The two upper chambers of the human heart are known as:
2 answers:
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Answer:
The rock must leave the cliff at a velocity of 28.2 m/s
Explanation:
The position vector of the rock at a time t can be calculated using the following equation:
r = (x0 + v0x · t, y0 + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.
When the rock reaches the ground, the position vector will be (see r1 in the figure):
r1 = (90 m, -50 m)
Then, using the equation of the vector position written above:
90 m = x0 + v0x · t
-50 m = y0 + 1/2 · g · t²
Since x0 and y0 = 0:
90 m = v0x · t
-50 m = 1/2 · g · t²
Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:
-50 m = 1/2 · g · t²
-50 m = -1/2 · 9.81 m/s² · t²
-50 m / -1/2 · 9.81 m/s² = t²
t = 3.19 s
Now, using the equation of the x-component of r1:
90 m = v0x · t
90 m = v0x · 3.19 s
v0x = 90 m / 3.19 s
v0x = 28.2 m/s
Answer:
F= 600 N
Explanation:
Given that
Initial velocity ,u= 0 m/s
Final velocity ,v= 30 m/s
mass ,m = 0.5 kg
time ,t= 0.025 s
The change in the linear momentum is given as
ΔP= m (v - u)
ΔP= 0.5 ( 30 - 0 ) kg.m/s
ΔP= 15 kg.m/s
We know that from second law of Newtons
Now by putting the values
F= 600 N