True or False: There is no object that can travel faster than the speed of light.

A bag of cement has a mass of 62 g. What is the mass of the bag of cement in S.I. units (kg)?

NNADVOKAT [17]

The mass of this bag of cement in S.I. units (kg) is equal to 0.062 kilograms.

Given the following data:

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:

Conversion:

1000 grams = 1 kilograms.

62 grams = X kilograms.

Cross-multiplying, we have:

X = $\frac{62}{1000}$

X = 0.062 kilograms.

Read more on mass here: brainly.com/question/13833323

8 0

1 year ago

WILL MARK BRAINLIEST. Which gas became part of Earth's atmosphere as a result of the evolution of life-forms?

Fittoniya [83]

Answer:

Earth's atmosphere was largely made up of nitrogen and carbon dioxide gases. After photosynthesizing organisms multiplied on Earth's surface and in the oceans, much of the carbon dioxide was replaced with oxygen.

Idk if this helps but plz mark brainliest if it does

5 0

2 years ago

What is the mass of an object that has a weight of 110N ?

fgiga [73]

Weight (W) = 110 N
Acceleration due to gravity (g) = 9.8 m/s^2
Let the mass of the object be m.
By using the formula, W = mg, we get,
110 N = 9.8 m/s^2 × m
or, m = 110 N ÷ 9.8 m/s^2
or, m = 11.2 Kg

Answer:

The mass of the object is 11.2 Kg.

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0

1 year ago

If this decay has a half-life of 10.2 years, what mass of 60.8g carbon-14 will remain after 20.4 years

OLEGan [10]

20.4 years is 20.4/10.2 = 2 half-life cycles, which means a quarter of the starting mass or 15.2 g will remain after this time.

5 0

2 years ago

a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance

Marta_Voda [28]

a) 10 m/s

b) 25 m

Explanation:

a)

The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

$v=u+at$

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the body in this problem:

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

t = 5 s is the time

So, the final velocity is

$v=0+(2)(5)=10 m/s$

b)

In this second part, we want to calculate the distance travelled by the body.

We can do it by using another suvat equation:

$v^2-u^2=2as$

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the distance travelled

Here we have

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

v = 10 m/s is the final velocity

Solving for s,

$s=\frac{10^2-0^2}{2(2)}=25 m$

3 0

2 years ago