Can u help me please

Physics

1 answer:

DanielleElmas [232]2 years ago

5 0

Answer:

Find answer below.

Explanation:

a. Group 2: alkaline earth metals. Some examples of alkaline earth metals are magnesium, calcium, barium, etc.

b. Transition metals: metals with unpredictable properties.

c. Chlorine: halogen.

d. Metalloid: make good semiconductors.

e. Group 1: alkali metals. These includes the soft, light and reactive chemical elements found in group 1 of the periodic table. Some examples are hydrogen, potassium, lithium, sodium, etc.

f. Noble gases: has a full outer energy level (shell). These are chemical elements with eight valence electrons and as such have a full octet. Some examples are argon, neon, etc.

h. Metal: loses electron in bonding. They usually lose electrons in order to attain a full outer energy level or noble gas electronic configuration.

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In the vertical jump, an athlete starts from crouch and jumps upward to reach as high as possible. Even the best athletes spend

Mumz [18]

Answer:

The ratio of the time he is above ymax /2 to the time it takes him to go from the floor to that height = 0.707

Explanation:

The mathematical derivation and steps is as shown in the attached file.

3 0

2 years ago

A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police

Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. $x=x_{0}+vt$

2. $x=x_{0}+v_{0}t+\frac{at^{2}}{2}$

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

$vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}$

We simplify the fraction present and rearrange for our formula so that it equals 0:

$0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0$

In the very last step we factored a common factor t. There is two possible solutions to the equation at $t=0$ and:

$0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s$

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at $t=0 s$ (when the SUV passed the police car) and $t=15.56s$ (when the police car catches up to the SUV)