total gas pressure= pressure of 1st gas+pressure of 2nd gas. ⇒1.22atm=0.58atm+unknown atm
1.22atm-0.58atm=unknown atm
unknown atm=0.64atm
Answer:
8 J and 2 J
Explanation:
Given that,
Mass of the rubber ball, m = 1 kg
Initial speed of the rubber ball, u = 4 m/s (in east)
Final speed of the rubber ball, v = -2 m/s (in west)
We need to find the kinetic energy of the ball before it hits the wall, the kinetic energy of the ball after it bounces off the wall.
Initial kinetic energy,
Final kinetic energy,
So, the initial kinetic energy is 8 J and the final kinetic energy is 2 J.
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:
Explanation:
This problem is based on conservation of rotational momentum.
Moment of inertia of rod about its center
= 1/12 m l² , m is mass of the rod and l is its length .
= 1 / 12 x 4.6 x .11²
I = .004638 kg m²
The angular momentum of the bullet about the center of rod = mvr
where m is mass , v is perpendicular component of velocity of bullet and r is distance of point of impact of bullet fro center .
5 x 10⁻³ x v sin60 x .11 x .5 where v is velocity of bullet
According to law of conservation of angular momentum
5 x 10⁻³ x v sin60 x .11 x .5 = ( I + mr²)ω , where ω is angular velocity of bullet rod system and ( I + mr²) is moment of inertia of bullet rod system .
.238 x 10⁻³ v = ( .004638 + 5 x 10⁻³ x .11² x .5² ) x 12
.238 x 10⁻³ v = ( .004638 + .000015125 ) x 12
.238 x 10⁻³ v = 55.8375 x 10⁻³
.238 v = 55.8375
v = 234.6 m /s