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4 years ago

Give an example of when your average speed would be higher than instantaneous speed

1 answer:

3 0

When I drive to the office, I drive through two school zones,
and four intersections that are controlled by traffic lights.

My average speed for the trip is higher than my instantaneous
speed is at any point in the school zones, or at any time when
I'm waiting for a red light to change.

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A 75-kg refrigerator is located on the 70th floor of a skyscraper (300meters a over the ground) What is the potential energy of

Nata [24]

Formula for potential energy is V=mgh, where m is mass in KG, g is earth acceleration (10 m/s^2), and h its height in meters. We know mass, acceleration is constant and also known, we know height also. Lets substitute
V=75*10*300=225000[J]=225[kJ] - its the answer

7 0

4 years ago

The real advantage to hydrostatic weighing is that it __________.

katovenus [111]

The real advantage to hydrostatic weighing is that it gives one of the most accurate measurements of body fat.

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3 years ago

Read 2 more answers

Convert 8 light years to Astronomical Units

marusya05 [52]

Answer:

505929 AU

Explanation:

As you may know, one light-year is equivalent to approximately 63241.1 Astronomical Units. To get your answer, simply multiply 63241.1 * 8 to get ≈505929 AU

5 0

3 years ago

A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h

timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

$F_n = \frac{mv^2}{r}$

$N+mgcos\theta = \frac{mv^2}{r}$

Here,

Normal reaction of the ring is N and velocity of the ring is v

$N+mgcos\theta = \frac{mv^2}{r}$

$N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})$

$N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}$

$N = 1.374lb$

PART B) Acceleration

$F_t = ma_t$

$-mgsin\theta = ma_t$

$-W sin\theta = \frac{W}{g} a_t$

$-2Sin30\° = (\frac{2}{32.2})a_t$

$a_T = -16.10ft/s^2$

Negative symbol indicates deceleration.

<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>

8 0

3 years ago

A total positive charge of 12.00 mC is evenly distributed on a straight thin rod of length 6.00 cm.

Gala2k [10]

It’s not in English can you change it please

3 0

3 years ago