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Novay_Z [31]
3 years ago
12

BB8 is an intergalactic robot (essentially a soccer ball with a half meter radius that can control how fast it rolls around) tha

t is trying out for a new role in the rebellion. One of the tests is how fast they can travel 100 m. They must start from rest and end at rest. BB8 has a maximum angular acceleration rate of 2 rad/s2, both when speeding up and slowing down. What is the minimum time BB8 can run the trial?
Physics
1 answer:
hram777 [196]3 years ago
6 0

Answer:

20 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

r = Radius of robot = 0.5 m

\alpha = Angular acceleration = 2 rad/s²

Linear acceleration is given by

a=r\alpha\\\Rightarrow a=0.5\times 2\\\Rightarrow a=1\ m/s^2

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 1\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{1}}\\\Rightarrow t=10\ s

While speeding up time taken is 10 seconds

v=u+at\\\Rightarrow v=0+1\times 10\\\Rightarrow v=10\ m/s

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=10t-\frac{1}{2}\times 1\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{1}}\\\Rightarrow 50=10t-0.5t^2\\\Rightarrow -5t^2+100t-500=0

t=\dfrac{-100\pm \sqrt{100^2-4\left(-5\right)\left(-500\right)}}{2\left(-5\right)}\\\Rightarrow t=10\ s

The time taken to slow down is 10 seconds

Total time the trial is rum is 10+10 = 20 seconds

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A jetliner rolls down the runway with constant acceleration from rest, it reaches its take off speed of 250 km/h in 1 min. What
nlexa [21]

Answer:

Acceleration, a=14970.05\ km/h^2

Explanation:

Given that,

Initially, the jetliner is at rest, u = 0

Final speed of the jetliner, v = 250 km/h

Time taken, t = 1 min = 0.0167 h

We need to find the acceleration of the jetliner. The mathematical expression for the acceleration is given by :

a=\dfrac{v-u}{t}

a=\dfrac{250}{0.0167}

a=14970.05\ km/h^2

So, the acceleration of the jetliner is 14970.05\ km/h^2. Hence, this is the required solution.

3 0
3 years ago
A robot arm that controls the position of a video camera in an automated surveillance system is manipulated by a servo motor tha
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Answer:

W = 270.9 J

Explanation:

given,  

F(x) = (12.9 N/m²) x²  

work = Force x displacement  

dW = F  .dx  

the push-rod moves from x₁= 1 m to x₂ = 4 m

integrating the above  

\int dW = \int_{x_1}^{x_2}F. dx

W = \int_{x_1}^{x_2}F. dx

W = \int_{1}^{4} (12.9 x^2) dx

W = 12.9\times [\dfrac{x^3}{3}]_1^4 dx

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W = 270.9 J

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3 years ago
What is the mass of 2.47 moles of<br> carbon dioxide CO2 ?
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Answer:

CO2 molecule is made up of Carbon (atomic mass =12) and oxygen (atomic mass=16).

So first finding the mass of 1 molecule of CO2 which is equals to

= mass of 1 carbon atom + masses of 2 oxygen atom, we get

= 12+(16*2)= 12+32= 44 a.m.u.

Now 1 molecule of CO2 has mass 44 amu so mass of 1 mole CO2 will be 44 grams.( 1 a.m.u.=1.6729*10^-33 grams. 1 mole = 6.022*10^23, so 44 a.m.u.=73.6076*10^-33 grams approx. For one mole CO2, 73.6076*10^-33*6.022*10^23 which is approximately equals to 44 grams. )

1 mole CO2= 44grams, so 2.5 moles = 44*2.5= 110 grams

So our answer is 110 grams

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