If a marble is released from a height of 10 m how long would it take for it to hit the ground?

Si el periodo de oscilación de resorte es de 0,44 segundos cuando oscila atado a una masa de 2 Kg. ¿Cuál será el valor de la con

boyakko [2]

Answer:

i d k h b u lol I wish I knew it sorry

7 0

2 years ago

4. What are 3 things that you can think of doing, that would reduce<br> your electricity bill?

Cloud [144]

Turn off standby appliances.
Install a smart thermostat.
Turn down your thermostat.
Buy efficient appliances.
Install a new boiler.
Wash clothes at a lower temperature.
Be smarter about water.
Invest in double glazing.

3 0

2 years ago

Which metal and nonmetal families are the most different and why

nevsk [136]

Answer:

since they have got specific and special metallic and non metallic characteristics I guess

3 0

3 years ago

An electric turntable 0.730 mm in diameter is rotating about a fixed axis with an initial angular velocity of 0.240 rev/srev/s a

Zolol [24]

Answer:

a) $\omega = 0.421\,\frac{rev}{s}$ , b) $\Delta \theta = 0.066\,rev$ , c) $v = 0.966\,\frac{mm}{s}$ , d) $a = 3.293\,\frac{mm}{s^{2}}$

Explanation:

a) The angular velocity of the turntable after $0.200\,s$ .

$\omega = \omega_{o} + \alpha\cdot \Delta t$

$\omega = 0.240\,\frac{rev}{s} + (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)$

$\omega = 0.421\,\frac{rev}{s}$

b) The change in angular position is:

$\Delta \theta = \omega_{o}\cdot t + \frac{1}{2} \cdot \alpha \cdot t^{2}$

$\Delta \theta = (0.240\,\frac{rev}{s} )\cdot (0.2\,s) + \frac{1}{2}\cdot (0.906\,\frac{rev}{s^{2}} )\cdot (0.2\,s)^{2}$

$\Delta \theta = 0.066\,rev$

c) The tangential speed of a point on the rim of the turn-table:

$v = r\cdot \omega$

$v = (0.365\times 10^{-3}\,m)\cdot (0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$v = 9.655\times 10^{-4}\,\frac{m}{s}$

$v = 0.966\,\frac{mm}{s}$

d) The tangential and normal components of the acceleration of the turn-table:

$a_{t} = (0.365\times 10^{-3}\,m)\cdot (0.906\,\frac{rev}{s^{2}})\cdot (\frac{2\pi\,rad}{1\,rev} )$

$a_{t} = 2.078\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{t} = 2.078\,\frac{mm}{s}$

$a_{n} = (0.365\times 10^{-3}\,m)\cdot \left[(0.421\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )\right]^{2}$

$a_{n} = 2.554\times 10^{-3}\,\frac{m}{s^{2}}$

$a_{n} = 2.554\,\frac{mm}{s^{2}}$

The magnitude of the resultant acceleration is:

$a = \sqrt{(2.078\,\frac{mm}{s} )^{2}+(2.554\,\frac{mm}{s} )^{2}}$

$a = 3.293\,\frac{mm}{s^{2}}$

8 0

3 years ago

Both segments of the wire are made of the same metal. Current I1 flows into segment 1 from the left. How does current I1 in segm

Andrei [34K]

Answer:

current I1 = current I2

Explanation:

since the wire is made up of the same material, from Kirchoff's current law sum of currents entering a particular node or segment of wire is equals to the sum of currents leaving that particular node or segment of wire

8 0

3 years ago