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scZoUnD [109]
3 years ago
5

Help

Physics
1 answer:
lakkis [162]3 years ago
7 0
Set up a unit conversion problem, arrange units to convert minutes to seconds, then cancel all time units.

8.3 minutes x (60 seconds/1 minute) x ( 3.00x10^8 meters/1 second) = 149,400,000,000 meters

Since the time is the term with the fewest number of significant figures (2), your answer should only be expressed with two sig figs as well.

1.5x10^11 m
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D the cells are very small
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3 years ago
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Lila is a track and field athlete. She must complete four laps around a circular track. The track itself is a 400 meter track an
fgiga [73]

Answer: 4.45m/s

Explanation:

I assume we're solving for her speed.

400m x 4 laps = 1600m

1600m / 6 minutes = 266.67m/min

(266.67m/min) x (1 minute/60 seconds) = 4.45m/s

266.67/60 = 4.45m/s

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Propose a situation where friction is undesirable.
Naddik [55]
Trying to move a large object such as a furniture piece across carpet.
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3 years ago
A 2.40 kg ball is attached to an unknown spring and allowed to oscillate. The figure below shows a graph of the ball’s position
irga5000 [103]

(a) The period of the oscillation is 0.8 s.

(b) The frequency of the oscillation is 1.25 Hz.

(c) The angular frequency of the oscillation is 7.885 rad/s.

(d) The amplitude of the oscillation is 3 cm.

(e) The force constant of the spring is 148.1 N/m.

The given parameters:

  • <em>Mass of the ball, m = 2.4 kg</em>

<em />

From the given graph, we can determine the missing parameters.

The amplitude of the wave is the maximum displacement, A = 3 cm

The period of the oscillation is the time taken to make one complete cycle.

T = 0.8 s

The frequency of the oscillation is calculated as follows;

f = \frac{1}{T} \\\\f = \frac{1 }{0.8} \\\\f = 1.25 \ Hz

The angular frequency of the oscillation is calculated as follows;

\omega = 2\pi f\\\\\omega = 2\pi \times 1.25\\\\\omega = 7.855 \ rad/s

The force constant of the spring is calculated as follows;

\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\ k = \omega ^2 m\\\\k = (7.855)^2 \times 2.4\\\\k = 148.1 \ N/m

Learn more about general wave equation here: brainly.com/question/25699025

4 0
3 years ago
How does a water well affect ground water?
Evgen [1.6K]

Answer:

Pumping a well lowers the water level around the well to form a cone of depression in the water table. If the cone of depression extends to other nearby wells, the water level in those wells will be lowered. The cone develops in both shallow water-table and deeper confined-aquifer systems.

Explanation:

6 0
3 years ago
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