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3 years ago

An object started at

Physics

1 answer:

nataly862011 [7]3 years ago

5 0

Answer:

According to me, I think this is the answer

Explanation:

initial position= -22 meters.

this indicates that object is in the negative direction as a negative sign is used to represent the location.

displacement is indicated as -10.4 meters. this is also in the negative direction.

initial + displacement= final

= -22 + (-10.4)

= -32.4 meters

this would be the final position, away from the origin.

displacement is a vector quantity; both magnitude and direction matter.

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Kim is ice-skating going 4.6 m/s. What is her velocity after 10 seconds ?

MArishka [77]

This is a uniform rectilinear motion (MRU) exercise.

To start solving this exercise, we obtain the following data:

v = 4.6 m/s
d = ¿?
t = 10 sec

To calculate distance, speed is multiplied by time.

We apply the following formula: d = v * t.

We substitute the data in the formula: the <u>speed is equal to 4.6 m/s,</u> the <u>time is equal to 10 s</u>, which is left as follows:

$\bf{d=4.6\dfrac{m}{\not{s}}*10\not{s} }$

$\bf{d=46 \ m}$

Therefore, the speed at 10 seconds is 46 meters.

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2 years ago

Bowl of water A bowl full of water is sitting out in a pouring rainstorm. Its surface area is 500 cm2. The rain is coming straig

vova2212 [387]

Answer: 4.9 x 10-3 N

Explanation:

A = 500cm^2 = 5 x 10^-2 m^2

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R = 10^-3 g/cm^2.sec = 10^-2kg/m^2 . sec

Prain water = R / V = 10^-2 / 5 = 2 x 10-3 kg/m^3

For the stationary bowl,

dm/dt =pAv= RA

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3 years ago

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wlad13 [49]

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3 years ago

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navik [9.2K]

Answer:

Im pretty sure its y!

Explanation:

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3 years ago

a disk with a radius of 0.1 m is spinning about its center with a constant angular speed of 10 rad/sec. What are the speed and m

padilas [110]

The speed of the bug at the rim of the circular disk is $\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the acceleration of the bug is $\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

Further Explanation:

Given:

The angular velocity of the disk is $10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ .

The radius of the circular disk is $0.1\,{\text{m}}$ .

Concept:

The linear speed of the bug present at the rim of the circular disk is given by.

$v = r \times \omega$

Here, $v$ is the linear speed, $r$ is the radius of the disk and $\omega$ is the angular speed of the disk.

Substitute $0.1\,{\text{m}}$ for $r$ and $10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $\omega$ in above expression.

$\begin{aligned}v &= 0.1\,{\text{m}} \times 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\\end{aligned}$

The magnitude of the centripetal acceleration of the bug cling to the rim of the disk is given as.

${a_c} = \dfrac{{{v^2}}}{r}$

Substitute $1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $v$ and $0.1\,{\text{m}}$ for $r$ in above equation.

$\begin{aligned}a_c&=\dfrac{(1\text{m/s})^2}{0.1\text{m}}\\&=\frac{1}{0.1}\text{m/s}^2\\&=10\,\text{m/s}^2\end{aligned}$

Thus, the speed of the bug at the rim of the circular disk is $\boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the acceleration of the bug is $\boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ .

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Answer Details:

Grade: College

Chapter: Uniform Circular motion

Subject: Physics

Keywords: Circular disk, circular motion, angular speed, linear speed, bug clinging, rim of the disk, acceleration, magnitude, constant, rad/s.

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3 years ago

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