Answer: 20 kgm/s
Explanation:
Given that M1 = M2 = 10kg
V1 = 5 m/s , V2 = 3 m/s
Since momentum is a vector quantity, the direction of the two object will be taken into consideration.
The magnitude of their combined
momentum before the crash will be:
M1V1 - M2V2
Substitute all the parameters into the formula
10 × 5 - 10 × 3
50 - 30
20 kgm/s
Therefore, the magnitude of their combined momentum before the crash will be 20 kgm/s
Answer:
Earth's water is always in movement, and the natural water cycle, also known as the hydrologic cycle, describes the continuous movement of water on, above, and below the surface of the Earth. Water is always changing states between liquid, vapor, and ice, with these processes happening in the blink of an eye and over millions of years.
Hope this helped!! :))
Explanation:
Answer:
A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
Explanation:
We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

Where:
v: is the tangential speed of the disk
t: is the time = 30 s
The tangential speed can be found as follows:

Where:
ω: is the angular speed = 100 rpm
r: is the radius = 50 cm = 0.50 m
Now, the distance traveled by the disk is:

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.
I hope it helps you!
Answer:
Stretch can be obtained using the Elastic potential energy formula.
The expression to find the stretch (x) is 
Explanation:
Given:
Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.
To find: Elongation in the spring (x).
We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).
The formula to find EPE is given as:

Rewriting the above expression in terms of 'x', we get:

Example:
If EPE = 100 J and spring constant, k = 2 N/m.
Elongation or stretch is given as:

Therefore, the stretch in the spring is 10 m.
So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.
Answer:
I would say the answer is A... but I'm not so sure ....