The mass of this bag of cement in S.I. units (kg) is equal to 0.062 kilograms.
<u>Given the following data:</u>
- Mass of cement = 62 grams.
To calculate the mass of this bag of cement in S.I. units (kg):
<h3>How to convert to
S.I. units.</h3>
In Science, kilograms (kg) is the standard unit of measurement or S.I. units of the mass of a physical object. Thus, we would convert the value of the mass of this bag of cement in grams to kilograms (kg) as follows:
<u>Conversion:</u>
1000 grams = 1 kilograms.
62 grams = X kilograms.
Cross-multiplying, we have:
X = 
X = 0.062 kilograms.
Read more on mass here: brainly.com/question/13833323
Answer:
A.) 42.7 m/s
B.) 0.33 m/s^2
C.) 90 kg
Explanation:
A.) If Justin races his Chevy S-10 down highway 37 north for 2,560 meters in 60 seconds, what is his velocity?
Velocity = displacement/time
Velocity = 2560/60
Velocity = 42.67 m/s
B.) The Chevy S-10 started rounding at 10 meters per hour. What is the acceleration at 30 seconds on the highway?
Acceleration = velocity/time
Acceleration = 10/30
Acceleration = 0.33 m/s^2
C.) The S-10 has a force of 30 N. What is the mass of the car?
Force = mass × acceleration
30 = mass × 0.33
Mass = 30/ 0.33
Mass = 90 kg
The mineral with Mohs hardness would be scratched because the mineral with Mohs 7 hardness is stronger than the Mohs 5 mineral. Eventually, that mineral would turn into dust if you kept rubbing it.
The force needed to accelerate an elevator upward at a rate of 
is 2000 N or 2 kN.
<u>Explanation:
</u>
As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.
As the object given here is an elevator with mass 1000 kg and the acceleration is given as , the force needed to accelerate it can be obtained by taking the product of mass and acceleration.
So 2000 N or 2 kN amount of force is needed to accelerate the elevator upward at a rate of .
Answer:
619.8 N
Explanation:
The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:
where
T is the tension
m is the mass of the rock
v is the speed
r is the radius of the circular path
At the beginning,
T = 50.4 N
v = 21.1 m/s
r = 2.51 m
So we can use the equation to find the mass of the rock:
Later, the radius of the string is decreased to
r' = 1.22 m
While the speed is increased to
v' = 51.6 m/s
Substituting these new data into the equation, we find the tension at which the string breaks:
