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2 years ago

A skateboarder has an acceleration of −1.9 m/s2. If her initial speed is 6 m/s, how long does it take her to stop?

1 answer:

6 0

You said that she's losing 1.9 m/s of her speed every second.

So it'll take

(6 m/s) / (1.9 m/s²) = 3.158 seconds (rounded)

to lose all of her initial speed, and stop.

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The answer is increase

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In the warm up, you reviewed the equation to calculate kinetic energy. What question could you ask about kinetic energy which wi

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2 years ago

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Choose the 200 kg refrigerator. Set the applied force to 400 N (to the right). Be sure friction is turned off.What is the net fo

White raven [17]

So, there should be two forces acting on the refrigerator: the applied force and the friction force.

The question mentioned that the friction force was set to zero, so the only effective force now would be the applied force.

We have an applied force of 400 N to the right, this means that:
<span>The magnitude of the net force is 400, directed to the right.</span>

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2 years ago

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Dumbo the elephant weights 13,000 pounds. What force would the Earths surface apply to Dumbo? 2.2 pounds = 1 kg

snow_lady [41]

Answer:

57908 N

Explanation:

Let's first convert Dumbo's mass into kg using the given relationship: 2.2 pounds =1 kg.

Then, 13000 lbs = 13000/2.2 kg = 5909 kg

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2 years ago

Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

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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2 years ago