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3 years ago

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____________occurs when an object travels in a curved path.

1 answer:

labwork [276]3 years ago

4 0

Answer:

Acceleration

Explanation:

can you mark me brainlies

So, if an object travels in a curved path, it changes velocity, and, thus, accelerates. This acceleration must be tied to a force. ... Therefore, whenever an object travels in a curved path, there must be an unbalanced force acting upon it. It is important to understand that all this may occur without a change in speed.t

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Consider this situation: A rope is used to lift an actor upward off the stage. Of the forces listed, identify which act upon the

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4 years ago

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Where should you place the values of the independent variable when constructing a data table?

kobusy [5.1K]

<span>On the y-axis (the bottom of the table) hope this helps</span>

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3 years ago

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a parent swings a 18.5 kg child in a circle of radius 1.05m, making 5 revolutions in 13.4s. what centripetal acceleration does t

Zolol [24]

Answer: $0.146 m/s^{2}$

Explanation:

The <u>centripetal acceleration</u> $a_{c}$ of an object moving in a uniform circular path is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$V$ is the tangential velocity

$r=1.05 m$ is the radius of the circle

On the other hand, the tangential velocity is expressed as:

$V=\omega r$ (2)

Where $\omega$ is the angular velocity, which can be found knowing the child makes 5 revolutions in 13.4s:

$\omega=\frac{5 rev}{13.4 s}=0.37 rev/s$ (3)

Substituting (3) in (2):

$V=(0.37 rev/s)(1.05 m)$ (4)

$V=0.39 m/s$ (5)

Substituting (5) in (1):

$a_{c}=\frac{(0.39 m/s)^{2}}{1.05 m}$ (6)

Finally:

$a_{c}=0.146 m/s^{2}$

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3 years ago

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If R = 12 cm, M = 520 g, and m = 20 g (below), find the speed of the block after it has descended 50 cm starting from rest. Solv

timofeeve [1]

Answer:

v = 0.84 m/s

Explanation:

given,

R = 12 cm

M (mass of pulley )= 520 g

m (mass of block)= 20 g

s = 50 cm = 0.5 m

using conservation of energy

Potential energy = Kinetic energy

$m g h = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

$I_{disk}= \dfrac{1}{2}MR^2$ and v = r ω

$m g h = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{1}{2}MR^2)(\dfrac{v}{R})^2$

$m g h = \dfrac{1}{2}mv^2 +\dfrac{1}{4}Mv^2$

$m g h = \dfrac{1}{2}v^2(m +\dfrac{1}{2}M)$

$v=\sqrt{\dfrac{2mgh}{m + 0.5 M}}$

$v=\sqrt{\dfrac{2\times 0.020 \times 9.8 \times 0.5}{0.02 + 0.5\times 0.52}}$

v = √0.7

v = 0.84 m/s

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3 years ago

A train of mass 5.6 × 10^5kg is at rest in a station.at time t=0s, a resultant force acts on the train and it starts to accelera

lana66690 [7]

Answer:

420000N

Explanation:

Given parameters:

Mass of the train = 5.6 x 10⁵kg

Acceleration = 0.75m/s²

Unknown:

Resultant force = ?

Solution:

According to newton's second law, force is the product of mass and acceleration;

Force = mass x acceleration

Resultant force that acts on the train is given below;

Force = 5.6 x 10⁵kg x 0.75m/s² = 420000N

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3 years ago