35 points! Will give brainliest!

An astronaut drops a hammer from 2.0 meters above the surface of the Moon. If the acceleration due to gravity on the Moon is 1.6

Elenna [48]

Answer:

option C

Explanation:

given,

height of the hammer, h = 2 m

acceleration due to gravity of the moon, g = 1.62 m/s²

time taken by the hammer to reach at bottom = ?

initial speed of the hammer, u = 0 m/s

using equation of motion for the time calculation

$s = ut + \dfrac{1}{2}gt^2$

$2 = 0+\dfrac{1}{2}\times 1.62\times t^2$

t² = 2.469

t = 1.57 s ≅ 1.6 s

hence, the time taken by the hammer to reach the bottom is equal to 1.6 s.

The correct answer is option C

8 0

4 years ago

The imaginary point in the sky directly above the North Pole is _____.

slavikrds [6]

The imaginary point in the sky directly above the North Pole is the <span>North celestial pole.</span>

7 0

4 years ago

A rigid tank initially contains 1.4 kg saturated liquid water at 200◦C. At this state, 25 percent of the volume is occupied by w

rjkz [21]

Answer:

(a) Volume of the tank is $6.47\times 10^{- 3}\ m^{3}$

(b) Temperature is $371^{\circ}C$

Pressure is 21.3 kPa

(c) The change in internal energy is 1373.54 kJ/kg

Solution:

As per the question:

Mass of liquid in the tank, m = 1.4 kg

Temperature, T = $200^{\circ}C$

Volume occupied by water, V = 25% $V_{t}$ = 0.25 $V_{t}$

Volume occupied by air, V' = 75% $V_{t}$

where

$V_{t}$ = Volume of tank

Now,

(a) In order to calculate the volume of the tank, we make use of the steam table for specific volume at as given temperature:

At $200^{\circ}C$ , the specific volume, v = 0.001157 $m^{3}/kg$

At $200^{\circ}C$ , the internal energy, u = 850.46 kJ/kg

Now, Volume of water, V = mv = $1.4\times 0.001157 = 1.62\times 10^{- 3} m^{3}$

Thus

$V = 0.25V_{t}$

$V_{t} = \frac{1.62\times 10^{- 3}}{0.25} = 6.47\times 10^{- 3}\ m^{3}$

(b) For the final temperature and pressure, we calculate the specific volume, v' and then find the corresponding temperature and pressure from the steam table:

v' = $\frac{V_{t}}{m} = \frac{6.47\times 10^{- 3}}{1.4} = 4.63\times 10^{- 3}\ m^{3}/kg$

The corresponding temperature to this specific volume, T' = $371^{\circ}C$

The corresponding pressure to this specific volume, P' = 21.3 kPa

The corresponding internal energy to this specific volume, u' = 2224 kJ/kg

(c) The change in the internal energy of water is given by:

$\Delta U = u' - u = 2224 - 850.46 = 1373.54 kJ/kg$

3 0

3 years ago

Calculate the number of molecules/m3 in an ideal gas at stp.

vivado [14]

STP (Standard Temperature and Pressure) has the following conditions:

Temperature = 273.15 K = 0°C
Pressure = 101325 Pa = 101.325 KPa = 1 atm

We also know that 1 mole = 6.022x10^23 molecules

Using the ideal gas equation: PV=nRT

n/V = P/RT

molecules/V = P*6.022x10^23/RT
molecules/V = 101325 Pa (6.022x10^23 molecules/mole)/ (8.314 Pa-m3/mol-K)(273.15K)

molecules/V = 7.339x10^27 molecules/m^3 - Final answer

4 0

4 years ago

the flame on a gas stove heats the bottom of a metal pot. (A) conduction (B)convection (C)radiation. the cold water in a metal p

Sedbober [7]

The flame on a gas stove heats the bottom of a metal pot is conduction.
The sun gives you a sunburn is a radiation

3 0

3 years ago

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