A rocket sled is tested at 5g

What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Volgvan

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

4 0

3 years ago

A star is estimated to be 4.8 km away from the earth. When we see this star in the night sky how old is the image. Let the speed

andre [41]

Time = distance / speed

Time = (4,800 meters) / (3 x 10⁸ m/s)

Time = 0.000016 second

This number is not one of the choices on the list. My hunch is that you copied the distance wrong.

If the estimated distance to the star is actually 4.8 x 10¹⁵ km, instead of 4.8 km, then the answer would be close to 500 years (B).

There's no way a star can be "4.8 km away from the Earth". You can walk that far in about an hour, and passenger jet airplanes fly twice as far as that away from the Earth !

5 0

3 years ago

A spring with a pointer attached is hanging next to a scale marked in millimeters. Three different packages are hung from the sp

SOVA2 [1]

solution;

$the expression for force applied on the spring due to the load is\\f=k\Delta x\\here,\Delta x is the extension in the spring due to appling force\\given three case as following\\110N=k(40-x_{o})----------1\\240N=k(60-x_{o})----------2\\w=k(30-x_{o})-------------3\\$ $To calculate the accrual length of the spring,solve th equation 1 and 2\\\frac{110N}{240N}=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458(60mm-x_{o})=(40mm-x_{o})\\x_{o}(1-0.458)=(40-60(0.458))mm\\x_{o}\frac{12.52}{0.542}\\=23.1mm\\to calculate the force on the spring in case,\\solve the equation 1 and 2\\\frac{110}{w}=\frac{k(40-x_{o})}{k(60-x_{o})}\\\frac{110}{w}=\frac{(40mm-23.1mm)}{30mm-23.1mm}\\w=\frac{110}{2.45}=44.9N$