Answer:
d = 10.076 m
Explanation:
We need to obtain the velocity of the ball in the y direction
Vy = 24.5m/s * sin(35) = 14.053 m/s
To obtain the distance, we use the formula
vf^2 = v0^2 -2*g*d
but vf = 0
d = -vo^2/2g
d = (14.053)^2/2*(9.8) = 10.076 m
Answer:
–50.96
Explanation:
The following data were obtained from the question:
Initial velocity (Vᵢ) = 0 m/s
Acceleration (a) = – 9.8 m/s²
Time (t) = 5.2 s
Final velocity (Vբ) =.?
Acceleration is simply defined as the change of velocity with time. Mathematically, it is expressed as:
Acceleration (a) = Final velocity (Vբ) – Initial velocity (Vᵢ) /Time (t) =
a = (Vբ – Vᵢ) / t
With the above formula, we can determine how fast the object is traveling after 5 s as follow:
Initial velocity (Vᵢ) = 0 m/s
Acceleration (a) = – 9.8 m/s²
Time (t) = 5.2 s
Final velocity (Vբ) =.?
a = (Vբ – Vᵢ) / t
– 9.8 = (Vբ – 0) / 5.2
– 9.8 = Vբ / 5.2
Cross multiply
Vբ = –9.8 × 5.2
Vբ = –50.96 m/s
Therefore, the object is traveling at
–50.96 m/s
Answer:
The magnification of the lens is 0.7
Explanation:
We have,
Height of an object is 5 cm
Object is placed 4.25 cm from a lens
Image is formed 3.00 cm behind the lens.
It means u = -4.25 cm
v = -3 cm (behind the lens)
Magnification of the lens is given by :
So, the magnification of the lens is 0.7
About 5 billion years ago