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3 years ago

What is an unbalanced force

1 answer:

7 0

There's no such thing as "an unbalanced force".

If all of the forces acting on an object all add up to zero, then we say that
the group of forces is balanced. When that happens, the group of forces
has the same effect on the object as if there were no forces on it at all.

An example:
Two people with exactly equal strength are having a tug-of-war. They pull
with equal force in opposite directions. Each person is sweating and straining,
grunting and groaning, and exerting tremendous force. But their forces add up
to zero, and the rope goes nowhere. The group of forces on the rope is balanced.

On the other hand, if one of the offensive linemen is pulling on one end of
the rope, and one of the cheerleaders is pulling on the other end, then their
forces don't add up to zero, because even though they're opposite, they're
not equal. The group of forces is unbalanced, and the rope moves.

A group of forces is either balanced or unbalanced. A single force isn't.

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A 250 g beach ball rolls across the sand with a speed of 11.16 km/h. First convert units to kg and m/ then determine the momentu

yuradex [85]

Answer:

0.775 kg-m/s

Explanation:

Convert the units to the right unit forms necessary

250 g -> 0.25 kg

11.16 km/h -> 3.1 m/s

Now use the formula:

velocity

mass /

momentum / /

\ / /

p = mv

p = 0.25 × 3.1 = 0.775 kg-m/s

Hope this helps you!

Bye!

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3 years ago

NEED HELP ASAP Which of the following chemical equations shows the Law of Conservation of Mass? (1 point)

natulia [17]

Answer:

OK + MgBR arrow KBR + MG

Explanation:

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1 year ago

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3 years ago

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3 years ago

A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo

m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 . t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

p₀ = m1 v₀₁

Moment after shock

$p_{f}$ = (m1 + m2) $v_{f}$

p₀ = $p_{f}$

m1 v₀₁ = (m1 + m2) $v_{f}$

$v_{f}$ = v₀₁ m1 / (m1 + m2)

$v_{f}$ = 4.4 600 / (600 + 500)

$v_{f}$ = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

Em₀ = K = ½ (m1 + m2) $v_{f}$ ²

After compressing the spring

$E_{mf}$ = Ke = ½ k x²

As there is no rubbing the energy is conserved

Em₀ = $E_{mf}$

½ (m1 + m2) $v_{f}$ ² = = ½ k x²

x = $v_{f}$ √ (k / (m1 + m2))

x = 2.4 √ (11/3000)

x = 0.396 m

7 0

3 years ago