Answer:
The automobile's acceleration in that time interval is -2 m/s^2
Explanation:
The acceleration is defined as the rate of change of the velocity.
The average acceleration in a given lapse of time is calculated as:
A = (final velocity - initial velocity)/time.
In this case, we have:
initial velocity = 31 m/s
final velocity = 15 m/s
time = 8 seconds.
Then the average acceleration is:
A = (15m/s - 31m/s)/8s = -2 m/s^2
Answer:
The body is said to be in static equilibrium if the net force acting on a body at rest is zero.As the net force is zero,the body will not undergo motion.
Explanation:
We have that valence electrons poses the three characteristics stated, as
Group 14 (carbon group) are identified by 4 valence electrons.
Valence electrons of atoms are used to form bonds.
Group 14 (carbon group) are identified by 4 valence electrons.
Option A,B,C
<h3>
Properties of Valence electrons</h3>
All elements in the same group or family have the same number of valence electrons: Yes, this is true as Group 14 (carbon group) are identified by 4 valence electrons.
Valence electrons are the only subatomic particles involved in forming bonds: Yes, Valence electrons of atoms are used to form bonds.
Carbon has 4 valence electrons because it is found in group 14:
True, Group 14 (carbon group) are identified by 4 valence electrons.
For more information on atoms visit
brainly.com/question/13981855
Answer:
mu = 0.56
Explanation:
The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v: final speed = 0m/s (the car stops)
v_o: initial speed in the interval of interest = 60km/h
= 60(1000m)/(3600s) = 16.66m/s
x: distance = 25m
BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

Furthermore, you use the relation between the friction force and the friction coefficient:

hence, the friction coefficient is 0.56