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kenny6666 [7]
3 years ago
15

Which is one way that scientists communicate the results of an experiment?

Physics
2 answers:
AURORKA [14]3 years ago
8 0
I would think D. Forming a hypothesis
Lemur [1.5K]3 years ago
3 0
C. Writing a procedure 
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A train moving at a constant speed on a surface inclined upward at 10.0° with the horizontal travels a distance of 400 meters in
Amiraneli [1.4K]
If the velocity of the train is v=s/t, where s is the distance and t is time, then v=400/5=80m/s. To get the vertical component of the velocity we need to multiply the velocity v with a sin(α): Vv=v*sin(α), where Vv is the vertical component of the velocity and α is the angle with the horizontal. So:

Vv=80*sin(10)=80*0.1736=13.888 m/s. 

So the vertical component of the velocity of the train is Vv=13.888 m/s.
7 0
3 years ago
What experimental evidence led scientists to change from the previous model to this one?
san4es73 [151]
The colors of light emitted from heated atoms had very specific energies.
6 0
3 years ago
A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li
Kitty [74]

Answer:

E_{net} = 6.44 \times 10^5 N/C

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

E = \frac{2k \lambda}{r}

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}

now plug in all data

\lambda_1 = 4.68 \muC/m

\lambda_2 = 2.48 \mu C/m

r = 0.200 m

now we have

E_{net} = \frac{2k}{r}(4.68 + 2.48)

E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})

E_{net} = 6.44 \times 10^5 N/C

8 0
3 years ago
-!!! URGENT !!!-
seraphim [82]

Try this option, the answers are marked with colour.

8 0
3 years ago
Two point charges, initially 2.0 cm apart, experience a 1.0 N force. If they are moved to a new separation of 0.25 cm, what is t
den301095 [7]

Explanation:

Th electric force between charges is inversely proportional to the square of distance between them. It means,

F\propto \dfrac{1}{r^2}

Initial distance, r₁ = 2 cm

Final distance, r₂ = 0.25 cm

Initial force, F₁ = 1 N    

We need to find the electric force between charges if the new separation of 0.25 cm. So,

\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2\\\\F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1\times 2^2}{(0.25)^2}\\\\F_2=64\ N

So, the new force is 64 N if the separation between charges is 64 N.

7 0
3 years ago
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