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elena-14-01-66 [18.8K]
3 years ago
14

16. A pug puppy is sitting on a skateboard. The pug has a mass of 2.4 kg, and the skateboard has a mass

Physics
1 answer:
kramer3 years ago
4 0

Answer:

12 N

Explanation:

Use Newton's second law:

∑F = ma

F = (2.4 kg + 1.3 kg) (3.2 m/s²)

F = 11.84 N

Rounded to two significant figures, the force is 12 N.

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If a pulley system with an ideal mechanical advantage of 2,000,000 is used in lifting a 2,000 lb. car, how far would the car mov
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C

Explanation:

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What is the next step if the data from an investigation do not support the original hypothesis? A. The data are revised to suppo
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I believe the answer is D. <span>The hypothesis is revised and another experiment is conducted.</span>
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An object is moving north with an initial velocity of 14 m/s accelerates 5m/s for 20 seconds. What is the final velocity of the
olga2289 [7]

Use the kinematic equation: Vf=Vi+at

Then plug;

Vi=14 m/s

a=5 m/s²

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3 years ago
Which equation is correct according to Ohm’s law? Which equation is correct according to Ohm’s law? A.) V = IR B.) I = R/V C.) R
vodomira [7]

Answer:

V = IR

Explanation:

Required

Which equation represents ohm's law?

Literally, ohm's law implies that current (I) is directly proportional to voltage (V) and inversely proportional to resistance (R).

Mathematically, this can be represented as:

I\ \alpha\ \frac{V}{R}

Convert the expression to an equation

I\ =\ \frac{V}{R}

Multiply both sides by R to make V the subject

I\ * R\ =\ \frac{V}{R} * R

I\ * R\ =V

Reorder

V = I\ * R

V = IR

<em>Option (a) is correct; Others are not</em>

6 0
2 years ago
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A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
2 years ago
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