Answer:
The speed of the 8-ball is 2.125 m/s after the collision.
Explanation:
<u>Law Of Conservation Of Linear Momentum</u>
The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:
P=mv
If we have a system of masses, then the total momentum is the sum of all the individual momentums:
When a collision occurs, the velocities change to v' and the final momentum is:
In a system of two masses, the law of conservation of linear momentum is simplified to:
The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.
After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.
The above equation is solved for v1':
The speed of the 8-ball is 2.125 m/s after the collision.
The amplitude of a wave corresponds to its maximum oscillation of the wave itself.
In our problem, the equation of the wave is
![y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])](https://tex.z-dn.net/?f=y%28x%2Ct%29%3D%20%280.750cm%29cos%28%5Cpi%20%5B%280.400cm-1%29x%2B%28250s-1%29t%5D%29)
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
and therefore this value corresponds to the amplitude of the wave.
Answer:
20cm
Explanation:
Hello!
remember that the condition for a body to be at rest is that the sum of its moments and its forces be zero,
To solve this problem you must draw the free body diagram of the stick (attached image) and sum up moments at point 0 (where the sharp is located), which results in the following equation
(100g)(40cm)=x(200g)
Answer:
<h2><em>
6000 counts per second</em></h2>
Explanation:
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;
2000 counts per second = 1 meter ... 1
In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;
x count per second = 3 meter ... 2
Solving the two expressions simultaneously for x we will have;
2000 counts per second = 1 meter
x counts per second = 3 meter
Cross multiply to get x
2000 * 3 = 1* x
6000 = x
<em></em>
<em>This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample</em>
