A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.796 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s

distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

Explanation:

Given that;

mass of block m = 0.200 kg

distance travelled d = 0.796 m

time t = 2.00 s

m₂ = 0.400 kg

If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?

Now, using the second equation of motion;

d = ut + ( × at²)

as the object started from rest, u=0

so, we substitute

0.796  = 0×2 + ( × a(2)²)

0.796  = 0 + ( × 4a)

0.796  = 2a

a = 0.796 / 2

a = 0.398 m/s²

using first equation of motion

= u + at

we substitute

= 0 + 0.398 × 2

= 0.796 m/s

now, average velocity is given as;

= ( 0.796 m/s  + 0 ) / 2

= ( 0.796 m/s  + 0 ) / 2

now, distance as the block moves in 2s will be;

D = [( 0.796 m/s  + 0 ) / 2 ] × 2

D = 0.796 m

Therefore, distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

Independent variable: the best method to get rid of them.

Dependent variable: washing with soap and water.

Hypothesis: Organic oils

Control group: water

Answer:

Step-by-step explanation:

D.
Heat each cube to the same temperature, place each cube into different containers with 500 grams of water at the same temperature, and measure the temperature of the water.

Answer:

Step-by-step explanation:

Time taken by the tomatoes to each the ground

using h = 1/2 g t^2 

t^2 = 2h/g = 2 x 50/ 9.8 = 10.2

t = 3.2 sec 

horizontal ditance = speed x time = 3 x 3.2 = 9.6 meters