Can you help me??? This is too hard.

Which statement best describes metallic bonding

horrorfan [7]

Answer:

It's a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons and positively charged metal bars. It can also be described as the sharing of free electrons among a structure of positively charged ions

4 0

3 years ago

An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.38 km. If the airplane rounds half

Andrej [43]

Answer:

a) 6076 m

b) 43.33 m/s

c) 68 m/s

Explanation:

(a) If the airplane rounds half the circle in 156s, its displacement is the circle diameter in 156s, or twice the circle's radius

s = 2r = 2* 3.38km = 6.76 km or 6760 m

(b) The average velocity would be displacement over unit of time

v = s/t = 6760 / 156 = 43.33 m/s

(c) The length of the chord it's swept in 156s is half of the circle perimeter

c = πr = π3.38 = 10.62 km or 10620 m

The airplane average speed is its chord length over a unit of time

c / t = c / 156 = 68 m/s

4 0

3 years ago

Which of the following is a derived unit? ampere newton second kilogram

Soloha48 [4]

Newton is your answer.

ampere, second, kilograms, are all base units.

hope this helps

5 0

3 years ago

An engine flywheel initially rotates counterclockwise at 6.77 rotations/s. Then, during 23.9 s, its rotation rate changes to 3.5

olganol [36]

Answer:

-2.70 rad/s²

Explanation:

Given that

ω1 = initial angular velocity of the flywheel, which is 6.77 rev/s

If we convert it to rad/s, we have

(6.77 x 2π) rad/s = 13.54π rad/s

ω2 = final angular velocity of the flywheel = -3.51 rev/s,

On converting to rad/s also, we have

(-3.51 x 2π) rad/s = 7.02π rad/s

α = average angular acceleration of the flywheel = ?

Δt = elapsed time = 23.9 s

Now, using the formula, α = (ω2 - ω1)/Δt. On substituting, we have

α = (-7.02π rad/s - 13.54π rad/s)/23.9 s

α = -20.56π rad/s / 23.9 s

α = -64.59 rad/s / 23.9 s

α = -2.70 rad/s²

Therefore, the average angular acceleration of the flywheel is -2.70 rad/s²

7 0

3 years ago

A force of 40N is applied to a 28 g mass, what is the acceleration? (round to the hundredths place)

Leno4ka [110]

Answer:

1428.6m/s²

Explanation:

Given parameters:

Force applied on the body = 40N

Mass of the body = 28g

1000g = 1kg

28g will therefore be 0.028kg

Unknown:

Acceleration = ?

Solution:

To solve this problem, we use the expression derived from Newton's second law of motion.

Force = mass x acceleration

Insert the parameters and solve;

40 = 0.028 x acceleration

Acceleration = $\frac{40}{0.028}$ = 1428.6m/s²

7 0

3 years ago