Solution :
Given weight of Kathy = 82 kg
Her speed before striking the water, = 5.50 m/s
Her speed after entering the water, = 1.1 m/s
Time = 1.65 s
Using equation of impulse,
Here, F = the force ,
dT = time interval over which the force is applied for
= 1.65 s
dP = change in momentum
dP = m x dV
= 82 x (1.1 - 5.5)
= -360 kg
∴ the net force acting will be
= 218 N
Answer:
Yes
Explanation:
The spring force is given as:
F = kd
F is the spring force
K is the spring constant
d is the magnitude of the stretch
Since k is a constant, therefore, doubling the stretch distance will double the force.
Both stretch distance and force applied can be said to be directly proportional to one another.
13.0m/s
1.2m/s
Explanation:
Given parameters:
Initial speed of the body = 7.1m/s
time taken = 2.23s
Acceleration = 2.64m/s²
Unknown:
Final speed = ?
Solution:
Acceleration is the rate of change of velocity with time.
a =
a = acceleration
V = final speed
U = initial speed
T = time taken
Input the variables and solve for V;
2.64 =
V - 7.1 = 5.9 expression 1
V = 5.9 + 7.1 = 13.0m/s
B
Using the same parameters, the speed after a uniform deceleration of -2.64m/s², the negative sign implies deceleration;
from expression 1;
V - 7.1 = -5.9
V = -5.9 + 7.1 = 1.2m/s
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Answer:
Yes is large enough
Explanation:
We need to apply the second Newton's Law to find the solution.
We know that,
And we know as well that
Replacing the aceleration value in the equation force we have,
Substituting our values we have,
The weight of the person is then,
<em>We can conclude that force on the ball is large to lift the ball</em>
Answer:
v = 29.4m/s
Explanation:
Since the ball is dropped at rest,
u = 0m/s
a = 9.81m/s²
Using
v = u + at
After 3 seconds,
v = 0 + (9.81)(3)
v = 29.4m/s