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2 years ago

6. A 50 N block is raised 2 m. If the net work done on the block is 50 J, what is the applied force on the block?

Physics

1 answer:

geniusboy [140]2 years ago

5 0

Answer:

F = 75[J]

Explanation:

We know that work is defined as the product of force by distance.

In this way we have two forces, the weight of the block down, and the force that bring about the block to rise.

$W = -(F_{weight*d})+(F_{upward}*d)$

where:

W = work = 50 [J]

d = distance = 2 [m]

Fweight = 50 [N]

Fupward [N]

Now replacing:

$50=-(50*2)+(F_{upward}*2)\\50+100=F_{upward}*2\\F_{upward}=150/2\\F_{upward}=75[J]$

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Refraction is the bending of a wave disturbance as it passes at an angle from one ____ into another.

levacccp [35]

Refraction is the bending of a wave disturbance as it passes at an angle from one medium into another

3 0

3 years ago

2. A common physics experiment involves lowering an open tube into a cylinder of water and moving the tube up and down to adjust

notsponge [240]

Answer:

Explanation:

This question pertains to resonance in air column. It is the case of closed air column in which fundamental note is formed at a length which is as follows

l = λ / 4 where l is length of tube and λ is wave length.

here l = .26 m

λ = .26 x 4 = 1.04 m

frequency of sound = 330 Hz

velocity of sound = frequency x wave length

= 330 x 1.04

= 343.2 m /s

b )

Next overtone will be produced at 3 times the length

so next length of air column = 3 x 26

= 78 cm

c )

If frequency of sound = 256 Hz

wavelength = velocity / frequency

= 343.2 / 256

= 1.34 m

= 134 cm

length of air column for resonance

= wavelength / 4

134/4

= 33.5 cm

7 0

2 years ago

How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .