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Dmitrij [34]
3 years ago
15

If an object has a mass of 20 grams and a volume of 40 cm3, what is its density in g/cm3?

Physics
1 answer:
sattari [20]3 years ago
3 0
Density = Mass/Volume 
So, given mass = 20 g and volume = 40 cm^3 
By substituting in above equation, Density = 20/40 = 0.5 g/cm^3
Hope it helps.

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A toy car starts from the rest and accelerates at 1.50m/s2 [E] for 2.25s. What is the final velocity, of the car
V125BC [204]

3.375m/s is the final velocity of the car.

<h3>How do you find final velocity?</h3>

The final velocity depends on how large the acceleration is and the distance over which it acts.

Initial velocity of an object, you can multiply the acceleration due to a force by the time the force is applied and add it to the initial velocity to get the final velocity.

According to the question,

A toy car starts from the rest and accelerates

So the acceleration = 1.50m/s²

Time =  2.25s

x=x_{0}  + vt

x = 0 + ( 1.50m/s^2*2.25s)

x = 3.375m/s

The final velocity, of the car is 3.375 m/s.

Learn more about velocity here:brainly.com/question/18084516

#SPJ1

8 0
1 year ago
A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e
mafiozo [28]

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

 ∑ M = 0

mass of the log will be concentrated at the center  

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)

200 \times 0.1 L = 53.5 \times (x - 0.6 L)

0.374 L =x - 0.6 L

       x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

7 0
3 years ago
A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski
Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0
3 years ago
What is the mass of a man who accelerates 4 m/s2 under the action of a 200 N net force?
Over [174]

Answer:

\huge  \boxed{ \boxed{50 \:   kg }}

Explanation:

The mass of the man can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{200}{4}  \\

We have the final answer as

<h3>50 kg</h3>

Hope this helps you

7 0
2 years ago
Anyone good with scientific notations? I sure am not
defon
B is the answer that I know of.
7 0
3 years ago
Read 2 more answers
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