Answer:
after 6 second it will stop
he travel 36 m to stop
Explanation:
given data
speed = 12 m/s
distance = 100 m
decelerates rate = 2.00 m/s²
so acceleration a = - 2.00 m/s²
to find out
how long does it take to stop and how far does he travel
solution
we will apply here first equation of motion that is
v = u + at ......1
here u is speed 12 and v is 0 because we stop finally
put here all value in equation 1
0 = 12 + (-2) t
t = 6 s
so after 6 second it will stop
and
for distance we apply equation of motion
v²-u² = 2×a×s ..........2
here v is 0 u is 12 and a is -2 and find distance s
put all value in equation 2
0-12² = 2×(-2)×s
s = 36 m
so he travel 36 m to stop
Answer:
The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.
Explanation:
Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

Dividing the second equation by the first one, we obtain:

And, since
, then:

It means that the velocity at the bottom of the ramp is 1.81m/s.
We could use this data, plus any of the two initial equations, to determine the acceleration:

So the acceleration is 3.30m/s^2.
Answer:
649kg/m^3
Explanation:
Let p be the density of this particular object.
Formula for density:

We can substitute the givenmass and volume to find density of the object.

Therefore the density of this object is 649kg/m^3.
The new oscillation frequency of the pendulum clock is 1.14 rad/s.
The given parameters;
- <em>Mass of the pendulum, = M </em>
- <em>Length of the pendulum, = L</em>
- <em>Initial angular speed, </em>
<em> = 1 rad/s</em>
The moment of inertia of the rod about the end is given as;

The moment of inertia of the rod between the middle and the end is calculated as;
![I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}](https://tex.z-dn.net/?f=I_f%20%3D%20%5Cint%5Climits%5EL_%7BL%2F2%7D%20%7Br%5E2%5Cfrac%7BM%7D%7BL%7D%20%7D%20%5C%2C%20dr%20%3D%20%5Cfrac%7BM%7D%7B3L%7D%20%5Br%5E3%5D%5EL_%7BL%2F2%7D%20%3D%20%20%5Cfrac%7BM%7D%7B3L%7D%20%5BL%5E3%20-%20%5Cfrac%7BL%5E3%7D%7B8%7D%20%5D%20%3D%20%5Cfrac%7BM%7D%7B3L%7D%20%5B%5Cfrac%7B7L%5E3%7D%7B8%7D%20%5D%3D%20%5Cfrac%7B7ML%5E2%7D%7B24%7D)
Apply the principle of conservation of angular momentum as shown below;

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.
Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129
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