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3 years ago

Going around a ferris wheel without stopping is an example of _____ circular motion(apex)

Physics

2 answers:

Natasha_Volkova [10]3 years ago

5 0

Explanation:

Going around a Ferris wheel without stopping is an example of uniform circular motion. In a circular motion, the objects moves with constant speed but its velocity is continuously changing because of changing direction.

The force acting when the object when it is moving in a circular motion is called centripetal force. The direction of both force and acceleration is inwards.

Hence, "going around a Ferris wheel without stopping" is an example of uniform circular motion.

victus00 [196]3 years ago

3 0

going around a ferris wheel without stopping is an example of uniform circular motion (apex).

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An object of mass m moving at a speed of v1 possesses kinetic energy that is equal to KE1. When the object's speed is doubled, t

Vera_Pavlovna [14]

Answer:

KE₂ = 4KE₁

Explanation:

KE₁ = ½mv₁²

KE₂ = ½mv₂²

KE₂ = ½m(2v₁)²

KE₂ = 4(½mv₁²)

KE₂ = 4KE₁

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2 years ago

2. A net force is _____.

Art [367]

The answer would be "the vector sum of forces acting on a particle or body."

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2 years ago

Suppose you have a 136-kg wooden crate resting on a wood floor. The coefficient of static friction is 0.4 and coefficient of kin

stealth61 [152]

Answer:

The the maximum force acting on the crate is 533.12 newtons.

Explanation:

It is given that,

Mass of the wooden crate, m = 136 kg

The coefficient of static friction, $\mu_s=0.4$

The coefficient of kinetic friction, $\mu_k=0.2$

We need to find the maximum force exerted horizontally on the crate without moving it. As the crate is not moving than the coefficient of static friction will act and the force is given by :

$F=\mu_s mg$

$F=0.4\times 136\ kg\times 9.8\ m/s^2$

F = 533.12 N

So, the maximum force acting on the crate is 533.12 newtons. Hence, this is the required solution.

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3 years ago

Convert 1.3 km to m

a_sh-v [17]

Answer: 1300m

Explain: from km to m times 1000

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2 years ago

What is the magnitude of the angular momentum relative to the origin of the 100 g particle in the figure(figure 1 ? express your

torisob [31]

Radius distance from origin to particle = √ (2²+1²) = √5 m = R
I = MR² = (0.200)(5) = 1.00 kg-m²
Θ = arctan 2/1 = 63.4° = R's angle CCW from horizontal
V = 3.0 m/s
V component that is at 90° to R = 3.0(sin 90°- 63.4°) = 3.0(sin 26.6°) = 1.3433 m/s
w = [V component / R] = 1.3433/√5 = 0.601 rad/s
size of angular momentum of particle relative to origin = Iw = (1.00)(0.601) = 0.601 kgm²/s

i hope I'm right

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3 years ago