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2 years ago

11

How much force is needed to push a 20kg at 0.2 m/s/s A)10 N B)20 N C)4 N D)40 N

1 answer:

ad-work [718]2 years ago

4 0

Answer:

4

Explanation:

20x.2=4

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The number of wave troughs that pass by every second is a wave's _________.

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A waves frequency (in Hertz) is how many crests pass by a point per second. easily confused with period, which is the amount of time it takes for a full wave to pass by a certain point

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3 years ago

Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative

SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

Explanation:

Given there are three blocks of masses $M_{a}$ , $M_{b}$ and $M_{c}$ (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *a ................(equation 1)

Also it is given that $M_{a}$ does not move with respect to $M_{c}$ , which gives tension T is exerted on pulley by $M_{a}$ only, Hence tension T is

T = $M_{a}$ *a ..........(equation 2)

There is also also tension exerted by $M_{b}$ . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = $M_{b}$ $\sqrt{a^{2} +g^{2} }$ ................(equation 3)

From equation 2 and 3, we get

$M_{a}$ *a = $M_{b}$ $\sqrt{a^{2} +g^{2} }$

Squaring both sides we get

$M_{a} ^{2}$ * $a^{2}$ = $M_{b} ^{2}$ * ( $a^{2}$ + $g^{2}$ )

$M_{a} ^{2}$ * $a^{2}$ = ( $M_{b} ^{2}$ * $a^{2}$ )+ ( $M_{b} ^{2}$ * $g^{2}$ )

( $M_{a} ^{2}$ - $M_{b} ^{2}$ ) * $a^{2}$ = $M_{b} ^{2}$ * $g^{2}$

$a^{2}$ = $M_{b} ^{2}$ * $g^{2}$ /( $M_{a} ^{2}$ - $M_{b} ^{2}$ )

Taking square root on both sides, we get acceleration a

a = $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ )

Hence substituting the value of a in equation 1, we get

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

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3 years ago

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3 years ago

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3 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$

velocity is constant, a = 0

$m g sin\theta - F_{rope} =0$

$F_{rope} = m g sin\theta$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0$

$F_{rope}= 102.73\ N$

b) now, when acceleration, a = 0.135 m/s²

$F_{rope}-m g sin\theta = ma$

velocity is constant, a = 0.135 m/s₂

$F_{rope} = m g sin\theta+ma$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135$

$F_{rope}= 112.19\ N$

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3 years ago