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3 years ago

9

Can someone help me with physics;((

1 answer:

vladimir2022 [97]3 years ago

4 0

Answer: vf= 51 m/s and d= 112 m

Explanation: solution attached

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Lightning flashed, and after 7 seconds there was a roll of thunder. To the nearest meter, determine how far lightning flashed fr

Nadya [2.5K]

Let’s come back to the equation of SPEED OF AN ECHO
SPEED =2XDISTANCE/ TIME
So 337*7/2=1179.5 nearest meter =1800 m away
Hope you will get it right

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3 years ago

Which type of border shows the division between Sonora and Chihuahua?

KonstantinChe [14]

The answer is the national border

7 0

3 years ago

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ki77a [65]

Answer:

a) $t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

b) $\theta_1=\frac{w_1^2}{2\alpha}rad$

c) $t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

$w=\frac{\Delat \theta}{\Delta t}$

Angular acceleration is the rate of change of the angular velocity respect to the time

$\alpha=\frac{dw}{dt}$

2) Part a

We can define some notation

$w_o=0\frac{rad}{s}$ ,represent the initial angular velocity of the wheel

$w_1=?\frac{rad}{s}$ , represent the final angular velocity of the wheel

$\alpha$ , represent the angular acceleration of the flywheel

$t_1$ time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

$w_1=w_o +\alpha t_1$

And solving for t1 we got:

$t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

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

Replacing the values for our case we got:

$\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2$

And we can replace $t_1$ from the result for part a, like this:

$\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2$

Since $\theta_o=0$ and $w_o=0$ then we have:

$\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}$

And simplifying:

$\theta_1=\frac{w_1^2}{2\alpha}rad$

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is $\alpha_1 =-5\alpha \frac{rad}{s}$

We have an initial angular velocity $w_1$ , and since at the end stops we have that $w_2 =0$

Assuming that $t_2$ represent the time in order to stop the wheel, we cna use the following formula

$w_2 =w_1 +\alpha_1 t_2$

Since $w_2=0$ if we solve for $t_2$ we got

$t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}$

And from part a) we can see that $w_1=\alpha t_1$ , and replacing into the last equation we got:

$t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

5 0

3 years ago

What is the speed of an aardvark that runs 10 meters every hour?

VashaNatasha [74]

Answer:

10m/hr

Explanation:

10m/3600 sec= 1m/360sec

8 0

3 years ago

A uniform metre rule of mass 100g balance the 40cm mark when a mass x is placed at the 20cm mark

vivado [14]

Answer:

X = 50 g

Explanation:

Please see attached photo for explanation.

From the attached photo,

Anticlock–wise moment = X × 20

Clockwise moment = 100 × 10

Anticlock–wise moment = clockwise moment

X × 20 = 100 × 10

X × 20 = 1000

Divide both side by 20

X = 1000 / 20

X = 50 g

Therefore, the value of X is 50 g

6 0

3 years ago