When the sun warms the sidewalk on a hot sunny day, what type of energy conversion occurs?

A particle has ~r(0) = (4 m) ˆ and ~v(0) = (2 m/s)ˆı. If its acceleration is constant and given by ~a = −(2 m/s 2 ) (ˆı +ˆ), a

Troyanec [42]

The particle has $\vec r(0)=(4\,\mathrm m)\,\vec\imath$ and $\vec v(0)=\left(2\frac{\rm m}{\rm s}\right)\,\vec\imath$ , and is undergoing a constant acceleration of $\vec a=-\left(2\frac{\rm m}{\mathrm s^2}\right)(\vec\imath+\vec\jmath)$ .

This means its position at time $t$ is given by the vector function,

$\vec r(t)=\vec r(0)+\vec v(0)t+\dfrac12\vec a t^2$

$\implies\vec r(t)=\left[4\,\mathrm m+\left(2\dfrac{\rm m}{\rm s}\right)t-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath-\left(1\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

The particle crosses the x-axis when the $\vec\imath$ component is 0 for some time $t>0$ , so we solve:

$4+2t-t^2=0\implies t^2-2t+1=5$

$\implies(t-1)^2=5$

$\implies t-1=\pm\sqrt5$

$\implies t=1\pm\sqrt5$

The negative square root introduces a negative solution that we throw out, leaving us with $\boxed{t=1+\sqrt5}$ or about 3.24 seconds after it starts moving.

3 0

4 years ago

An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu

shutvik [7]

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is $-1.74 rad/s^2$

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

a)

The initial angular speed of the wheel is

$\omega_i = 2000 rpm$

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

$1 rev = 2\pi rad$

$1 min = 60 s$

We find:

$\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s$

b)

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

$\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s$

Now we can find the angular acceleration, given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_i = 209.3 rad/s$ is the initial angular speed

$\omega_f = 104.7 rad/s$ is the final angular speed

t = 60 s is the time interval

Substituting,

$\alpha = \frac{104.7-209.3}{60}=-1.74 rad/s^2$

c)

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

$\omega' = \omega_i + \alpha t$

where we have

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 40 s, we find

$\omega' = 209.3 + (-1.74)(40)=139.7 rad/s$

d)

The angular displacement of the wheel in a certain time interval t is given by

$\theta=\omega_i t + \frac{1}{2}\alpha t^2$

where

$\omega_i = 209.3 rad/s$

$\alpha = -1.74 rad/s^2$

And substituting t = 60 s, we find:

$\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad$

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

$\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

8 0

3 years ago

Any1 here got twitter?

Marina86 [1]

Yeah I have Twitter!!

3 0

3 years ago

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A car drives over a hilltop that has a radius of curvature 120 m at the top of the hill. at what speed would the car be travelin

FinnZ [79.3K]

The speed of a car travelling over a hill that has a radius of curvature should not exceed a certain speed other it will topple. This speed is related to the radius of curvature and the gravitational acceleration as shown below:

V^2 = Rg, where V = maximum speed, R = Radius of curvature, g = gravitational acceleration.

Substituting;
V = Sqrt (Rg) = Sqrt (120*9.81) = 34.31 m/s

6 0

3 years ago

What is the potential energy of a one kilogram box of chocolates hidden on a

sveta [45]

Answer:

um it well it will be 19.6 d) 9.8 J

Explanation:

7 0

3 years ago

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