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Physics

2 answers:

bogdanovich [222]3 years ago

8 0

C. unbalanced is the correct answer for newton's first law.

UkoKoshka [18]3 years ago

7 0

An object will remain at rest or maintain constant, straight-line motion unless acted on by a force that is C. unbalanced. Balanced forces allows the object to maintain its position or movement in equilibrium. Hence, the answer to this problem is C.

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A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N

Gre4nikov [31]

Answer:

<h2>f₀ = 158.12 Hertz</h2>

Explanation:

The fundamental frequency of the string f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.

$V = \sqrt{\frac{T}{\mu} }$ where T is the tension in the string and $\mu$ is the density of the string

Given T = 600N and $\mu$ = 0.015 g/cm = 0.0015kg/m

$V = \sqrt{\frac{600}{0.0015} }\\ \\V = \sqrt{400,000}\\ \\V = 632.46m/s$

The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.

L = 2m

Substituting the derived values into the formula f₀ = V/2L

f₀ = 632.46/2(2)

f₀ = 632.46/4

f₀ = 158.12 Hertz

Hence the fundamental frequency of the string is 158.12 Hertz

5 0

3 years ago

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

Misha Larkins [42]

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.