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iragen [17]
3 years ago
14

Noble gases, such as argon and neon, are known for being extremely non-reactive. Neon and argon are non-reactive because they A.

have six electrons in their outer shell. B. have eight electrons in their outer shell. C. have ten electrons in their outer shell. D. have 18 electrons in their outer shell.
Physics
2 answers:
Nata [24]3 years ago
4 0

Answer: B. Have eight electrons in their outer shell.

Explanation:

Noble gases are known for their unreactive nature due to the presence of complete outermost electron. They possess eight electrons in their outermost shell (octet rule). Being unreactive means that they cannot share any of their outermost electron to any other atoms since they are completely filled up (stable).

Taking sodium(2 8 1) which is reactive as an example, its reactivity is due to the presence of a lone pair of electron in its outermost shell which it can easily share to other atom with incomplete outermost electron like chlorine (2 8 7).

Greeley [361]3 years ago
3 0
B. have eight electrons in their outer shell
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4 years ago
A car traveling at 28 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?
Maru [420]

Acceleration = (change in speed) / (time for the change)

change in speed = (speed at the end) minus (speed at the beginning)

change in speed = (zero) minus (28 m/s) = -28 m/s

Acceleration = (-28 m/s) / (13 sec)

Acceleration = -2.15 m/s²

5 0
3 years ago
A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1
Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain d=2.46\times10^{3}\ m

Height of islandh=1.80\times10^{3}\ m

Distance of the enemy ship from the mountain d'=6.10\times10^{2}\ m

Initial velocity v=2.55\times10^{2}\ m/s

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

v_{x}=v\cos\theta

Put the value into the formula

v_{x}=2.55\times10^{2}\cos74.9

v_{x}=66.42\ m/s

We need to calculate the vertical component of initial velocity

Using formula of vertical component

v_{y}=v\sin\theta

Put the value into the formula

v_{y}=2.55\times10^{2}\sin74.9

v_{y}=246.19\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{v_{x}}

t=\dfrac{2.46\times10^{3}}{66.42}

t=37.03\ sec

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

H= v_{y}t-\dfrac{1}{2}gt^2

Put the value in the equation

H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2

H=2397.4\ m

We need to calculate the distance close to the peak

Using formula of distance

H'=H-h

Put the value into the formula

H'=2397.4-1800

H'=597.4\ m

Hence, The distance close to the peak is 597.4 m.

6 0
3 years ago
If I wanted to generate a maximum emf of 20 V, what angular velocity (radians/sec, aka Hz) would be required given a circular co
sergij07 [2.7K]

Answer:

Angular velocity, \omega=35.36\ rad/s

Explanation:

It is given that,

Maximum emf generated in the coil, \epsilon=20\ V

Diameter of the coil, d = 40 cm

Radius of the coil, r = 20 cm = 0.2 m

Number of turns in the coil, N = 500

Magnetic field in the coil, B=9\times 10^{-3}\ T

The angle between the area vector and the magnet field vector varies from 0 to 2 π radians. The formula for the maximum emf generated in the coil is given by :

\epsilon=NBA\omega

\omega=\dfrac{\epsilon}{NBA}

\omega=\dfrac{20\ V}{500\times 9\times 10^{-3}\ T\times \pi (0.2\ m)^2}

\omega=35.36\ rad/s

So, the angular velocity of the circular coil is 35.36 rad/s. Hence, this is the required solution.

6 0
3 years ago
A 2500 N force accelerates a car at a rate of 3.0 m/s^2. What is the car’s mass? 250 kg
Ronch [10]

Apply Newton's second law to the car's motion:

F = ma

F = net force, m = mass, a = acceleration

Given values:

F = 2500N, a = 3.0m/s²

Plug in and solve for m:

2500 = m(3.0)

m = 830kg

5 0
4 years ago
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