Answer:
<h2> v= 21.47m/s </h2>
Explanation:
For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road
we will use the expression relating centripetal force and static friction below
let U represent the coefficient of static friction
Given that
U= 0.50
mass m= 1200-kg
radius r= 94.0 m
Assuming g= 9.81 m/s^2


substituting our given data in to expression we can solve for the speed V

making v the subject of formula we have

v= 21.47m/s
<em><u>hence the maximum velocity of the car is 21.47m/s</u></em>
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed

, and an accelerated motion on the y-axis, with initial speed

and acceleration

:


where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).
To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring

Therefore:

which has two solutions:

is the time of the beginning of the motion,

is the time at which the projectile hits the ground.
Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
Omnivores e<span>at both plants and meats. </span>
Yes
Explanation:
It is a reasonable result obtained.
Error = true value - measured value
true value = 24.5
measured value = 24.2
Error = 24.5 - 24.2 = 0.3g
The error reported in the reading is 0.3g
The reason why we had a disparity in the figures obtained from this measurement is primarily due to some erroneous scale.
The mixture at the end of the day is a solution.
We are expected to have the same mass but due to experimental or some form of random error introduced, we noticed a difference.
The value obtained is quite logical as we only had a deviation of 0.3g.
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Answer:
the can's kinetic energy is 0.42 J
Explanation:
given information:
Mass, m = 460 g = 0.46 kg
diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m
velocity, v = 1.1 m/s
the kinetic energy of the can is the total of kinetic energy of the translation and rotational.
KE =
I ω^2 + 
where
I =
and ω = 
thus,
KE =
(
)^2 + 
=
+ 
=
+ 
= 
=
= 0.42 J