Could anyone help with number 9?

What are meant by the statement relative density of gold are 19.3

Doss [256]

It means that 19.3g of gold is packed into 1cm^3.

7 0

3 years ago

At its closest approach, a moon comes within 200,000 km of the planet it orbits. At that point, the moon is 300,000 km from the

krok68 [10]

Answer:

s₁ = 240,000 km

Explanation:

The distance between both the focuses f₁ and f₂ will be the sum of distances of the moon from each focus at a given point. Therefore,

s = s₁ + s₂

where,

s = total distance between the focuses = ?

s₁ = distance between f1 and moon = 200,000 km

s₂ = distance between f₂ and moon = 300,000 km

Therefore,

s = 200,000 km + 300,000 km

s = 500,000 km

Now, when the distance from f₂ becomes 260,000 km, then the distance from f₁(planet) will become:

s = s₁ + s₂

500,000 km = s₁ + 260,000 km

s₁ = 500,000 km - 260,000 km

s₁ = 240,000 km

5 0

3 years ago

A weather balloon is designed to expand to a maximum radius of 24 m at its working altitude, where the air pressure is 0.030 atm

nlexa [21]

Answer:

Radius at liftoff 8.98 m

Explanation:

At the working altitude;

maximum radius = 24 m

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

radius = ?

First, we assume balloon is spherical in nature,

and that the working gas obeys the gas laws.

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere = $\frac{4}{3} \pi r^{3}$

volume of balloon = $\frac{4}{3}$ x 3.142 x $24^{3}$ = 57913.34 m^3

using the gas equation,

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.

imputing values, we have

$\frac{0.03*57913.34}{200}$ = $\frac{1*V2}{349}$

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = 3031.76 m^3 this is the volume occupied by the gas in the balloon at liftoff.

from the formula volume of a sphere,

V = $\frac{4}{3} \pi r^{3}$ = $\frac{4}{3}$ x 3.142 x $r^{3}$ = 3031.76

4.19 $r^{3}$ = 3031.76

$r^{3}$ = 3031.76/4.19

radius r of the balloon on liftoff = $\sqrt[3]{723.57}$ = 8.98 m

4 0

3 years ago

A 67 kg kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m m . What is the effective s

amid [387]

Point of correction, spring constant is 2.3×10−2 m not 2.3×10−2 m m

Answer:

28577 N/m

Explanation:

From Hooke's law, F=kx where F is force applied, k is spring constant and x is compression

F=mg=67*9.81

$k=\frac {mg}{x}$

$k=\frac {67*9.81}{2.3\times 10^{-2}}=28576.95652 N/m$

Approximately, 28577 N/m

8 0

3 years ago

Identify the physical property of a material that plays an important role in determining if the collision is elastic or not base

Nutka1998 [239]

Answer:

They are spherical and hollow (not compact or dense)

Explanation:

An elastic collision is a form of a collision where kinetic energy and momentum are conserved in the process. When there is zero loss of kinetic energy and momentum, it is called a perfectly elastic collision.

This form of collision is observed in atmospheric gases and colliding balls which happens to be spherical and hollow.

7 0

3 years ago