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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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A motorcycle is stopped at a traffic light. When the light turns green, the motorcycle accelerates to a speed of 91 km/h over a

zimovet [89]

Given :

Initial speed , u = 0 m/s .

Final speed , v = 91 km/h = 25.28 m/s .

To Find :

a) Average acceleration .

b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .

Solution :

a )

We know ,by equation of motion :

$v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{25.28^2-0^2}{2\times 47}\ m/s^2\\\\a=6.8\ m/s^2$

b)

Also , by equation of motion :

$s=ut+\dfrac{at^2}{2}\\\\s=0+\dfrac{6.8\times (3.3)^2}{2}\ m\\\\s=37.02\ m$

Hence , this is the required solution .

6 0

3 years ago

A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and

sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0

3 years ago

A diagram of a closed circuit with a power source on the left labeled 120 V. There are 3 resistors in parallel, separate paths,

Zina [86]

1) The equivalent resistance is $2.73\Omega$

2) The voltage in the circuit is 120 V

3) The total current in the circuit is 44.0 A

Explanation:

1)

Resistors are said to be in parallel when they have the same potential difference at their terminals.

The formula to calculate the equivalent resistance for three resistors in parallel is:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

where $R_1, R_2, R_3$ are the resistances of the three resistors.

In this problem, the resistance of each resistor is:

$R_1 = 5.0 \Omega$

$R_2 = 10.0 \Omega$

$R_3 = 15.0 \Omega$

Substituting,

$\frac{1}{R_T}=\frac{1}{5}+\frac{1}{10}+\frac{1}{15}=\frac{11}{30}$

So the equivalent resistance is

$R_T = \frac{30}{11}=2.73 \Omega$

2)

In this problem we are told that the three resistors are connected in parallel. This means that their terminals are connected to the same point of the circuit: therefore, this means that they also have the same potential difference across them.

Therefore, the voltage in each branch containing each resistor is the same, and it is equal to the voltage of the battery, which is

$V=120 V$

So, the voltage in the circuit is 120 V.

3)

The total current in the circuit can be found by using Ohm's law:

$V=RI$

where

V is the voltage

R is the equivalent resistance of the circuit

I is the current

In this circuit, we have:

V = 120 V

$R=2.73\Omega$

Re-arranging the equation for I, we find the current:

$I=\frac{V}{R}=\frac{120}{2.73}=44.0 A$

Learn more about current and potential difference:

brainly.com/question/4438943

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3 years ago

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AURORKA [14]

I believe that is true.

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3 years ago

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A 10 kg bear is running to the right at 10 m/s. He crashes into another 10 kg bear running to the left at -5 m/s. They collide a

faust18 [17]

Answer:

<em>Total momentum is conserved</em>

Explanation:

<u>Conservation of Momentum </u>

The momentum is a physical magnitude that measures the product of the object's velocity by its mass. The total momentum of a system is the sum of all its components' individual momentums. The two-bear system starts with a total moment of

$p=m_1v_1+m_2v_2=(10)(10)+(10)(-50)=50\ kg.m/s$

When both bears stick together, the total mass is 20 kg, and the new momentum is

$p'=(20)(2.5)=50 \ kg.m/s$

We have assumed both bears move to the right after the collision. In this situation, the total momentum is conserved

7 0

3 years ago