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S_A_V [24]
3 years ago
15

What do the group numbers mean on the periodic table?

Physics
1 answer:
Ad libitum [116K]3 years ago
8 0

Many periodic tables list numbers for element groups, which are columns of the periodic table. The elements in a group share the same number of valence electrons.

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2. Compare and Contrast A fault cuts through
Elenna [48]

Explanation:

The principle of cross-cutting relationships states that a fault or intrusion is younger than the rocks that it cuts through. The fault labeled "E" cuts through all three sedimentary rock layers (A, B,and C) and also cuts through the intrusion (D). So the fault must be the youngest formation that is seen and known of.

6 0
3 years ago
Light-rail passenger trains that provide transportation within and between cities are capable of modest accelerations. The magni
Mkey [24]
During the first phase of acceleration we have:
v o = 4 m/s;  t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?;  v o = 13 m/s; a = 1.125 m/s² ;  t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s 
3 0
3 years ago
An object is moving in a straight line at a constant speed. A resultant force begins to act upon the object. State the ways in w
sergiy2304 [10]

Answer:

accelerate / increase speed OR decelerate / decrease speed OR stop B1

change direction / move in a curve o.w.t.t.e

Explanation:

accelerate / increase speed OR decelerate / decrease speed OR stop B1

change direction / move in a curve o.w.t.t.e

8 0
3 years ago
A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff
postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

F_c=\frac{mv^2}{R}

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

N=mg

Therefore, frictional force, f=\mu mg

Now, frictional force = centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0
3 years ago
A moving small car has a head-on collision with a large stationary truck 7.3 times the mass of the car. Which statement is true
ad-work [718]
The car bounces off and moves in the opposite direction
8 0
3 years ago
Read 2 more answers
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