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2 years ago

Which of the following electron configurations for neutral atoms is correct?

Physics

2 answers:

Lynna [10]2 years ago

8 0

Answer:

nitrogen: 1s2, 2s2, 2p3 is correct answer.

Explanation: I got this answer correct on my quiz!!!

liq [111]2 years ago

6 0

Nitrogen: 1s2, 2s2, 2p3 is correct answer.

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A force of 50n is applied to an object at an angle of 50 degrees measured relative to the horizontal. Determine the horizontal a

kobusy [5.1K]

Answer:

Fx = 32.14 [N]

Fy = 38.3 [N]

Explanation:

To solve this problem we must decompose the force vector, for this we will use the angle of 50 degrees measured from the horizontal component.

F = 50 [N]

Fx = 50*cos(50) = 32.14 [N]

Fy = 50*sin(50) = 38.3 [N]

We can verify this result using the Pythagorean theorem.

$F = \sqrt{(32.14)^{2}+ (38.3)^{2}} \\F = 50 [N]$

3 0

2 years ago

Big Bob is on his Harley and moving at 14.0m/s. He then accelerates to a velocity of 25m/s over a distance of 0.250 km. What is

Xelga [282]

As we know that it has all given data given as

$v_i = 14 m/s$

$v_f = 25 m/s$

distance moved = 0.250 km = 250 m

now we can use kinematics to find acceleration

$v_f^2 - v_i^2 = 2 a d$

$25^2 - 14^2 = 2 * a * 250$

$a = 0.86 m/s^2$

so it will accelerate at rate of 0.86 m/s^2

8 0

3 years ago

A sample of gas with a volume of 750 mL exerts a pressure of 98 kPa at 30 °C.What pressure will the sample exert when it is comp

Flura [38]

Answer:

240 kPa

Explanation:

The ideal gas law states:

$pV=nRT$

where

p is the gas pressure

V is the gas volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

For a fixed amount of gas, n and R are constant, so we can rewrite the equation as

$\frac{pV}{T}=const.$

For the gas in the problem, which undergoes a transformation, this can be rewritten as

$\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$

where we have:

$p_1 = 98 kPa=9.8\cdot 10^4 Pa$ is the initial pressure

$V_1 = 750 mL=0.75 L=0.75\cdot 10^{-3} m^3$ is the initial volume

$T_1 =30^{\circ}C =303 K$ is the initial temperature

$p_2$ is the final pressure

$V_2=250 mL=0.25 L=0.25\cdot 10^{-3} m^3$ is the final volume

$T_2=-25^{\circ}C=248 K$ is the final temperature

Solving the formula for p2, we find the final pressure of the gas:

$p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(0.75\cdot 10^{-3}m^3)(248 K)}{(303 K)(0.25\cdot 10^{-3} m^3)}=2.4\cdot 10^5 Pa = 240 kPa$

3 0

2 years ago

15 points.

Trava [24]

Answer:

its 0.5 for all i beleive

Explanation:

8 0

2 years ago

An engine flywheel initially rotates counterclockwise at 5.03 rotations/s. Then, during 23.5 s, its rotation rate changes to 2.6

Sveta_85 [38]

Answer:

the flywheel's average angular acceleration is -2.05 rad/s²

Explanation:

<u>Note:</u> counterclockwise is positive

clockwise is negative

Given;

initial angular velocity, $\omega _i$ = 5.03 rev/s = $5.03\frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 31.61 \ rad/s$

final angular velocity, $\omega_f$ = -2.63 rev/s = $-2.63 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = -16.53 \ rad/s$

duration of the flywheel rotation, Δt = 23.5 s

The average acceleration of the flywheel is calculated as;

$a_r = \frac{\Delta \omega}{\Delta t} = \frac{\omega_f - \omega _i}{t_2-t_1} = \frac{-16.53 \ - \ 31.61}{23.5} = -2.05 \ rad/s^2$

Therefore, the flywheel's average angular acceleration is -2.05 rad/s²

5 0

2 years ago