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Alborosie
2 years ago
7

Directions: Using the T-chart below, compare balanced forces and unbalanced forces.

Physics
1 answer:
Salsk061 [2.6K]2 years ago
8 0

Explanation:

unbalanced: a turning vehicle, apple falling on the ground, kicking a ball

balanced: floating on water, fruit hanging from tree, tug of war equally balanced teams

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D would be it

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cause none of the rest Makes sense to the book of the picture and I'm 100% sure =)

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Gravity provides a centripetal force on the Moon, helping it stay in orbit around Earth.
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Experiments using "optical tweezers" measure the elasticity of individual DNA molecules. For small enough changes in length, the
GalinKa [24]

Answer:

Spring constant, k = 0.3 N/m

Explanation:

It is given that,

Force acting on DNA molecule, F=1.5\ nN=1.5\times 10^{-9}\ N

The molecule got stretched by 5 nm, x=5\times 10^{-9}\ m

Let k is the spring constant of that DNA molecule. It can be calculated using the Hooke's law. It says that the force acting on the spring is directly proportional to the distance as :

F=-kx

k=\dfrac{F}{x}

k=\dfrac{1.5\times 10^{-9}\ N}{5\times 10^{-9}\ m}

k = 0.3 N/m

So, the spring constant of the DNA molecule is 0.3 N/m. Hence, this is the required solution.

8 0
3 years ago
A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

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36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

P = \ro g h = 1000 * 10 * 20 = 200000 Pa

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

P_1V_1 = P_2V_2

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

V_1 = Ah

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

P_1Ah_1 = P_2Ah_2

h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m

b) If the temperatures changes, we can still reuse the ideal gas equation above:

\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}

h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m

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3 years ago
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