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3 years ago

9

An inclined plane reduces the effort force by _____.

2 answers:

bija089 [108]3 years ago

7 0

Answer:

<em>Increasing the distance.</em>

An inclined plane reduces the effort force by <u> increasing the distance</u> through which the force is applied.

Let us consider, work is the product of force and distance. And the Work on an inclined plane is defined as angle of force multiplied by distance. The angle may lies between sin or cos components. So the actual effort decreases and is less than the unity always.

Therefore, An inclined plane reduces the effort fort as well as work done.

xz_007 [3.2K]3 years ago

3 0

"increasing the distance through which the force is applied"

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Ilya [14]

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a) F = -1035.385 N

b) Backwards

c) s = 15.60 m

Explanation:

Given information

$u$ = Initial Speed of Car = 15.0 m/s

$v$ = Final Speed of Car = 9.00 m/s

$t_{b}$ = Breaking Time = 1.30 s

$m$ = Mass of Car = 1040 kg

Part (a)

To find the force exerted on the car we use the following formula

$F = ma$

Where

$F$ = Force = unknown

$m$ = Mass of Car = 1040 kg

$a$ = Acceleration of Car / Deceleration of Car = unknown

To find the force (F) we need to first find the deceleration rate (a)

To find the deceleration rate we use the following formula

$a = \frac{v - u}{t}$

Inputting the given values

$a = \frac{15 - 9}{1.30} \\ a = -4.615$

To find the force

$F = ma \\ F = (1040)(-4.615) \\ F = (1040)(-4.615) \\ F = -1035.385 N$

Part (b)

Since the value of F is negative this means the the force was opposite the direction of motion, hence the force was backwards.

Part (c)

To find the total distance the car moved while braking we use the following formula

$v^2 = u^2 + 2as$

Where

$s$ = distance traveled

Inputting the values given

$(9)^2 = (15)^2 + 2(-4.615)s$

$s = 15.60 m$

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3 years ago