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3 years ago

Why do we need the periodic table? First to answer get brainless!!!!!

2 answers:

3 0

The periodic table is the most important chemistry reference there is. It arranges all the known elements in an informative array. Elements are arranged left to right and top to bottom in order of increasing atomic number. Order generally coincides with increasing atomic mass.

EleoNora [17]3 years ago

3 0

It was used to predict the chemical and elemental properties.

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An electric current flows out of the page. What is the direction of the magnetic field?

Mademuasel [1]

When the electric current flows out of the page and the direction of the magnetic field is counter-clockwise. The magnetic field that is made by an electric current is said to be always leaning perpendicularly to the direction of the current flow.

3 0

4 years ago

A 2,100 kg car is lifted by a pulley. If the cable breaks at 4.50 m, what is the velocity of the car when it hits the ground? 88

sammy [17]

V^2-u^2=2as

v=final velocity=unkown
u=initial velocity=0 m/s, because freely falling
a=acceleration due to gravity=9.8 m/s^2
s=distance (here height) traveled=4.5m

therefore the final velocity,
v^2=2*9.8*4.5
v=9.39m/s

8 0

3 years ago

Which term names traces or remains of organisms that lived in the past? A. adaptations B. mutations C. traits D. fossils

grin007 [14]

The answer to this question is choice letter "D"

7 0

4 years ago

Read 2 more answers

At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r

Oksana_A [137]

Answer:

For proton:

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

For electron:

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s

Explanation:

For a proton we have:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

$W = \Delta Vq$ (1)

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C

V: is the potential

Also, the work is equal to:

$W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}$ (2)

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

$v_{a}$ : is the velocity in the point a

$v_{b}$ : is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

$\Delta Vq = \frac{1}{2}mv_{a}^{2}$

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}$

$v_{a} = 2.9 \cdot 10^{4} m/s$

For an electron we have:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

$\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}$

Since $v_{a}$ = 0 (starts from rest) and $m_{e}$ = 9.1x10⁻³¹ kg (electron's mass), we have:

$(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}$

$v_{b} = 1.2 \cdot 10^{6} m/s$

I hope it helps you!

6 0

4 years ago

At a busy intersection, a 1320-kg car traveling west with a speed of 12 m/s collides head-on with a minivan traveling east with

Elza [17]

p = mv so this question pertains to conservation of momentum.

The momentum of the car is 1320*12 = 15840kgms^-1
For the sake of easiness and the fact negatives are annoying, I will assume east to be positive.

So we know 9.4m - 15840 = (1320+m)*1.5
9.4m - 15840 = 1980+1.5m

Rearranging to solve for m
7.9m = 17820
m = 2255.70 = 2300kg to 2sf

7 0

4 years ago