You need to determine the density of an unknown liquid and decide to perform an experiment. You notice that a wooden block float
1 answer:
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Answer:
The volume of the larger cube is 5.08 g/cm³.
Explanation:
Given that,
Mass of smaller cube = 20 g
Density of smaller cube
Dylan has two cubes of iron.
The larger cube has twice the mass of the smaller cube.
Density is same for both cubes because both cubes are same material.
The density is equal to the mass divided by the volume.
Where, V = volume
m = mass
We need to calculate the volume of smaller mass
The volume of smaller mass
Now, We need to calculate the volume of large cube
Hence, The volume of the larger cube is 5.08 g/cm³.
Answer:
Option B, 93 cm
Explanation:
An diagram of the seed's motion is attached to this solution.
This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.
And this would be obtained from the equations of motion,
First of, the height of the plant is related to some quantities of the motion with this relation.
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)
(substituting the t = √(2H/g) derived from above
R = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2
R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.
QED!
Answer:
Explanation:
For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:
which can also be rewritten as
In our case, we have:
is the initial pressure
is the initial volume
is the final pressure
Solving for V2, we find the final volume: