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3 years ago

The bond order for a single covalent bond is. A. two B. four C. one D. three

Physics

1 answer:

IceJOKER [234]3 years ago

3 0

Answer:

I think it should be C, which is one

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If the resistance of an electric circuit is 12 ohms and the voltage in the circuit is 6 ov, the current flowing through the circ

Anestetic [448]

The current in the circuit =5 A

Explanation:

Voltage = 60 V

resistance= 12 ohm

using ohm's law V=i R

60= i (12)

i=60/12

i=5 A

Thus the current flowing in the circuit = 5 A

6 0

3 years ago

a car takes off from rest and covers a distance of 80m on a straight road in 10s.calculate the magnitude of its acceleration

olasank [31]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Here's the solution :

Let's find the final velocity :

$\dfrac{displacement}{time}$

$\dfrac{80}{10}$

$8 \: \: ms {}^{ - 1}$

Initial velocity (u) = 0 (cuz it started from rest)

Final velocity (v) = 8 m/s

Time taken (t) = 10 sec

now, we know that :

$acceleration = \dfrac{v - u}{t}$

$\dfrac{8 - 0}{10}$

$\dfrac{8}{10}$

$0.8 \: \: m {s}^{ - 2}$

Acceleration = 0.8 m/s²

7 0

3 years ago

In order to cool 1 ton of water at 20°C in an insulated tank, a person pours 140 kg of ice at –5°C into the water. The specific

salantis [7]

Answer:

$T = 6.31 ^0 C$

Explanation:

1 ton = 907.2 kg

so here heat given by water = heat absorbed by the ice

so by energy balance we will have

$Q_1 = Q_2$

$mL + ms\Delta T_1 + ms\Delta T_2 = ms\Delta T_3$

so we have

$140(2.11)(5) + 140(333.7) + 140(4.18)(T) = 907.2(4.18)(20 - T)$

$48195 + 585.2 T = 75841.9 - 3792.1 T$

$T = \frac{27646.92}{4377.3}$

$T = 6.31 ^0 C$

7 0

3 years ago

If lightning is caused by electricity in the air during a thunderstorm, what causes thunder?

NikAS [45]

Answer:

Since light travels faster than sound, the lightning arrives before the thunder. But thunder is the sound of lightning shooting through the sky in a thunderstorm.

Explanation:

7 0

3 years ago

Read 2 more answers

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

4 years ago