Answer:
y = 9.64 m
Explanation:
This exercise should be solved using kinematics in one dimension, let's write the equations for the two cases presented
The rock is released
y = y₀ + V₀₁ t₁ - ½ g t₁²
In this case the speed starts is zero
y = y₀ - ½ g t₁²
The rock is thrown up
y = y₀ + v₀² t₂ -½ g t₂²
The height that reaches the floor is zero
y₀ - ½ g t₁² = y₀ + v₀₂ t₂ - ½ g t₂²
We use the initial velocity with the equation
v₂² = v₀₂² - 2 g y
At the point of maximum height v₂ = 0
v₀₂ = √ (2 g
)
g (-t₁² + t₂²) = 2 √ (2 g
) t₂²
g (- 4.15² + 6.30²) = 2 √ (2 2 g) 6.3
g (22.4675) = 25.2 √ g
g² = 2²5.2 / 22.4675 g
g = 1.12 m / s²
Having the value of g we can use any equation to find the height
y = ½ g t₁²
y = ½ 1.12 4.15²
y = 9.64 m
Answer:
Explanation:
At the submicroscopic and microscopic levels temperature is proportional to the average degree of kinetic energy of the particles of a substance
Answer:
1980 kg m/s due south
8.2 m/s2 north-west
Explanation:
In order a quantity to be a vector, it should both has a magnitude and direction.
27 J/s --> Only magnitude. (Power)
1980 kg m/s due south --> Both magnitude and direction. (Momentum)
8.2 m/s2 north-west --> Both magnitude and direction. (Acceleration)
3.2 mi straight up --> Direction is not clear. (Position)
2.9 m/s2 --> Only magnitude. (Magnitude of acceleration)
293 K --> Only magnitude. (Temperature)
200 s --> Only magnitude. (Time)
Answer:
Heat is transfered via solid material (conduction), liquids and gases (convection), and electromagnetical waves (radiation).
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Answer:

Explanation:
Given:
Mass of the cannonball (M) = 20 kg
Mass of the marble (m) = 0.002 kg
Distance between the cannonball and marble (d) = 0.30 m
Universal gravitational constant (G) = 
Now, we know that, the gravitational force (F) acting between two bodies of masses (m) and (M) separated by a distance (d) is given as:

Plug in the given values and solve for 'F'. This gives,

The same force is experienced by both cannonball and marble.
Therefore, the gravitational force of the marble is 