Answer:
4.64m/s
Explanation:
We can use the formula [ v = √2gh ] to solve for this problem. We know that g is constant acceleration (9.8), and h is height (1.1).
v = √2(9.8)(1.1)
v ≈ 4.64m/s
Best of Luck!
Answer:
d.-10.3m
Explanation:
Note for short sightedness the focal length is negative
Let do be object distance=10m
And di= image distance=-300m
Using lens formula
F=do*di/do-di= 10*300/10-300=-10.3m
Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Answer:
Acceleration: 
Explanation:
The acceleration of an object is equal to the rate of change of velocity:
where
u is the initial velocity
v is the final velocity
t is the time taken for the velocity to change from u to v
For the space probe in this problem, we have:
u = 100 ft/s (initial velocity)
v = 5000 ft/s (final velocity)
t = 0.5 s (time taken)
Therefore, the acceleration is
A little confused by the wording of the problem, but it is true that an object can have a negative acceleration and be speeding up in the negative direction… so I’d go with True
