Calculate the weight in Newtons of a person who has a mass of 69 kg.

Physics

2 answers:

Advocard [28]2 years ago

6 0

Just multiply by 10,

69 x 10= 690N

BaLLatris [955]2 years ago

5 0

I believe the answer would be 676.65885 rounded to 676.7

You might be interested in

Which quantity best describes this description? Justin walks at 3 miles per hour north, and then speeds up to 4 miles per hour.

prisoha [69]

Personally, I agree with your answer, namely that the likely-intended event happening here is one of acceleration. Having said that, I also want to add: it pains me to see this type of wording because, clearly, it is vague and only invites confusion of the type you are talking about.

Good luck!

4 0

3 years ago

The volume of a piece of metal of a mass 6 gram is 15cm3. What is the density of the metal piece?

torisob [31]

Answer:

0.4 g/cm^3

Explanation:

The density of an object can be found using the following formula.

d= m/v

where m is the mass and v is the volume.

The mass of the metal is 6 grams and the volume is 15 centimeters^3

m=6 g

v= 15 cm^3

Substitute these into the formula.

d= 6 g/ 15 cm^3

Divide 6 g by 15 cm^3 (6/15=0.4)

d= 0.4 g/ cm^3

The density of the metal is 0.4 grams per cubic centimeter.

4 0

2 years ago

Read 2 more answers

A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat

Tanzania [10]

If a Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute then the acceleration of a passenger at his lowest point during the ride is 4.11 $\text{m/s}^{2}$ .

Calculation:

Step-1:

It is given that the radius of the Ferris wheel is r=15 m, and the angular speed of the wheel is $\omega$ =5rev/min.

It is required to find the angular acceleration of a passenger at his lowest point during the ride.

The formula required to calculate the angular acceleration is, $a=\omega ^2 r$ .

Step-2:

Now substituting the given values into the equation to get the value of the angular acceleration.

$\begin{aligned}a &=\omega^{2} r \\&=(5 \mathrm{rev} / \mathrm{min})^{2}(15 \mathrm{~m}) \\&=\left(5 \mathrm{rev} / \mathrm{min} \times \frac{\left(\frac{2 \pi}{60}\right) \mathrm{rad} / \mathrm{sec}}{1 \mathrm{rev} / \mathrm{min}}\right)^{2}(15 \mathrm{~m}) \\&=(0.2741 \times 15) \mathrm{m} / \mathrm{s}^{2} \\&=4.11 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}$