Answer:
<h2>
206.67N</h2>
Explanation:
The sum of force along both components x and y is expressed as;
The magnitude of the net force which is also known as the resultant will be expressed as 
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
Similarly,
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
Answer:
The final velocity of the thrower is 
and the final velocity of the catcher is 
.
Explanation:
Given:
The mass of the thrower, 
.
The mass of the catcher, 
.
The mass of the ball, 
.
Initial velocity of the thrower, 
Final velocity of the ball, 
Initial velocity of the catcher, 
Consider that the final velocity of the thrower is 
. From the conservation of momentum,
Consider that the final velocity of the catcher is 
. From the conservation of momentum,
Thus, the final velocity of thrower is 
and that for the catcher is 
.
Hélène de Pourtalès she was the first women
From the case we know that:
- The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
- The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
- The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².
Please refer to the image below.
We know from the case, that:
m = 2M
r = R
m2 = 1/2M
distance between the center of mass to point P = p = R
Distance of the point mass to point P = d = 2R
We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:
Icm = 1/2mr²
Icm = 1/2(2M)(R²)
Icm = MR² ... (i)
Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:
Ip = Icm + mp²
Ip = MR² + (2M)R²
Ip = 3MR² ... (ii)
Then, the total moment of inertia of the disk with the point mass is:
I total = Ip + I mass
I total = 3MR² + (1/2M)(2R)²
I total = 3MR² + 2MR²
I total = 5MR² ... (iii)
Learn more about Uniform Flat Disk here: brainly.com/question/14595971
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1 atm corresponds to 760 mmHg, so we can set up a simple proportion to find how many atmospheres correspond to 570 mmHg:
and from this, we find
