Answer:
Acceleration of the train is -3 m/s²
Explanation:
Since it stops,
final velocity (v) = 0 m/s
time (t) = 25 seconds
initial velocity(u) = 75 m/s
Now,
accelereation = (v-u)/t
or, a = (0-75)/25
or, a = -75/25
so, a = -3 m/s²
Answer:
- magnitude : 1635.43 m
- Angle: 130°28'20'' north of east
Explanation:
First, we will find the Cartesian Representation of the
and
vectors. We can do this, using the formula
![\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%3D%20%7C%20%5Cvec%7BA%7D%20%7C%20%5C%20%28%20%5C%20cos%28%5Ctheta%29%20%5C%20%2C%20%5C%20sin%20%28%5Ctheta%29%20%5C%20%29)
where
its the magnitude of the vector and θ the angle. For
we have:
![\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )](https://tex.z-dn.net/?f=%5Cvec%7BX%7D%3D%201430%20m%20%5C%20%28%20%5C%20cos%28%2042%20%5C%C2%B0%29%20%5C%20%2C%20%5C%20sin%20%2842%20%5C%C2%B0%29%20%5C%20%29)
![\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )](https://tex.z-dn.net/?f=%5Cvec%7BX%7D%3D%20%28%20%5C%201062.70%20m%20%5C%20%2C%20%5C%20956.86%20m%20%5C%20%29)
where the unit vector
points east, and
points north. Now, the
will be:
![\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )](https://tex.z-dn.net/?f=%5Cvec%7BY%7D%3D%20-%202200%20m%20%5Chat%7Bj%7D%20%3D%20%28%20%5C%200%20%5C%20%2C%20%5C%20-%202200%20m%20%5C%20%29)
Now, taking the sum:
![\vec{X} + \vec{Y} + \vec{Z} = 0](https://tex.z-dn.net/?f=%5Cvec%7BX%7D%20%2B%20%5Cvec%7BY%7D%20%2B%20%5Cvec%7BZ%7D%20%3D%200)
This is
![\vec{Z} = - \vec{X} - \vec{Y}](https://tex.z-dn.net/?f=%5Cvec%7BZ%7D%20%3D%20-%20%5Cvec%7BX%7D%20-%20%5Cvec%7BY%7D)
![(Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )](https://tex.z-dn.net/?f=%28Z_x%20%2C%20Z_y%29%20%3D%20-%20%28%20%5C%201062.70%20m%20%5C%20%2C%20%5C%20956.86%20m%20%5C%20%29%20-%20%28%20%5C%200%20%5C%20%2C%20%5C%20-%202200%20m%20%5C%20%29)
![(Z_x , Z_y) = ( \ - 1062.70 m \ , \ 2200 m \ - \ 956.86 m \ )](https://tex.z-dn.net/?f=%28Z_x%20%2C%20Z_y%29%20%3D%20%28%20%5C%20-%201062.70%20m%20%5C%20%2C%20%20%5C%202200%20m%20%5C%20-%20%5C%20956.86%20m%20%5C%20%29)
![(Z_x , Z_y) = ( \ - 1062.70 m \ , \ 1243.14 m\ )](https://tex.z-dn.net/?f=%28Z_x%20%2C%20Z_y%29%20%3D%20%28%20%5C%20-%201062.70%20m%20%5C%20%2C%20%20%5C%201243.14%20m%5C%20%29)
Now, for the magnitude, we just have to take its length:
![|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}](https://tex.z-dn.net/?f=%7C%5Cvec%7BZ%7D%7C%20%3D%20%5Csqrt%7BZ_x%5E2%20%2B%20Z_y%5E2%7D)
![|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}](https://tex.z-dn.net/?f=%7C%5Cvec%7BZ%7D%7C%20%3D%20%5Csqrt%7B%28-%201062.70%20m%29%5E2%20%2B%20%281243.14%20m%29%5E2%7D)
![|\vec{Z}| = 1635.43 m](https://tex.z-dn.net/?f=%7C%5Cvec%7BZ%7D%7C%20%3D%201635.43%20m)
For its angle, as the vector lays in the second quadrant, we can use:
![\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m})](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20180%5C%C2%B0%20-%20arctan%28%5Cfrac%7B1243.14%20m%7D%7B%20-%201062.70%20m%7D%29%20%20)
![\theta = 180\° - arctan( -1.1720)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20180%5C%C2%B0%20-%20arctan%28%20-1.1720%29%20%20)
![\theta = 180\° - 45\°31'40''](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20180%5C%C2%B0%20-%2045%5C%C2%B031%2740%27%27%20)
![\theta = 130\°28'20''](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20130%5C%C2%B028%2720%27%27%20%20)
Answer:
No answer available
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Answer:
a force that attracts matter to the earth
Explanation:
depends on where you are the gravity can be different in space there is no gravity on Earth there is , that's why when you jump you come back down