Elements<span> in the same </span>group<span> in the periodic table </span>have similar chemical properties<span>. This is because their atoms </span>have<span> the same number of electrons in the highest occupied energy level. </span>Group<span> 1 </span>elements<span> are reactive metals called the alkali metals.</span>Group<span> 0 </span>elements<span> are unreactive non-metals called the noble gases.
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Answer:
Vertical distance= 3.3803ft
Explanation:
First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:
Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h
Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h
time= 0.00012731h × (3600s/h)= 0.458316s
With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:
Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m
Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft
This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.
Answer:
d= 794.4 cmExplanation:
Given that
Speed ,V= 286 km/h

V=79.44 m/s
Given that time ,t= 100 ms
t= 0.1 s
We know that ( if acceleration is zero)
Distance = Speed x time
d= V t
Now by putting the values in the above equation
d = 79.44 x 0.1 m
d= 7.944 m
We know that 1 m = 100 cm
d= 794.4 cm
Answer:
160N
Explanation:
Moments must be conserved - so.


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