Answer:
A) 5 m/s/s
Explanation:
<u>Given the following data;</u>
Initial velocity = 10m/s²
Final velocity = 20m/s²
Time, t = 2 seconds.
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
Substituting into the equation, we have;
<em>Acceleration, a = 5m/s²</em>
Answer:
8.9
Explanation:
We can start by calculating the initial elastic potential energy of the spring. This is given by:
(1)
where
k = 35.0 N/m is the initial spring constant
x = 0.375 m is the compression of the spring
Solving the equation,
Later, the professor told the student that he needs an elastic potential energy of
U' = 22.0 J
to achieve his goal. Assuming that the compression of the spring will remain the same, this means that we can calculate the new spring constant that is needed to achieve this energy, by solving eq.(1) for k:
Therefore, Tom needs to increase the spring constant by a factor:
Answer:
Her angular velocity when tucked is greater than when straight by a factor of 0.23
Explanation:
Moment of inertia (I) = mr^2 = mv^2/w^2
m is mass of the diver
v is diver's linear velocity
w is her angular velocity
When straight, I = 14 kg.m^2
mv^2/w^2 = 14
w^2 = mv^2/14
w = sqrt(mv^2/14) = 0.27sqrt(mv^2)
When tucked, I = 4 kg.m^2
w^2 = mv^2/4
w = sqrt(mv^2/4) = 0.5sqrt(mv^2)
Her angular velocity when tucked is greater than when straight by 0.23 (0.5 - 0.27 = 0.23)
Answer:
30600kg
Explanation:
density=m/v make v subject of the formula
