PLEASE HELP PHYSICS
Wiley coyote pushes a 5kg boulder from an initial height of 10m, the boulder rolls down the hill, across the road, and up the hill on the left. The rock is able to reach a height of 17 m before rolling back down. How fast did Wiley have to push the rock to make this happen?
Answers
Answer:
v = 11.7 m / s
Explanation:
For this exercise we can use the conservation of energy
starting point. Where you drop the stone
Em₀ = K + U = ½ m v² + m g h₁
final point. Point where the stone arrives
Em_f = U = mgh₂
We are assuming that there is no friction, so the energy is conserved
Em₀ = Em_f
½ m v² + m g h₁ = m g h₂
v² = 2g (h₂-h₁)
let's calculate
v² = 2 9.8 (17 - 10)
v = √137.2
v = 11.7 m / s
Related Questions
If energy is conserved, then:
O A. initial (PE + KE) = final (PE + KE).
O B. the initial KE must be zero.
O C. the momentum does not change.
O D. PE(before) = KE(before).
Answers
If energy is conserved, then initial (PE + KE) = final (PE + KE). So, the correct option is A.
What is meant by Law of Conservation of Energy ?The law of conservation of energy states that, energy can neither be created nor be destroyed, but can be transformed from one form to another.
Here,
According to law of conservation of energy,
The total energy of an isolated system remains constant. That means, the total energy of the system in the initial state will be same as that in the final state.
The total mechanical energy is the sum of kinetic energy and potential energy.
TE = KE + PE
Therefore, the energy to be conserved in the system,
Initial TE = Final TE
So, Initial (KE + PE) = Final (KE + PE)
Hence,
If energy is conserved, then initial (PE + KE) = final (PE + KE).
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What is the net force acting on a 52 kg object that has a velocity of 8.0 m/s and is moving in a circle of radius 1.6 m? a. 4000N b. 20880N c. 2500N d. 3500N
Answers
Σf = m v^2 / r
Σf = 52 8^2 / 1.6
Σf = 2080 N
Given values are:
Mass, m = 52 kgVelocity, v = 8.0 m/sRadius, r = 1.6 mAs we know the formula,
→ [tex]\Sigma f = ma[/tex]
or,
→ [tex]\Sigma f = \frac{mv^2}{r}[/tex]
By putting the values, we get
[tex]= \frac{52\times 8^2}{1.6}[/tex]
[tex]= \frac{52\times 64}{1.6}[/tex]
[tex]= 2080 \ N[/tex]
Thus the response above is appropriate.
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A 1500 kg car is being lifted by a hydraulic jack attached to a flat plate. Underneath the plate is a pipe with radius of 24 cm at the top and 2cm at the bottom. To generate an upward acceleration for the car of 1.0 m/s2, how much force must be applied to the small end of the pipe? The answer is NOT 102 N btw
Answers
Answer:
The force that must be exerted at the other end is 102.08 N/m₂
Explanation:
Using Pascal's formula; P₁ = P₂
P₁ = F₁/A₁ = m*g/(πr²)
m = 1500 kg; g = 9.8 m/s², r = 24 cm = 0.24 m
P₁ = (1500 * 9.8)/(22/7 * 0.24 * 0.24)
P₁ = 81235 N/m²
P₂ = F₂/A₂
A₂ = πr² where π = 22/7; r = 2.00 cm = 0.02m
A₂ = 22/7 * 0.02 * 0.02 = 0.00126 m²
P₂ = F₂ / 0.00126 m₂
substituting in the formula P₁ = P₂
81235 N/m² = F₂ / 0.00126 m²
F₂ = 81235 N/m² * 0.00126 m²
F₂ = 102.08 N/m₂
Explanation:
Answer:Why does water stop flowing out of the top hole even before the water level falls ... Calculate pressure on the bottom of the pool due to the water. 3. ... 2. If the ratio of A2 to A1 was 100, what force would be required to lift a 10000 N car? ... A cylinder with a radius of 11 cm and a height of 3.4 cm has a mass of 10.0 kg. a.
Explanation: