Answer:
1.1 x 10⁵m/s²
Explanation:
Given parameters:
Velocity = 452m/s
distance = 0.93m
Unknown:
Acceleration of the bullet = ?
Solution:
To solve this problem, we use one of the kinematics equation which is given below:
V² = U² + 2aS
V is the final velocity
U is the initial velocity = 0m/s
a is the unknown acceleration
S is the distance traveled
So;
452² = 0² + (2 x a x 0.93)
204304 = 1.86a
a = 1.1 x 10⁵m/s²
Uhm.. based on what? Can you please specify/add an image or some more text and elaborate?
20 m away because it’s basically 60- 10 x 4 = ?
The answer is D. i got all the other ones wrong ._.
Explanation:
It is given that,
Initially, the jogger is at rest u₁ = 0
He accelerates from rest to 4.86 m, v₁ = 4.86 m
Time, t₁ = 2.43 s
A car accelerates from u₂ = 20.6 to v₂ = 32.7 m/s in t₂ = 2.43 s
(a) Acceleration of the jogger :
a₁ = 2 m/s²
(b) Acceleration of the car,
a₂ = 4.97 m/s²
(c) Distance covered by the car,
d₁ = 5.904 m
Distance covered by the jogger,
d₂ = 64.73 m
The car further travel a distance of, d = 64.73 m - 5.904 m = 58.826 m
Hence, this is the required solution.