Answer:
t = 10.4 s
Explanation:
given,
velocity of the police car = 18 m/s
reaction time of policeman = 0.8 s
acceleration of the policeman to catch up the speeder = 5 m/s²
time taken to catch the speeder be t₁
distance traveled by the police and the speeder will be same
t₁ = 9.6 s
total time taken by the policeman
= reaction time + t₁
= 0.8 + 9.6
t = 10.4 s
Total time taken by the policeman to catch the speeder is 10.4 s
Answer:
420.17 nm
Explanation:
Given:
f = 7.14 × 10¹⁴Hz
c = 3 × 10⁸ m/s
c = fλ
λ = c/f
= 420.17 nm
b) Explain which one(s) hit the ground last
Explanation:
Balls A and C have the same initial vertical velocity of zero. Therefore, they land at the same time.
Ball B has a positive initial vertical velocity, so it hits the ground last.
btw i'm not sure what subject to post this in since there's no astronomy subject.
Answer: the Moon.
Explanation:
Brightness of celestial object is given in terms of magnitude. Apparent magnitude reflects how bright an object appears from the Earth. The apparent magnitude of the Moon is (-12.7 )greater than the apparent magnitude of the Venus (-4.4). The more negative value, the brighter is the object. Thus, the brightness of the moon surpasses the brightness of Venus.
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
The acceleration of the bullet in the gun barrel can be calculated using the kinematic equation for motion. By substituying the given values into the equation, we find the acceleration to be approximately 1.095 x 10^5 m/s^2.
The subject of this question is Physics, specifically a topic under mechanics known as kinematics. The problem given can be solved using kinematic equations which are used to describe the motion of an object without considering the forces that cause it to move. In this case, the final velocity (vf) of the bullet is given as 452 m/s, the initial velocity (vi) is assumed to be 0 as it starts from rest, and the distance (d) is given as 0.93 m. We are asked to determine the value of acceleration (a).
Using the kinematic equation vf2 = vi2 + 2ad and substituting the given values, we get (452 m/s)2 = 0 + 2*a*0.93 m. We can rearrange to solve for acceleration to get: a = (452 m/s)2 / (2*0.93 m) = 109523.66 m/s2.
So, the acceleration of the bullet in the gun barrel is approximately 1.095 x 105 m/s2.
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