Answer:24
Explanation:
First you need to calculate the acceleration, for this use can use the following formula Force (F) = Mass (M) x Acceleration (A). If we extract A out of this, you can get the following equation A = F/M = 4/0.5 = 8 m/s^2. So, the object accelerates each second by 8 m/s, if the object does this for 3 second, the velocity would be 8 x 3 = 24
Given that,
Mass of a frog, m = 0.5 kg
The frog leaps pushing off of the rock with an acceleration of 5.0 m/s².
To find,
The magnitude of the net force exerted on the frog as it leaps.
Solution,
Let the net force exerted on the frog is given by the formula as follows :
F = ma
Putting the values of m and a to find F as follows :
F = 0.5 kg × 5 m/s²
F = 2.5 N
So, the magnitude of net force is 2.5 N.
Answer:
-4.8 m/s²
Explanation:
Apply Newton's second law:
∑F = ma
F − mg = ma
300 N − (60 kg) (9.8 m/s) = (60 kg) a
a = -4.8 m/s²
The acceleration five seconds after jumping is 4.8 m/s² downward.
Answer:
the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
Explanation:
This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.
Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.