Answer:
3 m/s squared
Explanation:
The formula you use is Vf= Vi + at. You rearrange it to a= Vf - Vi/t. The Vf is 27m/s. The Vi is 0m/s and the t is 9s. Cross out Vi since it’s zero and you’re left with a= 27m/s divided by 9s, which equals 3
To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.
In other words the acceleration can be described as

Where
G = Gravitational Universal Constant
M = Mass of Earth
r = Radius of Earth
This equation can be differentiated with respect to the radius of change, that is


At the same time since Newton's second law we know that:

Where,
m = mass
a =Acceleration
From the previous value given for acceleration we have to

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:




But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:


Therefore there is a weight loss of 0.3N every kilometer.
If F =m*a
and the question says how much force the s needed to accelerate a 68kg skier to a rate of 1.2ms^-2
Then F = 68*1.2
Answer: 4.speed
Explanation:
In this case, we know that the cart remains at a constant 20km/h.
Now, one could say that "the velocity remains constant, because it always is 20km/h"
But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.
But the module of the velocity, the speed, remains constant at 20km/h.
Then the correct option is 4, speed.
Answer:

Now when it will reach at point B then its normal force is just equal to ZERO


Explanation:
Since we need to cross both the loops so least speed at the bottom must be

also by energy conservation this is gained by initial potential energy


so we will have

now we have

here we have
R = 7.5 m
so we have


Now when it will reach at point B then its normal force is just equal to ZERO

now when it reach point C then the speed will be
![mgh - mg(2R_c) = \frac{1}{2]mv_c^2](https://tex.z-dn.net/?f=mgh%20-%20mg%282R_c%29%20%3D%20%5Cfrac%7B1%7D%7B2%5Dmv_c%5E2)


now normal force at point C is given as


