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2 years ago

9

Which statement(s) is/are indicative of a covalent bond?

1 answer:

Over [174]2 years ago

4 0

Do you have any answer options?

The most I can help is to tell you that a covalent bond is a type of bond that shares electrons between the bonded atoms.

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A ball is thrown into the air with an upward velocity of 20 feet per second. Its height, h, in feet after t seconds is given by

Margarita [4]

Thank you for posting your question here at brainly. Feel free to ask more questions.
<span>The best and most correct answer among the choices provided by the question is B. Reaches a max height of 8.25 feet after 0.63 seconds</span> . <span><span>

</span><span>Hope my answer would be a great help for you. </span> </span>

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3 0

2 years ago

An atom that has an excess positive or negative electrical charge caused by the loss or addition of an electron is called a(n)

maria [59]

Answer:

ion

Explanation:

4 0

2 years ago

Read 2 more answers

The acceleration of gravity is -9.8 m/s2. A sling shot fires a rock straight up into the air with a speed of +39.2 m/s. 1. what

Vanyuwa [196]

Given that,

The acceleration of gravity is -9.8 m/s²

Initial velocity, u = 39.2 m/s

Time, t = 2 s

To find,

The final velocity of the shot.

Solution,

Let v is the final velocity of sling shot. Using first equation of motion to find it.

v = u +at

Here, a = -g

v = u-gt

v = (39.2)-(9.8)(2)

v = 19.6 m/s

So, its velocity after 2 seconds is 19.6 m/s.

4 0

2 years ago

A ball droped from a building. How fast is it traveling after falling 3.55s

algol [13]

Answer:

d = 61.75 m

Explanation:

Given that,

A ball droped from a building.

We need to find how fast is it traveling after falling 3.55 s.

As it is dropped, its initial velocity is equal to 0.

Let d is the distance it covers after falling 3.55 s.

We can use second equation of motion to find d.

$d=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a =g

$d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times (3.55)^2\\\\d=61.75\ m$

So, it will cover 61.75 m after falling 3.55 seconds.

4 0

2 years ago

An object of mass 0.400.40 kg, hanging from a spring with a spring constant of 8.08.0 N/m, is set into an up-and-down simple har

sineoko [7]

Answer:

Acceleration will be equal to $2m/sec^2$

Explanation:

We have given mass of the object m = 0.4 kg

Spring constant k = 8 N/m

Maximum displacement of the spring is given x = 0.1 m

From newton's law force is equal to $F=ma$ .....eqn 1

By hook's law spring force is equal to $F=kx$ .....eqn 2

From equation 1 and equation 2

$ma=kx$

$0.4\times a=8\times 0.1$

$a=2m/sec^2$

So acceleration will be equal to $2m/sec^2$

8 0

2 years ago