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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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Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth,
jeka57 [31]

Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = -\frac{GMm}{R}

Therefore: W = Energy at earth surface - \frac{GMm}{R}

Energy at earth surface (E) at an altitude of 5R = -\frac{GMm}{5r} +\frac{1}{2}mV^2

But V=\sqrt{\frac{GM}{5R} }

Therefore: E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2=  -\frac{GMm}{5R}+\frac{GMm}{10R}  = -\frac{GMm}{10R}

W = E - PE

W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}

7 0
3 years ago
The Andromeda galaxy is the closest major galaxy to our own. Andromeda shows a distinct blue-shift of light when we analyze it.
saul85 [17]
At the present time, the only way we know of that light can get shifted
toward the blue end of the spectrum is the Doppler effect ... wavelengths
appear shorter than they should be when the source is moving toward us. 

IF that's true in the case of the Andromeda galaxy, it means the galaxy is
moving toward us.

We use the same reasoning to conclude that all the galaxies whose light is red-shifted are moving away from us.  That includes the vast majority of all galaxies that we can see, and it strongly supports the theory of the big bang
and the expanding universe.

If somebody ever comes along and discovers a DIFFERENT way that light
can get shifted to new, longer or shorter wavelengths, then pretty much all
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4 0
3 years ago
Rock formed from harden magma that rises to the surface is called ​
salantis [7]

Answer:

Igneous rocks

Explanation:

3 0
3 years ago
A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn
kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>

<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>

<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite (\vec{v}_{S}), measured in meters per second, is:

\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}

Where:

v_{o,x} - Initial velocity in +x direction, measured in meters per second.

a_{x} - Acceleration in +x direction, measured in meter per square second.

t - Time, measured in seconds.

v_{y} - Velocity in +y direction, measured in meters per second.

If we know that v_{o,x} = 0\,\frac{m}{s}, a_{x} = 0.250\,\frac{m}{s^{2}}, t = 45\,s and v_{y} = 21.4\,\frac{m}{s}, the final velocity of the satellite is:

\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}

\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}

\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }

\tan \alpha = 1.902

\alpha = \tan^{-1}1.902

\alpha \approx 62.266^{\circ}

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

4 0
3 years ago
A stone sphere of radius 7.00 m rests in a flat field. Relative to the ground, what is the gravitational potential energy of a 9
Molodets [167]

Answer:

(B)  1.23 x 10⁴ J

Explanation:

Given;

radius of the sphere, r = 7.0 m

diameter of the sphere, d = 2r = 14.0 m

mass of the person sitting on the sphere, m = 90.0 kg

The gravitational potential energy of the person is given by;

P.E = mgh

where;

g is acceleration due to gravity = 9.8 m/s²

h is the height above the ground level = d = 14.0 m

P.E = mgh

P.E = (90)(9.8)(14)

P.E = 12348 J

P.E = 1.2348 x 10⁴ J

Therefore, the gravitational potential energy of the person is 1.2348 x 10⁴ J

8 0
2 years ago
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