From the answer choices provided the one that is not an example of this is the process that give us day and night. This is because, the day/night cycle occurs as the Earth spins on it's axis. This cause one side of the Earth to be facing the Sun, which is the side that is currently experiencing day time while the other side is experiencing night time. As the Earth spins the cycles repeat. This is not an example of revolving.

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3 years ago

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let $F_{e}$ be the force of gravity exerted by the earth

and let $F_{m}$ be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d, is given by:

$F = \frac{Gm_{1} m_{2} }{d^{2} }$

Mass of the earth, $m_{e} = 5.97 * 10^{24} kg$

Mass of the moon, $m_{m} = 7.348 * 10^{22} kg$

Mass of the satellite, $m_{s} = ?$

$F_{e} = \frac{G*5.97 * 10^{24} M }{d^{2} }$ ...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

$F_{m} = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }$ ............................(2)

Equating equations (1) and (2)

$\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }$

$(5.97 * 10^{24})(14.78 * 10^{16} -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d + 5.97 * 10^{24}d^{2} = 7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\$

Factorising out $10^{24}$

$1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0$

Solving for d in the quadratic equation above:

d = 346 * 10⁶ m

4 0

2 years ago