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V125BC [204]
4 years ago
5

In a collision, a 15 kg object moving with a velocity of 3 m/s transfers some of its momentum to a 5 kg object. What would be th

e velocity of the 5 kg object after the collision if the 15 kg object is still moving at 1 m/s?
Physics
1 answer:
Misha Larkins [42]4 years ago
4 0

The key to solve this problem is the conservation of momentum. The momentum of an object is defined as the product between the mass and the velocity, and it's usually labelled with the letter p:

p=mv

The total momentum is the sum of the momentums. The initial situation is the following:

m_A=15,\quad v_A=3,\quad m_B=5,\quad v_B=0

(it's not written explicitly, but I assume that the 5-kg object is still at the beginning).

So, at the beginning, the total momentum is

p=m_Av_A+m_Bv_B=15\cdot 3+5\cdot 0=45

At the end, we have

m_A=15,\quad v_A=1,\quad m_B=5,\quad v_B=x

(the mass obviously don't change, the new velocity of the 15-kg object is 1, and the velocity of the 5-kg object is unkown)

After the impact, the total momentum is

p=m_Av_A+m_Bv_B=15\cdot 1+5\cdot x=15+5x

Since the momentum is preserved, the initial and final momentum must be the same. Set an equation between the initial and final momentum and solve it for x, and you'll have the final velocity of the 5-kg object.

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Cells are small because they need to keep a surface area to volume ratio that allows for adequate intake of nutrients while being able to excrete the cells waste.

Explanation:

That is why the cell needs to be small

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3 years ago
What would we see looking across a flat land of the earth was curved?
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the modulus of the vector product of two vector is 0.577 times their scalar product the angle between vectors​
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3 years ago
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o
Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

Learn more about Tension here brainly.com/question/2287912

#SPJ1

8 0
2 years ago
A 600 N force acts on an object with a mass of 50 kg. What is the resulting acceleration of the object?
DochEvi [55]

Answer:

<h3>The answer is 12 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{600}{50}  =  \frac{60}{5}  \\

We have the final answer as

<h3>12 m/s²</h3>

Hope this helps you

4 0
3 years ago
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