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timurjin [86]
3 years ago
11

The drawing shows a golf ball passing through a windmill at a miniature golf course. The windmill has 8 blades and rotates at an

angular speed of 1.25 rad/s. The opening between a successive blades is equal to the width of a blade. A golf ball (diameter 4.50 x 10^-2 m) is just passing by one of the rotating blades. What must be the minimum speed of the ball so that it will not be hit by the next blade?
Physics
1 answer:
Naily [24]3 years ago
5 0

<em>To understand the passage between two blades, it is required </em>

<em>to travel the distance of the circumference equivalent to the </em>

<em>segment of the diameter that exists between them,</em>

v = \frac{d_{ball}}{\Delta t}

Where

d_{ball} =Ball diameter

\Delta t= Space time

So the angle swept out by either a blade or a space is:

\theta = 2\pi / 16 = \pi / 8 rad.

Through the angular velocity

\omega = \frac{\theta} {t}

t = \frac{\theta}{\omega}

t= \frac{\pi /8}{1.25} = 0.3141s

So,

v = 4.50*10^-2m / 0.3141 s = 0.1432m/s

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a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T
wlad13 [49]

Explanation:

Given:F=m\ddot{x}=Fe^{-\frac{t}{T}}

Solving for \ddot{x}:

\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}

where:

T=\sqrt{\frac{m}{F}}

Integrating to get \dot{x} with initial conditions \dot{x}(0)=0:

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}

Integrating to get x with initial conditions x(0) = 0:

x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}

When t=T:

x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}

\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\sqrt{\frac{F}{m}}(1-\frac{1}{e})

4 0
3 years ago
Which part of the ear receives the signal from the eardrum?​
Aleksandr [31]

Answer:

Ossicles

Explanation:

Sound causes eardrums to vibrate. These vibrations are then passed on to the ossicles, which is made up of 3 small bones-- the malleus, incus, and stapes. The stapes are connected to the inner ear, specifically to the cochlea which transforms sound waves into electrical signals that are sent to the brain.

7 0
3 years ago
The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves pa
kipiarov [429]

Answer:

1.04 s

Explanation:

The computation is shown below:

As we know that

t = t' × 1 ÷ (√(1 - (v/c)^2)

here

v = 0.5c

t = 1.20 -s

So,

1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)

1.20 = t' × 1 ÷ (√(1 - (0.5)^2)

1.20 = t' ÷ √0.75

1.20 = t' ÷ 0.866

t' = 0.866 × 1.20

= 1.04 s

The above formula should be applied

8 0
3 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km.A)If you and your spacesuit ha
WITCHER [35]

A) 0.189 N

The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

F=\frac{GMm}{R^2}

where

G is the gravitational constant

8.7×10^13 kg is the mass of the asteroid

m = 130 kg is the mass of the man

R = 2.0 km = 2000 m is the radius of the asteroid

Substituting into the equation, we find

F=\frac{(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)(130 kg)}{(2000 m)^2}0.189 N=

B) 2.41 m/s

In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

\frac{GMm}{R^2}=\frac{mv^2}{R}

where

v is the speed of the astronaut

Solving the formula for v, we find the minimum speed at which the astronaut should launch himself and then orbit the asteroid just above the surface:

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)}{2000 m}}=2.41 m/s

3 0
3 years ago
What does gravitational potential energy depend on?
jolli1 [7]
It depends on the mass of an object and acceleration because of the gravity and the height of an object
3 0
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