Assuming the object is on earth the objects weight would be equal to its mass multiplied by the gravitational field constant
mass=22kg
g=9.80665N/kg
weight=(22 kg) (9.80665 N/kg)=215.7463N
generally g is rounded to be 10 N/kg so for any question where it asks the weight given the mass just multiply by 10 and that should suffice. In this case the answer would be 220 N
Answer:
6.86 × 10²⁴ kg
Explanation:
The mass of the earth m = density of earth, ρ × volume of earth, V
m = ρV
The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m
Since m = ρV
m = ρ4πr³/3
So, substituting the values of the variables into the equation for the mass of the earth, m, we have
m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3
m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3
m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3
m = 6546105.64378π × 10¹⁸ kg/3
m = 20565197.400122 × 10¹⁸ kg/3
m = 6855065.8 × 10¹⁸ kg
m = 6.8550658 × 10²⁴ kg
m ≅ 6.86 × 10²⁴ kg
Answer:
v = ![\left[\begin{array}{c}0.66&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0.66%260%5Cend%7Barray%7D%5Cright%5D)
m/s
Explanation:
The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:
The velocity vector v is the first derivative of the position vector r with respect to time:
![\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv%7D%3D%5Bvcos%28%5Comega%20t%29-%5Comega%20vtsin%28%5Comega%20t%29%5D%5Chat%7Bx%7D%2B%5Bvsin%28%5Comega%20t%29%2B%5Comega%20vtcos%28%5Comega%20t%29%5D%5Chat%7By%7D)
The given values are:
Answer:
Final volumen first process 
Final Pressure second process 
Explanation:
Using the Ideal Gases Law yoy have for pressure:
where:
P is the pressure, in Pa
n is the nuber of moles of gas
R is the universal gas constant: 8,314 J/mol K
T is the temperature in Kelvin
V is the volumen in cubic meters
Given that the amount of material is constant in the process:
In an isobaric process the pressure is constant so:
Replacing : 
Replacing on the ideal gases formula the pressure at this piont is:
For Temperature the ideal gases formula is:
For the second process you have that 
So:
The stopwatch will be the most useful in determining the kinetic energy of a 50 g battery- powered car traveling a distance of 10 m.
<h3>What is kinetic energy?</h3>
Kinetic energy is the energy of a body possessed due to motion.
This means that for an object to possess kinetic energy, it must be in motion.
The kinetic energy is measured in Joules, which is a product of the mass of the substance and the time taken to travel a distance.
A stopwatch is an instrument used to measure time as one of the components of kinetic energy.
Therefore, the stopwatch will be the most useful in determining the kinetic energy of a 50 g battery- powered car traveling a distance of 10 m.
Learn more about kinetic energy at: brainly.com/question/12669551
