Ask question

Ask question

All categories

4 years ago

13

You are traveling through the forest and encounter an ancient well. You want to know how deep the well is. If you drop a coin an

d hear it hit the bottom of the well in 3 seconds, how deep is the well?

2 answers:

eimsori [14]4 years ago

4 0

Answer:

About 30 ft

Explanation:

horrorfan [7]4 years ago

3 0

Depth = 1/2 x 9,8 x 9 = 44.1 m

You might be interested in

15 . A scientist who studies the whole environment as a working unit .

soldier1979 [14.2K]

Answer:

<em>Ecologist</em><em>.</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em>Your</em><em> </em><em>answer </em><em>is</em><em> </em><em>Ecologist</em><em>.</em>

3 0

3 years ago

Read 2 more answers

Un libro del peso di 12 N è in equilibrio su un tavolo. Sapendo che il coefficiente di attrito statico vale 0,5, la forza di att

Tanzania [10]

Answer:

60

Explanation:

Translation -

A book weighing 12 N is balanced on a table. Knowing that the static friction coefficient is 0.5, how much is the friction force worth?

Friction force is

f = u * n

f = 0.5 * 12N

f = 60

4 0

4 years ago

Which material would you take, steel or bricks/stone to design a bridge. Why? ( Question related to poisson's ratio)

Ann [662]

i would choose steel because the steel used for normal construction have several hundred Mega Pascal strength. one of the special feature of steel is its ductility. it is the deformation capability before the final breakage.

8 0

4 years ago

A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t

Talja [164]

Answer:

$P = 133.13 Watt$

Explanation:

Initial angular speed of the ferris wheel is given as

$\omega_i = 2\pi f$

$\omega_i = 2\pi(8.5/3600)$

$\omega_i = 0.015 rad/s$

final angular speed after friction is given as

$\omega_f = 2\pi f$

$\omega_f = 2\pi(7.5/3600)$

$\omega_f = 0.013 rad/s$

now angular acceleration is given as

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

$\alpha = \frac{0.015 - 0.013}{15}$

$\alpha = 1.27 \times 10^{-4} rad/s^2$

now torque due to friction on the wheel is given as

$\tau = I \alpha$

$\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})$

$\tau = 8875.3 N m$

Now the power required to rotate it with initial given speed is

$P = \tau \omega$

$P = 8875.3 \times 0.015$

$P = 133.13 Watt$

8 0

3 years ago

СРОЧНО ПОМОГИТЕ ПОЖАЛУЙСТА!!!!!!!!!!<br>

Deffense [45]

Боже, как это сложно! Ну ладно.

Между прочим это ты сам должен делать, а то не куда не поступишь!

3 0

3 years ago