PHYSICS MIDDLE SCHOOL

On earth’s surface, how much does a 100 kg mass weigh?

Answers

Answer 1

Answer:

Weight = (mass) x (gravity)

Weight = (100kg) x (9.8 m/s^2)

Weight = 980 kg-m/s^2

Weight = 980 Newtons.

(about 220.5 pounds)

Answer 2

Answer:

Final answer:

The weight of a 100 kg mass on Earth's surface is 980 Newtons(N). This is calculated using the formula 'w=mg', where 'w' is weight, 'm' is mass, and 'g' is the acceleration due to gravity on Earth (9.80 m/s²).

Explanation:

On earth, the weight of an object is determined by the gravitational force acting on it. This can be calculated using the formula w = mg, where 'w' is the weight of the object, 'm' is its mass, and 'g' is the acceleration due to gravity on Earth. Given that the mass 'm' is 100 kg and the acceleration due to gravity 'g' on Earth's surface is approximately 9.80 m/s², we can determine the weight of the object by substituting these values into the formula.

Applying these values into our equation, we get: w = (100 kg)(9.80 m/s²) = 980 N. Therefore, a 100 kg mass weighs 980 Newtons (N) on Earth's surface.

It is pertinent to note that an object's weight can vary based on its location as the value of 'g' changes with location. For instance, the value of 'g' on the moon is only about 1.67 m/s², which is far less compared to Earth, resulting in a lighter weight for the same mass.

Learn more about Weight on Earth here:

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Figure 1.18 (Chapter 1) shows the Hoover Dam Bridge overthe Colorado River at a height of 271 m. If a heavy object is
dropped from the bridge, how much time passes before the
object makes a splash?

Answers

Answer: 7.436 s

Explanation:

This situation is related to vertical motion, specifically free fall and can be modelled by the following equation:

$y=y_(o)+V_(o) t+(gt^(2))/(2)$

Where:

$y= 0m$ is the final height of the object (when it makes splash)

$y_(o)=271 m$ is the initial height of the object

$V_(o)=0 m/s$ is the initial velocity of the object (it was dropped)

$g=-9.8m/s^(2)$ is the acceleration due gravity (directed downwards)

$t$ is the time since the objecct is dropped until it makes splash

$0=y_(o)+0+(gt^(2))/(2)$

Clearing $t$ :

$t=\sqrt{(-2y_(o))/(g)}$

$t=\sqrt{(-2(271 m))/(-9.8m/s^(2))}$

Finally:

$t=7.436 s$

Explain why free fall acceleration near earths surface is constant

Answers

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Answers

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Answers

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A circuit has a known resistance of 5 Ohms. An ammeter in the circuit gives a reading of 5 Amps. What is the Voltage of the circuit?

Answers

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