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ad-work [718]
3 years ago
14

What distance does a plate with an average speed of 1.95 cm/year move in 1000 years? help pls

Physics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

1950cm

Explanation:

You should multiply 1.95 cm by 1000. This will result in the answer.

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An object accelerates from rest with a constant acceleration of 7.5 m/s2. How fast will it
Serhud [2]

Answer:

13.4 m/s

Explanation:

Given:

Δx = 12 m

v₀ = 0 m/s

a = 7.5 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (7.5 m/s²) (12 m)

v ≈ 13.4 m/s

8 0
3 years ago
Please I really need this !! Potassium loses electrons when it reacts with oxygen. Which statement is true of potassium in this
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D. It oxidises /////
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2 years ago
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A helicopter flies 25 km north, 5 km east, then 5 km S, then 15 km W. What is the resultant displacement and direction of the he
ch4aika [34]

Answer:

Explanation:

Plotting the original location of the helicopter before it flies 25 km north, it would be at the origin, (0,0) then after it flies north, the y vertex gains 25 points, so it would be (0,25)

After it flies east, the x coordinate gains 5 points, so it would now be (5,25)

After it flies south, the y coordinate loses or is subtracted by 5 points. so it would now be (5,20)

After flying west, the x coordinate loses 15 points. So the final vertex would be at (-10,20)

East = Right

West = Left

South= Down

North = Up

I used mainly mathematical methods by adding and subtracting the x and y coordinate values, but this could be graphed easily since I gave the coordinates just incase!

Hope this helps!

7 0
2 years ago
Mercury is in the 80th position in the periodic table. How many protons does it have?
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Mercury has 80 protons. Ironic? 
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2 years ago
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Two electrons are initially at rest separated by a distance of 2nm. At time t=0, they start to move apart due to Coulombic repul
Gnom [1K]

Answer:

t=2.5\times 10^{-14}\ s

Explanation:

We know that charge on electron

q=1.6\times 10^{-19}\ C

r= 2 nm

We know that force between two charge given

F=K\dfrac{Q_1Q_2}{r^2}

Now by putting the value

F=9\times10^9\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(2\times 10^{-9})^2}

F=5.67\times 10^{-11}\ N

We know that mass of electron

The mass of electron

m=9.1\times 10^{-31}\ kg

F= m a

a= Acceleration of electron

a= F/m

a=\dfrac{5.67\times 10^{-11}}{9.1\times 10^{-31}}\ m/s^2

a=6.2\times 10^{19} m/s^2

S=ut+\dfrac{1}{2}at^2

initial velocity given that zero ,u=0

20\times 10^{-9}=\dfrac{1}{2}\times 6.2\times 10^{19} t^2

t=\sqrt {\dfrac{40\times 10^{-9}}{6.2\times 10^{19}}}

t=2.5\times 10^{-14}\ s

3 0
3 years ago
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