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8_murik_8 [283]
3 years ago
12

A ball is launched horizontally at 150 m/s from a cliff. What is its vertical speed after 3 seconds?

Physics
1 answer:
Setler79 [48]3 years ago
3 0

Answer:

v_f = 30 m/s

Explanation:

v_f = v_I + a*t

v_f = 0m/s + (9.81m/s²)*(3s)

v_f = 29.43m/s

v_f = 30 m/s

Since the object is being launched horizontally off the cliff the only vertical force acting on the ball is gravity which is an acceleration of 9.81m/s². So we have all the flowing information to use the the kinematic equation: v_f = v_I + a*t.

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E = \frac{2k\lambda sin\frac{\theta}{2}}{R}

here we know that

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\frac{\pi}{2}R = L = 2m

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also we know that

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now from above formula

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4 0
3 years ago
A 40kg mass is pulled by a horizontal force of 200N, with a constant velocity. What is h^2?
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What is the relationship between the center of gravity and the support base for an object that is in stable equilibrium?
Norma-Jean [14]

Answer:

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Explanation:

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When we analyze the balance of a body that is the torque it is the one that defines the balance

      τ = F xd

If the torque tends to restore the body to the initial position the balance is stable, but if the torque has to increase the body's rotation the balance is unstable

. When the body tends to rotate the torque with respect to the pivot point at the base it decreases because the distance from the center of gravity to the end of the base decreases in value, but it has to return it to the initial position, the balance is stable. The critical point of this process is when the center of gravity is at the limit of the body base area in this case the torque is zero; If the body rotates a little more the center of gravity is outside the base, the torque changes sign and has to increase the turn, going to an unstable balance.

In summary, when the center of gravity is within the washing area, the torque returns in the body to its initial position and is in a stable equilibrium.

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Explanation:

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