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1 year ago

In general, how do you find the average velocity of any object falling in a vacuum?

1 answer:

5 0

In general, how do you find the average velocity of any object falling in a vacuum? (Assume you know the final velocity.) Multiply the final velocity by final time. 3. Calculate : Distance, average velocity, and time are related by the equation, d = v • t A

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You live on a planet far from ours. "Based on extensive communication with a physicist on earth", you have determined that all l

Ugo [173]

Answer:

8.56 m/s2

Explanation:

Using law of energy conservation while taking into account of the rotational and translation kinetic energy, when the solid cylinder rolls down the incline we have the potential energy converted to kinetic energy:

$E_p = E_{kv} + E_{k\omega}$

$mgh = mv^2/2 + I\omega^2/2$

where m is the mass, $I = mr^2/2$ is the moments of inertia of the solid cylinder $\omega = v / r$ is the angular speed of the cylinder

$mgh = mv^2/2 + \frac{mr^2}{2}\frac{(v/r)^2}{2}$

$mgh = mv^2/2 + mv^2/4 = 3mv^2/4$

$h = 3v^2/(4g)$

So if you plot a liner chart of h vs $v^2$ and get a slope of 6.42 then that means 3/(4g) = 6.42 so $g = 6.42*4/3 = 8.56 m/s^2$

The gravitational acceleration on this planet is 8.56 m/s2

3 0

2 years ago

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

Alja [10]

Answer:

Terminal velocity of object = 12.58 m/s

Explanation:

We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.

Gravitational force = mg = 80 * 9.8 = 784 N

Drag force = $12.0v+4.00v^2$

Equating both, we have

$784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0$

So v = 12.58 m/s or v = -15.58 m/s ( not possible)

So terminal velocity of object = 12.58 m/s

7 0

2 years ago

Use scientific (exponential) notation to express the following quantities in terms of the SI base units in

Anna007 [38]

Answer:

When a number is written in scientific notation (representing the number using powers of base ten) it is expressed so that it contains a digit in the place of the units and all other digits after the decimal point, multiplied by the respective exponent. For example, the number $4.232(10)^{3}$ .

On the other hand, it is known the units in the SI for mass, length, time and temperature are kilogram (kg), meter (m), second (s) and Kelvin (K), respectively. In addition, thera are prefixes of the International System (SI) that indicate a specific factor of 10.

For example:

-Giga (G) is a prefix that indicates a factor of $10^{6}$

-Pico (p) is a prefix that indicates a factor of $10^{-12}$

-Mili (m) is a prefix that indicates a factor of $10^{-3}$

-Micro ( $\mu$ ) is a prefix that indicates a factor of $10^{-6}$

-Tera (T) is a prefix that indicates a factor of $10^{12}$

-Kilo (K) is a prefix that indicates a factor of $10^{3}$

Knowing this, let's express these quantities in terms of the SI base units:

$0.13 g=0.00013 kg=1.3 (10)^{-4}kg$

$232 Gg=232(10)^{6}g=232 (10)^{3}kg$

$5.23 pm=5.23(10)^{-12}m$

$86.3 mg=86.3(10)^{-3}g=8.63 (10)^{-5}kg$

$37.6 cm=0.376 m=3.76 (10)^{-1}m$

$54 \mu m=54 (10)^{-6}m$

$1 Ts=1 (10)^{-12}s$

$27 ps=27 (10)^{-12}s$

$0.15 mK=0.15(10)^{-3}K$

8 0

2 years ago

The free-body diagram of a crate is shown. What is the net force acting on the crate? 352 N to the left 176 N to the left 528 N

Umnica [9.8K]

As per given conditions there are two directions along which forces are acting

1. Net force along left direction is given as

$F_{left} = 352N + 176 N = 528 N$

2. Net force towards right direction is given as

$F_{right} = 528N + 440 N = 968 N$

now since the two forces here in opposite direction so here we will have net force given as

$F_{net} = F_{right} - F_{left}$

$F_{net} = 968 - 528$

$F_{net} = 440 N$

so here net forces must be 440 N towards right

7 0

2 years ago

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You are driving on a road where rain has left large pools of water, and you have driven through water that was several inches de

makvit [3.9K]

In would say that you may have water in your brakes which may have gotten in the brake lines or in the brake discs so that could cause the brakes to malfunction due to driving through the pools of water so the brakes should be examined as soon as possible.

3 0

2 years ago

Read 2 more answers