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3 years ago

Forces that are equal in strength and opposite indirection are

Physics

2 answers:

blsea [12.9K]3 years ago

8 0

They are called balanced forces.

Otrada [13]3 years ago

4 0

Hey there! My name is Christy and I'm gladly to help you out!

The three main forces that stop moving objects are friction, gravity and wind resistance. Equal forces acting inopposite directions are called balanced forces. Balanced forcesacting on an object will not change the object's motion. When you add equal forces in opposite direction, the netforce is zero.

Hope this helped!

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A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb

NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

$\sum \tau = F*d$

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

$F*(27-6)= 6*600$

$F = \frac{6*600}{21}$

$F= 171.42 lb$

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

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3 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

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Sidana [21]

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The galaxy's diameter is a mere 1. 0 ly.

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