Answer:
The acceleration of the snowball is 0.3125
Explanation:
The initial speed of the snowball up the hill, u = 0
The speed the snowball reaches, v = 5 m/s
The length of the hill, s = 40 m
The equation of motion of the snowball given the above parameters is therefore;
v² = u² + 2·a·s
Where;
a = The acceleration of the snowball
Plugging in the values, we have;
5² = 0² + 2 × a × 40
∴ 2 × 40 × a = 5² = 25
80 × a = 25
a = 25/80 = 5/16
a = The acceleration of the snowball = 5/16 m/s².
The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .
yes is is true cuz when you hold the thermometer by a bulb the heat coming from the bulb will increase the thermometer
Answer:
a) time t1 = 2.14s
b) initial angular speed w1 = 6 rad/s
Explanation:
Given that;
Initial Angular velocity = w1
Angular distance = s = 65 rad
time = t = 5 s
Angular acceleration a = 2.80 rad/s^2
Using the equation of motion;
s = w1t + (at^2)/2
w1 = (s-0.5(at^2))/t
Substituting the values;
w1 = (65 - (0.5×2.8×5^2))/5
w1 = 6rad/s
Time to reach w1 from rest;
w1 = at1
t1 = w1/a = 6/2.8 = 2.14s
a) time t1 = 2.14s
b) initial angular speed w1 = 6 rad/s