The grandfather clock will now run slow (Option A).
<h3>What is Time Period of an oscillation?</h3>
- The time period of an oscillation refers to the time taken by an object to complete one oscillation.
- It is the inverse of frequency of oscillation; denoted by "T".
Now,
, where L is the length and g is the gravitational constant, is the formula for a pendulum's period. - The period will increase as one climbs a very tall mountain because g will slightly decrease.
- Due to this and the previous issue, the clock runs slowly and it seems that one second is longer than it actually is.
Hence, the grandfather clock will now run slow (Option A).
To learn more about the time period of an oscillation, refer to the link: brainly.com/question/26449711
#SPJ4
Answer:
12mph in 2hrs and 3mph in 0.5hrs the total distance would be 12*2 and 3*0.5 which would be 24 and 1.5 so we add those 24+1.5= 25.5. The answer would be 25.5
Answer:
remote control, neon and fluorescent lights, wireless transmission, computers, smartphones
Explanation:
Answer:

Explanation:
Given:
- length of a steel-string,

- area of the string,

- Young's modulus of the steel,

- force of tension on the string,

We have the relation for change in length:




Answer:
0.12
Explanation:
The acceleration due to gravity of a planet with mass M and radius R is given as:
g = (G*M) / R²
Where G is gravitational constant.
The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg
The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m
Therefore:
g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²
g(planet) = 1.18 m/s²
Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:
g(planet)/g(earth) = 1.18/9.8 = 0.12