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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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ANSWER

$3.30\text{ m/s}$

EXPLANATION

Parameters given:

Mass of car, mc = 1103 kg

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Inital velocity of truck = 0 m/s

To solve this problem, we have to apply the law of conservation of momentum, which states that the total momentum of a system is constant.

This implies that:

$m_cu_c+m_tu_t=m_cv_c+m_tv_t$

Since the car and the truck stick together after the collision, they will have the same final velocity.

Hence:

$m_cu_c+m_tu_t=(m_c+m_t)v_{}_{}$

Substitute the given values and solve for v (final velocity):

$\begin{gathered} (1103\cdot18)+(4919\cdot0)=(1103+4919)v \\ \Rightarrow19854=6022v \\ \Rightarrow v=\frac{19854}{6022} \\ v=3.30\text{ m/s} \end{gathered}$

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1 year ago

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3 years ago

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An object moving with a speed of 5m/s comes to rest in 10s after the brakes are applied . What is the initial velocity?

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Initial velocity is 5m/s.

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2 years ago

A crane lifts a 1,750 kg mass using a steel cable whose mass per unit length is 0.88 kg/m. What is the speed of transverse waves

Sauron [17]

Answer:

139.6m/s

Explanation:

Calculate the tension first, T=m*g

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3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

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3 years ago