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4 years ago

14

Bob and Lily are riding on a typical carousel. Bob rides on a horse near the outer edge of the circular platform, and Lily rides

on a horse near the center of the circular platform. When the carousel is rotating at a constant angular speed, Bob's angular speed is

1 answer:

alukav5142 [94]4 years ago

6 0

Answer:

Bob's angular speed is the same as that of lily

Explanation:

Because for a carousel the angular speed remains the same since velocity at center and edge are the same

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This ball is technology too! It can be<br> rolled, kicked, or thrown. Is that a need<br> or a want?

Inga [223]

Answer:

want

Explanation:

people always want the new thing especially if its technology

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3 years ago

Read 2 more answers

Solve for work when

BlackZzzverrR [31]

So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.

<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

$\boxed{\sf{\bold{W = F \times s}}}$

If the Angle Effect on Work

$\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}$

With the following condition:

W = work that done by object (J)
F = force that applied (N)
s = shift or distance (m)
$\sf{\theta}$ = angle of elevation (°)

<h3>Solution</h3>

We know that :

F = force that applied = $\sf{1.41 \times 10^4}$ N
s = shift or distance = 84.9 m
$\sf{\theta}$ = angle of elevation = 45°

What was asked ?

W = work that done by object = ... J

Step by step :

$\sf{W = F \times s \times \cos(\theta)}$

$\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}$

$\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}$

$\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}$

$\sf{W = 59.8545 \sqrt{2}}$

$\boxed{\sf{W \approx 84.65 \: J}}$

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

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1 year ago

How do you find the molarity of: 48.2 g of sodium carbonate in 0.500 L of solution

romanna [79]

The answer is why lani

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3 years ago

The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan

PSYCHO15rus [73]

Answer:

volume is 700 mL

Explanation:

pressure = 2 atm

volume = 350 mL = 0.350 L

to find out

volume

solution

we will apply here equation that is

P1×V1 = P2×V2 ..............1

here P1 = 2 and V1 = 0.350 and P2 = 1 for standard atmospheric pressure

so put all value here in equation 1 and get V2 volume

2 × 0.350 = 1 × V2

V2 = 0.700 L

V2 = 700 mL

so volume is 700 mL

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3 years ago

A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package hits the ground, how high

Lelu [443]

Answer:

The package was released at a height of 1015.296 meters.

Explanation:

The package is dropped at an initial velocity different of zero, decelerated and later accelerated by gravity. Let assume that final height is equal to zero, the final height is given by the following equation of motion:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$

Where:

$v_{o}$ - Initial velocity, measured in meters per second.

$y$ - Final height, measured in meters.

$y_{o}$ - Initial height, measured in meters.

$t$ - Time, measured in seconds.

$g$ - Gravitational constant, measured in meters per square second.

(Positive sign - Package is moving upward, Negative sign - Package is moving downward)

The initial height is now cleared:

$y_{o} = y - v_{o}\cdot t - \frac{1}{2}\cdot g \cdot t^{2}$

Given that $y = 0\,m$ , $v_{o} = 15\,\frac{m}{s}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t = 16\,s$ , the final height of the package is:

$y_{o} = 0\,m - \left(15\,\frac{m}{s} \right)\cdot (16\,s) - \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (16\,s)^{2}$

$y_{o} = 1015.296\,m$

The package was released at a height of 1015.296 meters.

7 0

3 years ago