An object is moving with constant speed in a circular path. The object's centripetal acceleration remains constant in

Which is most likely a covalent compound? LiF MgS NH3 CaCl2

Artemon [7]

Answer:

NH₃

Explanation:

The compound that is covalent from the given choices is NH₃.

Covalent compounds are usually formed between two atoms with similar values of electronegativities such that the difference is very small or zero.

This bond type involves the sharing of electrons between two atoms with similar electronegativities.
Nitrogen and hydrogen forms stable configuration that are isoelectronic with noble gases by sharing their valence electrons.
The 3 hydrogen electrons are enough to make nitrogen isoelectronic with neon.
Also, the nitrogen with 3 lone pairs of electrons provides the bonding hydrogen with needed electrons to attain a structure similar to helium.

8 0

3 years ago

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Traveler A starts from rest at a constant acceleration of 6 m/s^2. Two seconds later, traveler B starts with an initial velocity

Troyanec [42]

Answer:

3. 3.5 s

Explanation:

The position of traveller A is given by the equation:

$x_A(t) = \frac{1}{2}a t^2$

where

$a = 6 m/s^2$ is the acceleration of A

t is the time measured from when A started the motion

The position of traveller B instead is given by

$x_B(t) = u_B (t-2) + \frac{1}{2}a(t-2)^2$

where a (acceleration) is the same as traveller A, and

$u_B = 20 m/s$

is B's initial velocity. We can verify that the formula is correct by substituting t=2, and we get $x_B=0$ , which means that B starts its motion 2 seconds later.

Traveller B overtakes traveller A when the two positions are the same, so:

$x_A = x_B\\\frac{1}{2}at^2 = u_B (t-2) + \frac{1}{2}a(t-2)^2\\\frac{1}{2}at^2 = u_B t - 2u_B +\frac{1}{2}at^2 +2a-2at\\u_Bt-2at = 2u_B-2a\\t=\frac{2u_B-2a}{u_B-2a}=\frac{2(20)-2(6)}{20-2(6)}=3.5 s$

4 0

3 years ago

How is work involved in stopping a car?

vichka [17]

You have to move your foot to stop the car so I guess that would be considered work by moving your foot

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3 years ago

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What is the difference between an atom in the ground state and an atom in an excited state

SIZIF [17.4K]

The atom in an excited state has more energy and is less stable than the atom in the ground state.

7 0

4 years ago

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If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

5 0

3 years ago