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Aleksandr [31]
3 years ago
13

A penny rolls along the table for a distance of 1.3 meters. Jackie pushes it 40 centimeters further in the same direction.

Physics
1 answer:
brilliants [131]3 years ago
7 0
40 meters times 1 meter over 100 centimeters equals 0.4 meters. 1.3 meters + 40 centimeters =. 1.3 m + 0.4 m = 1.7 m. The answer is 1.7 meters
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This is the number of complete movements of a wave per second.
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A unit for measuring frequency, equal to one cycle per second. If a sound wave has a frequency of 20,000 Hz, this means that 20,000 waves are passing through a given point during the interval of one second.

5 0
3 years ago
Which is a true statement about a series circuit?
vaieri [72.5K]

the answer is B your welcome

6 0
3 years ago
Where on the Earth would you have to be to see the North Celestial Pole exactly halfway between your horizon and zenith?
mario62 [17]

Answer:

The point straight overhead on the celestial sphere for any observer is called the zenith and is always 90 degrees from the horizon. The arc that goes through the north point on the horizon, zenith, and south point on the horizon is called the meridian.

From any location on Earth you see only half of the celestial sphere, the half above the horizon.

If you stood at the North Pole of Earth, for example, you would see the north celestial pole overhead, at your zenith. The celestial equator, 90° from the celestial poles, would lie along your horizon.

4 0
3 years ago
A man 6 feet tall walks at a rate of 6 feet per second away from a light that is 15 feet above the ground.
Tems11 [23]

Answer:(a)10 ft/s

(b)4 ft/s

Explanation:

Given

height of light =15 feet

height of man=6 feet

\frac{\mathrm{d} x}{\mathrm{d} t}=6 ft/s

From diagram

\frac{15}{y}=\frac{6}{y-x}

5(y-x)=2y

3y=5x

differentiate both sides

3\times \frac{\mathrm{d} y}{\mathrm{d} t}=5\times \frac{\mathrm{d} x}{\mathrm{d} t}

Tip of shadow is moving at the rate of

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{5}{3}\times 6=10 ft/s

(b)rate at which length of his shadow  is changing

Length of shadow is y-x

differentiating w.r.t time

\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} t}-\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=10-6=4 ft/s

7 0
3 years ago
How far did the object travel by the end of eight seconds, according to the graph above?
Ulleksa [173]

Answer:

<em>The object traveled 4 cm by the end of eight seconds.</em> Correct: A)

Explanation:

<u>Speed vs Time Graph</u>

In a speed-time graph, speed is plotted on the vertical axis and time is plotted on the horizontal axis. If the graph is a horizontal line, the speed is constant, if the line is sloped up, the speed is increasing and the acceleration is positive and constant, and if the line is sloped down, the speed is decreasing and the acceleration is negative and constant.

The distance traveled by the object can be found by calculating the area under the graph and above the x-axis.

The graph provided shows two different zones: the first 4 seconds, the speed is constant at 1 cm/s, and the last 4 seconds, the speed is zero, i.e. the object is not moving.

The area behind the first zone is a rectangle of height 1 cm/s and base 4 sec, thus the distance is 1 * 4 = 4 centimeters.

The second zone corresponds to an object at rest, thus no distance is traveled.

The object traveled 4 cm by the end of eight seconds.

A) Correct. As shown above

B) The distance traveled is 4 cm. Incorrect

C) The distance traveled is 4 cm. Incorrect

D) The distance traveled is 4 cm. Incorrect

6 0
3 years ago
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