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oee [108]
3 years ago
8

Stat the law of conservation of momentum​

Physics
1 answer:
Alex787 [66]3 years ago
3 0

Answer:

The law has three definitions

Firstly,

The law of conservation of linear momentum states that in any system of colliding objects the total momentum is always conserved provided that there is no net external force acting on the system.

Secondly

If two or more bodies collide in a closed system,the total momentum after the collision is equal to the total momentum before the collision

Thirdly,

The total momentum of an isolated or closed system of colliding bodies remains constant

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sattari [20]
On the Newtonian theory of gravity, gravitation affects anything with mass. Assuming that none of the answer choices is the only thing that exists in the universe, all of the answer choices are subject to the law of universal gravitation (hence “universal”).

Satellites, water, frogs, and stars all have mass as they are all composed of matter. Thus, all four answer choices should be circled.
3 0
3 years ago
If there are 50 people per square kilometer in a city, and the area of the city is 1.5 × 10 square kilometers. What is the total
qaws [65]

Answer:

750 people

Explanation:

From the question,

Number of people in the city = population density×Area of the city

N = D×A.......................... Equagtion 1

Where N = Number of people in the city, D = population density, A = Area of the city.

Given: D = 50 people per square kilometer, A = 1.5×10 square kilometer.

Substitute into equation 1

N = 50(1.5×10)

N = 750 people.

Hence the total number of people in the city is 750 people.

6 0
3 years ago
If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m. s
Sladkaya [172]
 the orbital period is 5170 s
6 0
3 years ago
How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to
finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is x=\sqrt{\frac{2\times EPE}{k}}

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:

EPE=\frac{1}{2}kx^2

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

x=\sqrt{\frac{2\times EPE}{k}}

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0
3 years ago
Do anyone k ow the answers?
OLga [1]
Hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
7 0
3 years ago
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