The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.
<h3>
Acceleration of the car </h3>
The acceleration of the car before stopping at the given distance is calculated as follows;
v² = u² + 2as
when the car stops, v = 0
0 = u² + 2as
0 = 15² + 2(76.5)a
0 = 225 + 153a
-a = 225/153
a = - 1.47 m/s²
<h3>Distance traveled when the speed is 32 m/s</h3>
If the same force is applied, then acceleration is constant.
v² = u² + 2as
0 = 32² + 2(-1.47)s
2.94s = 1024
s = 348.3 m
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Answer:
Ruko zara kuch Time dedo na please
Answer:
The coefficient of static friction between the puppy and the floor is 0.7273.
Explanation:
The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

Where
is the static friction force,
is the coefficient of static friction and
is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore,
, to make the puppy moving we need to use a force of 80 N, therefore,
, so we can solve for the coefficient as shown below:

The coefficient of static friction between the puppy and the floor is 0.7273.
A. THE ENERGY THAT IS RELEASED IN A NUCLEAR REACTION.
M =MASS
C^2= SPEED OF LIGHT SQUARED