Answer:
Time needed: 2.5 s
Distance covered: 31.3 m
Explanation:
I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by
v2f=v2i−2⋅a⋅d
Isolate d on one side of the equation and solve by plugging your values
d=v2i−v2f2a
d=(15.02−10.02)m2s−22⋅2.0ms−2
d=31.3 m
To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation
vf=vi−a⋅t, which will get you
t=vi−vfa
t=(15.0−10.0)ms2.0ms2=2.5 s
A is the answer I really don’t know the answer to that but if u can u can help me on my work
The acceleration of gravity is inversely proportional to
the square of the distance from Earth's center.
The acceleration of gravity is 9.8 m/s² on the Earth's surface ...
6380 km from the center.
If the acceleration of gravity at 'h' is 4.9 m/s² ... 1/2 of what it is
on the surface, then the distance from the center is
(6380 x √2) = 9,023 km (rounded) ,
and 'h' is the distance above the surface
= (9,023 - 6,380) = 2,643 km (rounded) .
Pitch is related to frequency