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3 years ago

a bicycle travels at a velocity of -2.33 m/s and has a displacement of -58.3 m. how much time did it take?

1 answer:

5 0

We know that velocity is equal to the total displacement of an object over time.
$velocity = \frac{displacement}{time}$
Deriving from that equation, we can say that:
$t = \frac{d}{v}$
Okay, so here it goes:
$t = \frac{58.3m}{2.33 \frac{m}{s} } \\ t = 25.02s$
The bicycle took 25.02 seconds to displace at 58.3 meters.

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3 years ago

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A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

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

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

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3 years ago

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2 years ago

Earth orbits once around the sun every 365.25 days at an average radius of 1.5X10^11 m. Earth's mass is 6X10^24 kg. What distanc

Katyanochek1 [597]

(a) The distance the Earth travels every year is the perimeter of the orbit; since we have the average radius of the orbit, we can calculate the perimeter:
$d=2\pi r = 2 \pi (1.5 \cdot 10^{11}m)=9.4 \cdot 10^{11}m$

(b) The average centripetal acceleration is given by
$a_c = \frac{v^2}{r}$
where v is the average speed of the Earth, that we can find by dividing the distance covered by the Earth in one year (=the perimeter of the orbit) by the the time corresponding to 1 year:
t=1 year=365.25 days=3.2 \cdot 10^7 s
So the velocity is:
$v= \frac{d}{t}= \frac{9.4 \cdot 10^{11}m}{3.2 \cdot 10^7 s} =2.94 \cdot 10^4 m/s$
And so the centripetal acceleration is:
$a_c = \frac{v^2}{r}= \frac{(2.94 \cdot 10^4 m/s)^2}{9.4 \cdot 10^{11}m} =9.2 \cdot 10^{-4} m/s^2$

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3 years ago