<span>The number of the group identifies the column of the standard periodic table in which the element appears.</span>
Group 1 contains the alkali metals ( lithium<span> (</span>Li<span>), </span>sodium<span> (</span>Na<span>), </span>potassium<span> (</span>K<span>), </span>rubidium<span> (</span>Rb<span>), </span>caesium<span> (</span>Cs<span>), and </span>francium(Fr).)<span>
Group 2 contains the alkaline earth metals (</span> beryllium<span> (</span>Be),magnesium<span> (</span>Mg<span>), </span>calcium<span> (</span>Ca<span>), </span>strontium<span> (</span>Sr<span>), </span>barium<span> (</span>Ba<span>) and </span>radium<span> (</span>Ra<span>) )
Group 3: </span><span> Scandium (Sc) and yttrium (Y) </span>
Complete question
A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.
Answer:
The car must travel 68.94 meters.
Explanation:
First, we are going to find the acceleration of the car using Newton's second Law:
(1)
with m the mass , a the acceleration and
the net force forces that is:
(2)
with F the force provided by traction and f the resistive force:
(2) on (1):

solving for a:

Now let's use the Galileo’s kinematic equation
(3)
With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and
the time took to achieve that velocity, solving (3) for
:


The level of greenhouse gases in our atmosphere would decrease, due to less automobiles.
Answer:
Speed at which the ball passes the window’s top = 10.89 m/s
Explanation:
Height of window = 3.3 m
Time took to cover window = 0.27 s
Initial velocity, u = 0m/s
We have equation of motion s = ut + 0.5at²
For the top of window (position A)

For the bottom of window (position B)


We also have

Solving

So after 1.11 seconds ball reaches at top of window,
We have equation of motion v = u + at

Speed at which the ball passes the window’s top = 10.89 m/s
Answer:
(A) L = 115.3kgm²/s
(B) dL/dt = 94.1kgm²/s²
Explanation:
The magnitude of the angular momentum of the rock is given by the foemula
L = mvrSinθ
We have been given θ = 36.9°, m = 2.0kg, v = 12.0m/s and r = 8.0m.
Therefore L = 2.00 × 12 × 8.0 × Sin 36.9° =
115.3 kgm²/s
(B) The magnitude of the rate of angular change in momentum is given by
dL /dt = d(mvrSinθ)/dt = mgrSinθ = 2.00 × 9.8 × 8.0× Sin36.9 = 94.1kgm²/s²