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mrs_skeptik [129]

3 years ago

12

During their physics field trip to the amusement park, Leslie and Maria took a ride on the Whirligig. The Whirligig ride consist

s of long swings which spin in a circle at relatively high speeds. As part of their lab, Leslie and Maria estimate that the riders travel through a circle with a radius of 6.5 m and make one circle every 5.8 seconds. Determine the frequency, the period, the speed and acceleration of the riders on the Whirligig

1 answer:

grandymaker [24]3 years ago

6 0

Answer:

a) frequency = 0.1724 Hz

b) Period = 5.8 sec

c) speed = 7.04 m/s

d) acceleration = 7.62 m/s²

Explanation:

Given that;

radius = 6.5m

time period = 5.8 sec every circle

a) the frequency

frequency is the number of rotation in unit time

frequency = 1 / time period = 1/5.8

frequency = 0.1724 Hz

b) the period

period is time taken in one rotation

period = total time / rotation = 5.8 / 1

Period = 5.8 sec

c) the speed

speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8

speed = 7.04 m/s

d) acceleration

To find the acceleration we take the linear velocity squared divided by the radius of the circle.

so

acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5

acceleration = 7.62 m/s²

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MArishka [77]

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4 0

3 years ago

Calcular la longitud del faldón de una Rampa de Acceso , que en planta tiene una longitud de 20 m y la pendiente es 27%.

seraphim [82]

La longitud del faldón de la rampa es de 5.4 m.

La pendiente expresada en porcentaje sigue la siguiente ecuación:

$m=\frac{y}{x}*100$ (1)

Donde:

y es la elevacion de la rampa (faldón)
x es la longitud de la ramapa (20 m)

Sabemos que la pendiente es de 27%. Por lo tanto, usando la ecuación 1, despejamos y.

$27=\frac{y}{20}*100$

$y=\frac{27*20}{100}$

$y=5.4\: m$

La longitud del faldón es 5.4 m

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5 0

3 years ago

Please help me!! I can't figure the answer out, AND haven't gotten past this (being the block test) for a couple weeks now!!

ololo11 [35]

The density of the nugget is $19 g/cm^3$ and is made of gold

Explanation:

The density of an object can be calculated as

$d=\frac{m}{V}$

where

d is the density

m is the mass

V is the volume of the object

We have to note that density of an object actually depends on the material the object is made of (therefore, two objects made of the same material can have different mass and different volume, but they have same density).

For the nugget in this problem, we have:

mass: m = 38 g

volume: $V=2 cm^3$

So, its density is

$d=\frac{38}{2}=19 g/cm^3$

And by looking at the table, we see that this value corresponds approximately to the density of gold, so the nugget is made of gold.

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5 0

3 years ago

A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely i

blsea [12.9K]

Answer:

7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

= 7066kg/m³

Explanation:

8 0

3 years ago

Read 2 more answers

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

3 years ago