a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )
2 = ( v(2) - 10 ) ÷ ( 6 - 0 )
2 × 6 = v(2) - 10
v(2) = 12 + 10
v(2) = 22 m/s
Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.
Answer: force = mass x accelegation
Explanation: 15x3 = 45 N
The velocity of the canoe is 1.7 m/s.
<h3>What is momentum?</h3>
Momentum in physics is the products of mass and velocity. Now we have to find momentum with the formula; p = mv
a) Initial momentum = (15)8 m/s + 135 = 255 Kgms-1
b) Since momentum is conserved, the total momentum after throwing the anchor is still 255 Kgms-1
c) The final velocity of the boat is obtained from;
255 Kgms-1 = (15Kg + 135 Kg) v
v = 255 Kgms-1/(15Kg + 135 Kg)
v = 1.7 m/s
Learn more about momentum: brainly.com/question/904448
4*5=20; ration of 20:25,
20/25= %80