Ask question

Ask question

All categories

3 years ago

10

Which of these period three elements from the periodic table would you predict to be the MOST metallic and why? A) Al B) Si C) P

D) S E) Cl F) Ar

1 answer:

Sauron [17]3 years ago

6 0

Sorry I don’t even know

You might be interested in

I am pretty sure its the last one.

CaHeK987 [17]

The answer is has no moons. Mars has two moons

4 0

3 years ago

Read 2 more answers

What is the momentum of an 80 kg runner moving at the speed of 2.5 m/s? Use the

Vladimir79 [104]

Answer:

200 kgm/s

Explanation:

momentum = mass x velocity

8 0

3 years ago

VMariaS [17]

Answer:

a.proton, proton

hope this helped :) have a goodday

6 0

4 years ago

A jet liner must reach a speed of 82 m/s for takeoff. If the

SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial ( $v_{o}$ ) and final speeds ( $v_{f}$ ), measured in meters per second, of the aircraft and likewise the runway length ( $d$ ), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed ( $a$ ), measured in meters per square second:

$a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 82\,\frac{m}{s}$ and $d = 1500\,m$ , then the acceleration that the jet must have is:

$a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}$

$a = 2.241\,\frac{m}{s^{2}}$

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0

3 years ago

A person travels by car from one city to another with differem constan1 speeds between pairs of cities. She drives for 30.0 min

Montano1993 [528]

Answer:

a.52.9 km/h

b.90 km

Explanation:

We are given that

$v_1=89km/h$

$t_1=30min$

$v_2=100km/h$

$t_2=12min$

$v_3=40km/h$

$t_3=45 min$

Time spend on eating lunch and buying ga=15 min.

a.Total time=30+12+45+15=102 minute= $\frac{102}{60}=1.7 hour$

1 hour=60 minutes

Distance= $speed\times time$

$d_1=v_1\times t_1=80\times\frac{30}{60}=40km$

$d_2=100\times \frac{12}{60}=20 km$

$d_3=40\times \frac{45}{60}=30 km$

Total distance= $d_1+d_2+d_3=40+20+30=90km$

Average speed= $\frac{total\;speed}{total\;time}$

Using the formula

Average speed= $\frac{90}{1.7}=52.9Km/h$

b.Total distance between the initial and final city lies along the route=90 km

4 0

3 years ago