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3 years ago

Help me please!! 15points

Physics

1 answer:

Romashka-Z-Leto [24]3 years ago

4 0

Explanation:

Acceleration is the change in speed over change in time.

a = Δv / Δt

a. The car's acceleration is:

a = (80 km/h − 0 km/h) / 10 s

a = 8 km/h/s

So every second, the speed increases by 8 km/h.

b. The cyclist's acceleration is:

a = (16 m/s − 4.0 m/s) / 5.6 s

a = 2.1 m/s²

c. The stone's speed is:

10.0 m/s² = (v − 0 m/s) / 3.5 s

v = 35 m/s

d. The time is:

1.6 m/s² = (10 m/s − 0 m/s) / t

t = 6.3 s

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Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

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