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AveGali [126]
2 years ago
6

A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of

2 m and its period is measured to be 2 s. The value of g obtained in this investigation is most nearly:.
Physics
1 answer:
Dafna1 [17]2 years ago
5 0

Answer:

<em>-</em><em> </em><em>p</em><em>e</em><em>n</em><em>d</em><em>u</em><em>l</em><em>u</em><em>m</em><em> </em><em>u</em><em>s</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>d</em><em>e</em><em>t</em><em>e</em><em>r</em><em>m</em><em>i</em><em>n</em><em>e</em><em> </em><em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em>

Explanation:

<u>b</u><u>y</u><u> </u><u>c</u><u>a</u><u>l</u><u>c</u><u>u</u><u>l</u><u>a</u><u>t</u><u>i</u><u>n</u><u>g</u><u> </u><u>n</u><u>e</u><u>e</u><u>d</u><u> </u><u>t</u><u>o</u><u> </u><u>u</u><u>n</u><u>d</u><u>e</u><u>r</u><u>s</u><u>t</u><u>a</u><u>n</u><u>d</u><u> </u><u>f</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>p</u><u>r</u><u>o</u><u>b</u><u>l</u><u>e</u><u>m</u>

<em>t</em><em>h</em><em>e</em><em> </em><em>l</em><em>e</em><em>n</em><em>g</em><em>t</em><em>h</em><em> </em><em>i</em><em>s</em><em> </em><em>m</em><em>e</em><em>a</em><em>s</em><em>u</em><em>r</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>b</em><em>e</em><em> </em><em>2</em><em>s</em>

<em>t</em><em>h</em><em>e</em><em> </em><em>l</em><em>e</em><em>n</em><em>g</em><em>h</em><em>t</em><em> </em><em>i</em><em>s</em><em> </em><em>a</em><em>l</em><em>s</em><em>o</em><em> </em><em>2</em>

<em>2</em><em> </em><em>o</em><em>b</em><em>t</em><em>a</em><em>i</em><em>n</em><em>e</em><em>r</em><em> </em><em>o</em><em>n</em><em> </em><em>i</em><em>n</em><em>v</em><em>i</em><em>s</em><em>t</em><em>i</em><em>g</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>m</em><em>o</em><em>s</em><em>t</em><em> </em><em>n</em><em>e</em><em>a</em><em>r</em><em>l</em><em>y</em><em> </em><em>i</em><em>s</em><em> </em><em>8</em><em>.</em>

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Within a vacuum, the property to all electromagnetic waves is their
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The answer is D) Velocity
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How does an electromagnet become permanent
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6 0
3 years ago
Suppose a car is traveling at +20.3 m/s, and the driver sees a traffic light turn red. After 0.207 s has elapsed (the reaction t
olga nikolaevna [1]

Answer:

33.6371 m

Explanation:

t = Time taken

u = Initial velocity = 20.3 m/s

v = Final velocity

s = Displacement

a = Acceleration = -7 m/s²

Distance traveled in the 0.207 seconds

Distance = Speed × Time

⇒Distance = 20.3×0.207 = 4.2021 m

Equation of motion

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-20.3^2}{2\times -7}\\\Rightarrow s=29.435\ m

Distance traveled by the car while braking is 29.435 m

Total distance measured from the point where the driver first notices the red light is 29.435+4.2021 = 33.6371 m

7 0
3 years ago
A coach shouts "Go!" from the finish line of a 100.0-m track to the runners.
irina [24]

Answer:

a. Speed = 342.5 meters per seconds.

b. Wavelength = 2.0 meters

Explanation:

Given the following data;

Distance = 100m

Time = 292 milliseconds to seconds = 292/1000 = 0.292 seconds

Frequency = 171 Hz

a. To find the speed of sound in air;

Speed = distance/time

Speed = 100/0.292

Speed = 342.5 m/s

b. To find the wavelength;

Wavelength = speed/frequency

Wavelength = 342.5/171

Wavelength = 2.0 m

5 0
2 years ago
At 600.0 k the rate constant is 6.1× 10–8 s–1. what is the value of the rate constant at 785.0 k?
photoshop1234 [79]
Missing details. Complete text is:"The following reaction has an activation energy of 262 kJ/mol:
C4H8(g) --> 2C2h4(g)
At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 785.0 K?"
To solve the exercise, we can use Arrhenius equation:
\ln( \frac{K_2}{K_1} ) =  \frac{Ea}{R} ( \frac{1}{T_1}- \frac{1}{T_2}  )
where K are the reaction rates, Ea is the activation energy, R=8.314 J/mol*K and T are the temperatures. Using T1=600 K and T2=785 K, and Ea=262 kJ/mol = 262000 J/mol, on the right side of the equation we have
\frac{Ea}{R}( \frac{1}{T_1}- \frac{1}{T_2}  )=12.38
And so
\ln( \frac{K_2}{K_1})=12.38
And using K_1=6.1\cdot 10^{-8} s^{-1} , we find K2:
K_2=K_1 e^{12.38}=0.0145 s^{-1}


5 0
3 years ago
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