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Helen [10]
2 years ago
7

Differentiate between loadstone and bar magnet​

Physics
1 answer:
MakcuM [25]2 years ago
7 0
As nouns the difference between magnet and lodestone




is that magnet is a piece of material that attracts some metals by magnetism while lodestone is a naturally occurring magnet.
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A grandfather clock depends on the period of a pendulum to keep correct time.(ii) Suppose a grandfather clock is calibrated corr
ANEK [815]

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,

  • \sqrt{\frac{L}{g}}, 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

6 0
2 years ago
Jim rode at an average speed of 12mph in 2 hours the he walked at an average of speed of 3 mph 0.5 hours what was the total dist
mrs_skeptik [129]

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

6 0
4 years ago
Five things that were invented by tesla?<br><br> please answer!!!!
algol13

Answer:

remote control, neon and fluorescent lights, wireless transmission, computers, smartphones

Explanation:

6 0
3 years ago
Read 2 more answers
Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m
ira [324]

Answer:

\Delta L=15\,mm

Explanation:

Given:

  • length of a steel-string, L=1m
  • area of the string, A=0.5\,mm^2
  • Young's modulus of the steel, Y=2\times 10^{11} Pa
  • force of tension on the string, F=1500\,N

We have the relation for change in length:

\Delta L=\frac{F.L}{A.Y}

\Delta L=\frac{1500\times 1000}{0.5\times 10^{-6}\times 2\times 10^{11}}

\Delta L=0.015m

\Delta L=15\,mm

6 0
3 years ago
. A newly discovered planet has three times the mass and five times the radius of Earth. What is the ratio of the acceleration d
NikAS [45]

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

3 0
3 years ago
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