Between which types of elements do ionic bonds occur, and how do electrons act within the bond?(1 point)

A rock is thrown downward from an unknown height above the ground with an initial speed of 6.1 m/s. It strikes the ground 1.7 s

insens350 [35]

Answer:

24.531 m

Explanation:

t = Time taken = 1.7 s

u = Initial velocity = 6.1 m/s

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=6.1\times 1.7+\dfrac{1}{2}\times 9.8\times 1.7^2\\\Rightarrow s=24.531\ m$

The initial height of the rock above the ground is 24.531 m

7 0

3 years ago

On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass

Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

$\theta=53^{\circ}$

$\mu_k=0.16$

Speed,v=1.5m/s

Net force in x direction must be zero

$F_{net}=0$

$Fsin\theta-f=0$

$Fsin\theta=f$

Net force in y direction

$N-mg-Fcos\theta=0$

$N=mg+Fcos\theta$

$f=\mu_kN=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_kmg+\mu_kFcos\theta$

$Fsin\theta-\mu_kFcos\theta=\mu_kmg$

$F(sin\theta-\mu_kcos\theta)=\mu_kmg$

$F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}$

Power,P=Fv

$P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v$

Where $g=9.8m/s^2$

B.Substitute the values

$P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5$

$P=40.17W$

6 0

4 years ago

The speed of light (c) is:

Radda [10]

Answer:

A

Explanation:

Speed of light is 299 792 458 m / s. So option A is answer.

7 0

2 years ago

Two +1 C charges are separated by 30000 m, what is the magnitude of the force?

Kipish [7]

Answer:

The magnitude of the force is 10 N

Explanation:

Coulomb's Law

The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

We have two identical charges of q1=q2=1 c separated by d=30000 m, thus the magnitude of the force is:

$\displaystyle F=9\cdot 10^9\frac{1*1}{30000^2}$

F = 10 N

The magnitude of the force is 10 N

7 0

3 years ago

1. A car travels along a straight road to the east for 100 meters in 4 seconds, then go the west for 50 meters in 1 second.

Alla [95]

Answer:

a.Distance = 150 m

b. Displacement = 50 m

Time lapsed = 5 seconds

Explanation:

a. Distance is the change in the position of an object.

The distance covered by the car = 100 + 50

= 150 m

b. Since displacement is a vector quantity,

Displacement of the car = 100 - 50

= 50 m due east

c. Time elapsed is the time taken for the motion of the car starting from when its starts to when it stops.

Thus, the time elapsed = 4 + 1

= 5 seconds

4 0

3 years ago