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Andrews [41]
3 years ago
10

Why are atoms in a covalent bond usually a certain distance away from each other?

Physics
1 answer:
ANTONII [103]3 years ago
4 0
I think it is because the electrons repel each other

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soldier1979 [14.2K]
Earth Spheres. Earth's Spheres. Everything in Earth's system can be placed into one of four major subsystems: land, water, living things, or air. These four subsystems are called “spheres.” Specifically, they are the lithosphere (land), hydrosphere (water), biosphere (living things), and atmosphere (air).
6 0
3 years ago
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⦁ A car going 50 m/s is brought to rest in a distance of 20.0 m as it strikes a pile of dirt. How large an average force is exer
gtnhenbr [62]

Answer:

the average force exerted by seatbelts on the passenger is 5625 N.

Explanation:

Given;

initial velocity of the car, u = 50 m/s

distance traveled by the car, s = 20 m

final velocity of the after coming to rest, v = 0

mass of the passenger, m = 90 kg

Determine the acceleration of the car as it hit the pile of dirt;

v² = u² + 2as

0 = 50² + (2 x 20)a

0 = 2500 + 40a

40a = -2500

a = -2500/40

a = -62.5 m/s²

The deceleration of the car is 62.5 m/s²

The force exerted on the passenger by the backward action of the car is calculated as follows;

F = ma

F = 90 x 62.5

F = 5625 N

Therefore, the average force exerted by seatbelts on the passenger is 5625 N.

8 0
3 years ago
Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a
andrey2020 [161]

Explanation:

It is given that,

Force, F=(-8\ N)i+(6\ N)j

Position vector, r=(3i+4j)\ m

(a) The torque on the particle about the origin is given by :

\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m

(b) To find the angle between r and F use dot product formula as :

F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
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A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr
Molodets [167]

Answer:

Explanation:

Given

P_1=150 kPa

T_1=20^{\circ}C

V_1=0.5 m^3

T_2=140^{\circ}C

P_2=400 kPa

R for Helium R=2.076

c_v=3.115 kJ/kg-K

mass of gas m=\frac{P_1V_1}{RT_1}

m=\frac{150\times 0.5}{2.076\times 293}

m=0.123 kg

Similarly V_2 can be found

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=0.264 m^3

Work done W=\int_{V_1}^{V_2}PdV

W=\frac{P_2V_2-P_1V_1}{n-1}

W=\frac{mR(T_2_T_1)}{n-1}

Since it is a polytropic Process

therefore PV^n=c

P_1V_1^n=P_2V_2^n

(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}

(\frac{0.5}{0.264})^n=\frac{400}{150}

n=\frac{\ln 2.66}{\ln 1.893}

n=1.533

W=\frac{0.123\times 2.076(140-20)}{1.533-1}

W=57.48 kJ    

From Energy balance

E_{in}-E_{out}=\Delta E_{system}

Neglecting kinetic and Potential Energy change

Q_{in}+W_{in}=change\ in\ Internal\ Energy

Change in Internal Energy \Delta U=u_2-u_1

\Delta U=mc_v(T_2-T_1)

\Delta U=0.123\times 3.115(140-20)

\Delta U=45.977 kJ

Q_{in}+57.48=45.977

Q_{in}=-11.50 kJ  

i.e. Heat is being removed

3 0
3 years ago
What temperature does water freeze at in celsius and fahrenheit?
andre [41]
32 degrees Fahrenheit
And
0 degrees Celsius
4 0
3 years ago
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