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7nadin3 [17]
3 years ago
8

Ashley decides to enter her pet turtle in a race. She knows her turtle can travel at a rate of 2 meters per hour. The race track

measures 5 meters from start to finish. How long will it take Ashley’s turtle to finish the race?
A) 3 hours
B) 10 hours
C) .4 hours
D) 2.5 hours
Thanks for the help! And if it’s not a bother could you please explain how you got the answer? However it’s not necessary if you don’t want to.
Physics
1 answer:
Genrish500 [490]3 years ago
7 0
It would probably be B
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3 years ago
A skater rotates with her arms crossed at an angular speed of 8.0 rad/s and she has a moment of inertia of 100 kg/m2. She extend
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Answer:

When her hands extends, her momen of inertia is 4.28\ kg-m^2.

Explanation:

Given that,

Initial angular speed, \omega_i=8\ rad/s

Initial moment of inertia, I_1=100\ kg-m^2

Final angular speed, \omega_f=7\ rad/s

Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

I_1\omega_1=I_2\omega_2

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I_2=\dfrac{I_1\omega_1}{\omega_2}\\\\I_2=\dfrac{100\times 8}{7}\\\\I_2=114.28\ kg-m^2

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3 years ago
3. An astronaut on a space-walk swings a 0.5 kg metal tool in a circular path at the end of a cable 3.6 m long. The astronaut th
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3 years ago
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
xz_007 [3.2K]

Answer:

a) \Delta{t} = 5.39s

b) the motorcycle travels 155 m

Explanation:

Let t_2-t_1 = \Delta{t}, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}

where:

v_{m2} is the speed of the motorcycle at time 2

v_{c} is the velocity of the car (constant)

v_{0} is the velocity of the car and the motorcycle at time 1

d is the distance between the car and the motorcycle at time 1

x is the distance traveled by the car between time 1 and time 2

Solving the system of equations:

\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]

v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933

3 0
3 years ago
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