Answer:
The jp2003parker guy is extremely wrong
So he says that the size wont matter and a physical change should occur, but how would the size change without having a physical change occur.
Explanation:
Answer: 42.49
Explanation:
To solve this, we need to keep in mind the following:
While the sphere hangs it is under the effect of gravity. It is creating a Angle of 90° taking the roof as a reference.
Gravity can be noted as a Acceleration Vector. The magnitud for Earth's Gravity is a constant: 9.81 
The acceleration of the Van will affect the sphere also, but this accelaration will be on the X-axis and perpendicular to the gravity. Because this two vectors are taking action under the sphere they will create a angle. This angle can be measured as a relation of the two magnitudes.
Tangent (∅) = Opossite Side / Adyacent Side
By trigonometry, we know the previous formula. This formula allows us to find the Tangent of a angle as a relation between the two perpendiculars magnitudes. In this case the Opossite Side will be the Gravity Accelaration, while the Adyancent Side is the Van's Acceleration.
(1) Tangent (∅) = Gravity's Acceleration (G) / Van's Acceleration (Va)
Searching for the Va in (1)
Va = G/Tan(∅)
Where ∅ in this case is equal to 13.0°
Va = 9.81
/ Tan(13.0°)
Va = 42.49
The vans acceleration need to be 42.49
to create an angle of 13° with the Van's Roof
Answer:
d ≈ 7,6 g/cm³
Explanation:
d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³
V = l³ = (1.74cm)³ ≈ 5,27 cm³
Mass of the object is given as

now the speed of object is given as

here we know that


now we will have

now we will have kinetic energy of the object as



now the power is defined as rate of energy
so here we can find power as


so above is the power used for the object
The highest elevation reached by the ball in its trajectory is 16.4 m.
To find the answer, we need to know about the maximum height reached in a projectile.
What's the mathematical expression of the maximum height reached in a projectile motion?
- The maximum height= U²× sin²(θ)/g
- U= initial velocity, θ= angle of projectile with horizontal and g= acceleration due to gravity
What's the maximum height reached by a block that is thrown with an initial velocity of 30.0 m/s at an angle of 25° above the horizontal?
- Here, U = 30.0 m/s and θ= 25°
- Maximum height= 30²× sin²(25)/9.8
= 16.4m
Thus, we can conclude that the highest elevation reached by the ball in its trajectory is 16.4 m.
Learn more about the projectile motion here:
brainly.com/question/24216590
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