Answer:
1.73 m/s²
3.0 cm
Explanation:
Draw a free body diagram of the yo-yo. There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.
Sum of forces in the y direction:
∑F = ma
T cos 10° − mg = 0
T cos 10° = mg
T = mg / cos 10°
Sum of forces in the x direction:
∑F = ma
T sin 10° = ma
mg tan 10° = ma
g tan 10° = a
a = 1.73 m/s²
Draw a free body diagram of the sphere. There are two forces: weight force mg pulling down, and air resistance D pushing up. At terminal velocity, the acceleration is 0.
Sum of forces in the y direction:
∑F = ma
D − mg = 0
D = mg
½ ρₐ v² C A = ρᵢ V g
½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g
3 ρₐ v² C = 8 ρᵢ r g
r = 3 ρₐ v² C / (8 ρᵢ g)
r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))
r = 0.030 m
r = 3.0 cm
Ofcoure the elephante as the elephant has more weight
The food substance being used which turned Fehling's solution to brick red is a reducing sugar.
<h3>What is Fehling's solution?</h3>
This is an indicator which is used to test for the presence of reducing sugar or aldehydes in a solution.
Brick red precipitate of copper(I) oxide is formed when it tests positive to compounds such as glucose etc.
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Gravity
The moon doesn't smash into the earth because the gravity from the earth keeps the moon in orbit around it.
-GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.
Given
A particle of mass m moving under the influence of a fixed mass's M, gravitational potential energy of formula -GMm/r, where r is the separation between the masses and G is the gravitational constant of the universe.
As the Gravity Potential energy of particle = -GMm/r
Total energy of particle = Kinetic energy + Potential Energy
As we know that
Kinetic energy = 1/2mv²
Also, v is equals to square root of GM/r
v = √GM/r
Put the value of v in the formula of kinetic energy
We get,
Kinetic Energy = GMm/2r
Total Energy = GMm/2r + (-GMm/r)
= GMm/2r - GMm/r
= -GMm/2r
Hence, -GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.
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