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3 years ago

For each one of this, what kinds of observation could you record on a calendar?

Physics

2 answers:

jek_recluse [69]3 years ago

5 0

Answer:

ummmmmmmmm

Explanation:

hello

3241004551 [841]3 years ago

5 0

The correct answer to this open question is the following.

The complete question asks us to record in a calendar the events of the moon cycle and during storming weather.

So in the case of the moon, we can record the time that passes from the last new moon to the next new moon, and also from the last full moon and the next full moon. In the case of the weather and storms, we can record the amount of rain that falls and can compare it to another rainy season in other years.

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What is the density of iron if 5.0 cm3 has a mass of 39.5g

weqwewe [10]

Answer : $\rho = 7.9\ g/cm^3$

Explanation :

It is given that

Mass of iron, m = 39.5 g

Volume of iron, $V = 5\ cm^3$

So, density is :

density, $\rho =\dfrac{mass}{volume}$

$\rho =\dfrac{39.5g}{5cm^3}$

$\rho = 7.9\ g/cm^3$

7 0

4 years ago

A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Accelerati

JulsSmile [24]

$m=852 \ kg \\ h=3,5 \ m \\ g=9,8 \ m/s^2 \\ \boxed{P_e-?} \\ \bold{Solving:} \\ \boxed{P_e=m \cdot g \cdot h} \\ P_e=852 \ kg \cdot 9,8 \ m/s^2 \cdot 3,5 \ m =8 \ 349,6 \ N \cdot 3,5 \ m \\ \Rightarrow \boxed{P_e=29 \ 223,6 \ J}$

7 0

3 years ago

Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2.

IrinaVladis [17]

Answer:

option C

Explanation:

Let mass of the bullet be m and velocity be v

mass of gun be M and bullet be V

now,

using conservation of momentum for gun 1

(M+m) V' = 2 mv + 3 MV

V' = 0

3 M V = - 2 mv

momentum of gun 1 =- 2 mv---------(1)

now for gun 2

(M+m) V' = mv + MV

V' = 0

M V = - mv

momentum of gun 1 = -mv-----------(2)

dividing equation (1) by (2)

$\dfrac{P_m1}{P_m2} = \dfrac{- 2mv}{-mv}$

$\dfrac{P_m1}{P_m2} = \dfrac{2}{1}$

the correct answer is option C

7 0

3 years ago

A diode has a power rating of 5W. If the diode voltage is 3.6V and the diode current is 1.75A, what is the power dissipation?

son4ous [18]

Answer:

Explanation:

Power = current * voltage

Power = 3.6 * 1.75

Power = 6.3 watts.

I hope this diode does not have to do this for very long. The power dissipation is above the diode's rated power -- a condition that won't last forever.

The power dissipation is 6.3 watts.

4 0

3 years ago

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Ksenya-84 [330]

I don't know if my answer is right or not but on the test I picked ( D)

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4 years ago