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katrin [286]
3 years ago
8

What is your hypothesis (or hypotheses) for this experiment?

Physics
2 answers:
Ad libitum [116K]3 years ago
6 0

Answer: In that case, Put something like "My hypothesis is that the car will take (x) seconds to get to checkpoint 1, (x) seconds to get to checkpoint 2, (x) seconds to get to checkpoint 3, and (x) seconds to get to checkpoint 4" Replacing x with random numbers, or number close to the actual found numbers, but honestly as long as the guesses arent outrageous you should get the question right

Explanation:

Alex787 [66]3 years ago
4 0

Answer:

My hypothesis is that the car will take (x) seconds to get to checkpoint 1, (x) seconds to get to checkpoint 2, (x) seconds to get to checkpoint 3, and (x) seconds to get to checkpoint 4.

This above is a good starter the hypothesis.

Explanation:

I am literally here struggling to write a lab report because I didn't pay any attention while doing this lab project.

I didn't even record the seconds it took for the race cars to get to each checkpoint. ;; very d u m b move, correct, I know it was. I really regret it now. I never knew we had to write a whole lab report for it.

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Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge
Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃=  2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

3 0
3 years ago
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