Answer:
The final speed of the two carts is 1 m/s.
Explanation:
let m = 0.5 kg being the mass of the cart moving at an intial speed of v = 4 m/s and M = 1.5 kg is the mass of the cart at rest V = 0 m/s.
then by conservation of linear momentum, the final speed of the objects Vf is:
m×v + M×V = Vf×(M + m)
Vf = [m×v + M×V]/(M + m)
Vf = [(0.5 kg)×(4 m/s) + (1.5 kg)×(0 m/s)]/(1.5 + 0.5 kg)
= 1 m/s
Therefore, the final speed of the two carts is 1 m/s.
We know the equation for calculation of speed of sound at temperature:
<span>v = 331 m/s + (0.6 m/s/C) * T
</span>Here, T = 30 C
v = 331 m/s + (0.6 m/s/C) * 30 C
v = 331 m/s + 18 m/s
v = 349 m/s
In short, Your Answer would be 349 m/s
Hope this helps!
Work= force (N) x distance (m)
F= mass (kg) x acceleration (gravity; m/s^2)
for this question, your formula would be
Work= mass (kg) x acceleration (gravity) x distance (m)
a. F=20kg x 9.81 m/s^2 x6 m = 1177.2 J
b. F=25kg x 9.81 m/s^2 x3 m = 735.75 J
c. F=25kg x 9.81 m/s^2 x6 m = 1471.5 J
d. F=50kg x 9.81 m/s^2 x1 m = 490.5 J
Tip:
- If this was a timed test, you could save some time by just multiplying the mass (kg) in the question by the distance because 9.81 is a constant in the formula, you can ignore it (+you're not asked for the final answer). c would still be the largest number
- it would also help if you noticed:
- the distance in c is half that of b but the mass is the same. eliminate b because c is obviously bigger
- a and c have the same distance but 25 is greater than 20; eliminate a
Given; Mass m = 65.0 Kg; acceleration a = 1.25 m/s²
Required: Force F = ?
Formula: F =ma
F = (65.0 Kg)(1.25 m/s²)
F = 81.25 N
The sound gets louder as it gets closer and when it passes is gets softer