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goblinko [34]
2 years ago
3

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cmcm . The explorer finds th

at the pendulum completes 103 full swing cycles in a time of 132 s.What is the value of the acceleration of gravity on this planet?
Physics
1 answer:
Pavel [41]2 years ago
5 0

Answer:

13.01 m/s²

Explanation:

The period of a simple pendulum is given as

T = 2π√(L/g) .......................... Equation 1

Where T = Period of the simple pendulum, L = Length of the pendulum, g = acceleration due to gravity of the planet.

Given; T = 132/103 = 1.28 s, L = 54 cm = 0.54 m, π = 3.14

Substitute into equation 1

1.28 = (2×3.14)√(0.54/g)

1.28 = 6.28√(0.54/g)

Making g the subject of the equation,

√(0.54/g) = 1.28/6.28

√(0.54/g) = 0.2038

0.54/g = (0.2038)²

0.54/g = 0.0415

g = 0.54/0.0415

g = 13.01 m/s²

Hence the value of the acceleration due to gravity on the planet = 13.01 m/s²

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Find The Magnitude Of The Displacement

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Answer:

- Problem-recognition.

Explanation:

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As per the given description, Gilbert's evaluation of his need that his current phone is not able to meet up his requirement and realization that he requires a phone with a longer-battery life exemplifies that he is in the '<u>problem-recognition</u>' phase of the consumer decision-making process. This stage would be followed by the assessment of alternatives(various phones with longer battery life) and reaching the conclusion to buy a specific chosen product after the evaluation.

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10×5 hope this helps

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The Doppler effect refers to the frequency as perceived by _________. A. the source of the waves B. an observer of the waves who
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1 year ago
A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w
FinnZ [79.3K]

Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

vf^{2} -vo^{2} = 2*a*x

As the proton is coming at rest after travelling 0.200 m to the right,  vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

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mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C

⇒ E = 1.04*10⁻¹ N/C

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1 year ago
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