Ask question

Ask question

All categories

photoshop1234 [79]

3 years ago

15

What is the most important element they keep in mind for the missions? of apollo

1 answer:

s344n2d4d5 [400]3 years ago

4 0

They keep the astronauts safety as a priority. They also try to stay in budget, but mainly the astronauts safety is most important.

You might be interested in

How do hydrogen bonds affect boiling points?

wel

Answer:

I would say A but am not sure

7 0

3 years ago

A sky diver, mass 90 kg, reaches a terminal velocity of 60 m/s. What is the approximate magnitude of the force of air resistance

Komok [63]

Answer:

The approximate magnitude of the force of air resistance is 540 N.

Explanation:

6 0

2 years ago

Metre is the unit of dash.

djverab [1.8K]

Answer:

Length is the unit of metre

7 0

3 years ago

Read 2 more answers

A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t0 = 0 with a v

ch4aika [34]

Answer

given,

Time period= T = 1.5 s

If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s

v_max=1.00 m/s

we know,

v_ max=A ω

v = A sin (ωt)

-0.50= -1.00 sin (ωt)

sin (ωt) = 0.5

$\omega t = sin^{-1}(0.5)$

$\dfrac{2\pi}{T}\times t =0.524$

$\dfrac{2\pi}{1.5}\times t =0.524$

t = 0.125 s

we have time period T=1.5 it is the time to complete one oscillation

means from eq to right,then left,then eq,then left,then from right to eq

time taken for left = t/4 = 0.125/4 = 0.375 s

smallest value of time

=0.375 + 0.125

= 0.50 sec

7 0

3 years ago

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of

Elan Coil [88]

Answer: 230.50 m

Explanation:

We have the following information:

$h_{Hg-TOP}=675mmHg=0.675m$ the barometric reading at the top of the building

$h_{Hg-BOT}=695mmHg=0.695m$ the barometric reading at the bottom of the building

$\rho _{air}=1.18 kg/m^{3}$ density of air

$\rho _{Hg}=13600 kg/m^{3}$ density of mercury

$g=9.8/m^{2}$ gravity

And we need to find the height of the building.

In order to approach this problem, we will firstly use the following equations to find the pressure at the top of the building $P_{TOP}$ and the perssure at the bottom $P_{BOT}$ :

$P_{TOP}=\rho _{Hg} g h_{Hg-TOP}$ (1)

$P_{BOT}=\rho _{Hg} g h_{Hg-BOT}$ (2)

From (1): $P_{TOP}=(13600 kg/m^{3})(9.8/m^{2})(0.675m)=89964 Pa$ (3)

From (2): $P_{BOT}=(13600 kg/m^{3})(9.8/m^{2})(0.695m)=92629.6 Pa$ (4)

Having the pressures at the top and the bottom of the building, we can calculate the variation in pressure $\Delta P$ :

$\Delta P=P_{BOT} - P_{TOP}$ (5)

$\Delta P=92629.6 Pa - 89964 Pa=2665.6 Pa$ (6)

On the other hand, we have a column of air with a cross-section area $A$ and the same height of the building, lets name it $h_{air}$ .

As pressure is defined as the force $F$ exerted on a specific area $A$ , we can write:

$\Delta P=\frac{F}{A}$ (7)

If we isolate $F$ we have:

$F= A \Delta P$ (8)

Also, the force gravity exerts on this column of air (its weight) is:

$F=m_{air} g$ (9)

Knowing the density of air is: $\rho_{air}=\frac{m_{air}}{V_{air}}$ (10)

where the volume of air can be written as: $V_{air}=(A)(h_{air})$ (11)

Substituting (1) in (10):

$\rho_{air}=\frac{m_{air}}{(A)(h_{air}}$ (12)

Isolating $m_{air}$ :

$m_{air}=(\rho_{air}) (A) (h_{air})$ (13)

Substituting (13) in (9):

$F=(\rho_{air}) (A) (h_{air}) (g)$ (14)

Matching (8) and (14)

$A \Delta P=(\rho_{air}) (A) (h_{air}) (g)$ (15)

Isolating $h_{air}$ :

$h_{air}=\frac{\Delta P}{g \rho_{air}}$ (16)

Substituting the known and calculated values:

$h_{air}=\frac{2665.6 Pa}{(9.8m/s^{2}) (1.18 kg/m^{3})}$ (17)

Finally:

$h_{air}=230.50 m$ This is the height of the building

8 0

3 years ago