Answer:
Try and depend more on renewable energy sources. Use products that are more energy efficient. Make use of lighting control measures. Maintain climate change
A. The acceleration of the ball while it is in flight?
magnitude is 0 m/s² (magnitude is zero)
B. The velocity of the ball when it reaches its maximum height is 0 m/s (magnitude is zero)
C. The initial velocity of the ball 8.036 m/s upward
D. The maximum height reached by the ball is 3.29 m
<h3>A. How to determine the acceleration in the flight</h3>
Considering that the ball came to rest after 1.64s, it means the entire acceleration of the flight is zero as the ball was not moving in any form again.
<h3>B. How to determine the velocity at maximum height</h3>
At maximum height, the velocity of the ball is zero as it no longer has magnitude to keep going upwards. Hence the ball begins to ball down.
<h3>C. How to determine the initial velocity</h3>
- Acceleration due to gravity (g) = 9.8 m/s²
- Final velocity (v) = 0 m/s
- Time of flight (T) = 1.64 s
- Time to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 s
- Initial velocity (u) =?
v = u - gt (since the ball is going against gravity)
0 = u - (9.8 × 0.82)
0 = u - 8.036
Collect like terms
u = 0 + 8.036
u = 8.036 m/s upward
<h3>D. How to determine the maximum height reached by the ball</h3>
- Time to reach maximum height (t) = T / 2 = 1.64 / 2 = 0.82 s
- Acceleration due to gravity (g) = 9.8 m/s²
- Maximum height (h)
h = ½gt²
h = ½ × 9.8 × 0.82²
h = 3.29 m
Learn more about motion under gravity:
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143m/s if you just perhaps by what you know you'll figure it out
Explanation:
LD₁ = 10⁵ mm⁻²
LD₂ = 10⁴mm⁻²
V = 1000 mm³
Distance = (LD)(V)
Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m
Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m
Conversion to miles:
Distance₁ = 10×10⁴ m / 1609m = 62 miles
Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.
<span>Answer:
Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by
h = 0.5*g*T^2
However, the speed of the package-rocket system must be sufficient to cross the river in that time
v2 = L/T
Conservation of momentum says that
m1*v1 = (m1 + m2)*v2
where m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system.
Expressing v2 in terms of v1
v2 = m1*v1/(m1 + m2)
and then expressing the time in terms of v1
T = (m1 + m2)*L/(m1*v1)
substituting T in the first expression
h = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2
solving for v1, the speed before impact is given by
v1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1</span>