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Alexus [3.1K]
3 years ago
8

How many of the elements of the periodic table are considered solids

Physics
1 answer:
Alex17521 [72]3 years ago
7 0

Answer:

There are two liquids, bromine and mercury, as well as eleven gases. The rest are solids, which is 105.

Explanation:

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A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon
Anettt [7]
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed v_x=30 m/s, and an accelerated motion on the y-axis, with initial speed v_y=20 m/s and acceleration g=9.81 m/s^2:
S_x(t)=v_xt
S_y(t)=v_y t- \frac{1}{2} gt^2
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
S_y(t)=0
Therefore:
v_y t -  \frac{1}{2}gt^2=0
which has two solutions:
t=0 is the time of the beginning of the motion,
t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m
4 0
3 years ago
Symbion pandora has a symbiotic relationship with the Norway lobster. True False
pashok25 [27]
<em>The answer you are looking for is: </em>
<em><u>True</u>
Hope that helps!! </em>
<em>Have a wonderful day!! </em>
<em>If you ever need anything feel free to ask me!!</em>
8 0
2 years ago
A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential
castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge q=+10.5\times10^{-6}\ C

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

\Delta U=-W

\Delta U=-F\cdot d

\Delta U=-q(E\cdot d)

Put the value into the formula

\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}

\Delta U=-3.32\times10^{-3}\ J

The change in electric potential energy  is -3.32\times10^{-3}\ J

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

\Delta U=-W

\Delta U=-F\cdot (-d)

\Delta U=-q(E\cdot (-d))

Put the value into the formula

\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})

\Delta U=3.32\times10^{-3}\ J

The change in electric potential energy  is 3.32\times10^{-3}\ J

Hence, This is the required solution.

3 0
3 years ago
A certain field line diagram illustrates the electric field due to three particles that carry charges 5.0 μC, -3.0 μC, and -2.0
4vir4ik [10]

Answer:

6

Explanation:

Number of lines emanate from + 5 micro coulomb is 15 .

They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.

the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.

So the lines terminating at - 3 micro coulomb

                                    = \frac{3}{5}\times 15 = 9

So the lines terminating at - 2 micro coulomb

                                    = \frac{2}{5}\times 15 = 6

So, the number of filed lines terminates at - 2 micro Coulomb are 6.

4 0
3 years ago
Can enter through any cavity lined with mucus membranes
Gre4nikov [31]

Answer:

mucus

Explanation:

7 0
3 years ago
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