Answer:
v = 1.28 m/s
Explanation:
Given that,
Maximum compression of the spring, 
Spring constant, k = 800 N/m
Mass of the block, m = 0.2 kg
To find,
The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.
Solution,
When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,
Initial energy = Final energy
Substituting all the values in above equation,
v = 1.28 m/s
Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.
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:
what do u need help with
Explanation:
Calculate the escape velocity for the spacecraft. [G= 6.67×10^-11Nm^2kg^-2, mass of the Earth= 5.97×10^24kg, radius of the Earth= ...
A and C Im pretty sure :)
A. logic, would be your answer i believe!
