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4 years ago

9

We’re measuring the velocity, so that we can find the change in kinetic energy. Where should you place the photogate to measure

the cart’s velocity? Briefly explain your reasoning.

1 answer:

7nadin3 [17]4 years ago

4 0

Answer:

minimum two photogate

Explanation:

The photogate allows us to activate the measurement of time, so we must place at least two of them, one at the beginning of the movement to measure the initial time and another at the end of the movement to measure the final time, of course we must have measured the distance. This gives us the average speed between the two photocells

For measurements with more changes curves, slopes, one photogate must be placed at the beginning of the curve or slope and another at the end.

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Determine the magnitude, fda, acting along member da, due to the applied force f=10.0 iâ20.0 j+27.2 k n.

VashaNatasha [74]

Force vector is given as

$F = 10\hat i + 20\hat j + 27.2\hat k$

now we need to find the magnitude of this force

as we know that

$F_x = 10$

$F_y = 20$

$F_z = 27.2$

so here we will find the net force

$F_{net} = \sqrt{F_x^2 + F_y^2 + F_z^2}$

$F_{net} = \sqrt{10^2 + 20^2 + 27.2^2}$

$F_{net} = 35.2 N$

so it is having magnitude as 35.2 N

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3 years ago

What is the difference between real and apparent weightlessness?

MrRissso [65]

Answer:

In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.

Explanation:

Hope this helps:)

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3 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )

Also at equilibrium the moment about the point where the string holding the bar (where ( $m_1 +m_2$ ) and $m_3$ are hanged ) is attached is zero

So basically

$(m_1 + m_2 ) * 20 = m_3 * 30$

$(0.42 + 0.47) * 20 = 30 * m_3$

Making $m_3$ subject

$m_3 = \frac{(0.42 + 0.47) * 20 }{30 }$

$m_3 = 0.59 kg$

3 0

3 years ago

Consider the following three objects, each of the same mass and radius: (1) a solid sphere (2) a solid disk (3) a hoop All three

Vinil7 [7]

Answer:

The correct answer is

a) 1, 2, 3

Explanation:

In rolling down an inclined plane, the potential energy is Transferred to both linear and rotational kinetic energy thus

PE = KE or mgh = 1/2×m×v² + 1/2×I×ω²

The transformation equation fom potential to kinetic energy is =

m×g×h = $\frac{1}{2} mv^{2} + \frac{1}{2} (\frac{2}{5} mr^{2} )(\frac{v}{r}) ^{2}$

$v_{Sphere}$ = $\sqrt{\frac{10}{7} gh}$

$v_{Hoop}$ = $\sqrt{gh}$

$v_{Disc}$ = $\sqrt{\frac{4}{3} gh}$

Therefore the order is with increasing rotational kinetic energy hence

the first is the sphere 1 followed by the disc 2 then the hoop 3

the correct order is a, 1, 2, 3

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4 years ago

How much current does a 10.0 Ω resistor draw from a 12 V battery?

Amiraneli [1.4K]

The answer would be B.
(V=IR, 12-10i, 12/10=1.2)

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3 years ago