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4 years ago

What is responsible for the apparent motion of the Moon in the sky from East to West?

1 answer:

6 0

As a result, although the Moon is moving to the east relative to the stars, the much faster westward motion of the sky is carrying it to the west, so despite its eastward motion relative to the center of the Earth, it rises in the east and sets in the west, just like any other celestial body.

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Draw the vector c⃗ =1.5a⃗ −3b⃗

Debora [2.8K]

The magnitude of a is 1.5
The magnitude of b is -3
The magnitude of the vector is
√(1.5² + (-3)²) = 3.35
The angle is
θ = tan⁻¹ (-3/1.5) = 63.43°
The vector is drawn with a magnitude of 3.35 and an angle of 63.43°.

3 0

4 years ago

The basic unit of electric current is the .

Masja [62]

The Ampere (A). You could literally google to be more efficient, all the same to me tho

4 0

3 years ago

Read 2 more answers

A bike race against the clock takes place on a straight road. Yan drives at 37 km / h and he starts the course 30s before Christ

joja [24]

Given data:

Yan speed;

$u_1=37\text{ km/h}$

Christopher speed;

$u_2=38.9\text{ km/h}$

Christophe starts 30 s later than Yan. Therefore, Christophe takes 30 s less than Yan to reach the same distance.

Part (A)

The distance is given as,

$d=ut$

Let both Yan and Christophe meet at d distance from the start position. Therefore,

$u_1t=u_2(t-30)$

Substituting all known values,

$\begin{gathered} (37\text{ km/h})t=(38.9\text{ km/h})\times(t-30) \\ \frac{(37\text{ km/h})}{(38.9\text{ km/h})}=\frac{(t-30)}{t} \\ 0.95=1-\frac{30}{t} \\ \frac{30}{t}=1-0.95 \\ \frac{30}{t}=0.05 \\ t=\frac{30}{0.05} \\ t=600\text{ s} \end{gathered}$

Therefore, 600 s after Yan's departure Christophe will join him.

Part (B)

The distance is given as,

$d=u_1t$

Substituting all known values,

$\begin{gathered} d=(37\text{ km/h})\times(600\text{ s}) \\ =(37\text{ km/h})\times(600\text{ s})\times(\frac{1\text{ hr}}{3600\text{ s}}) \\ \approx6.17\text{ km} \end{gathered}$

Therefore, Christophe joins Yan after 6.17 km from the start.

3 0

1 year ago

A body of mass 400 kg is suspended at a lower end of a light vertical chain and is being pulled up vertically. Initially the bod

yKpoI14uk [10]

Answer:31.62 m/s

Explanation:

Given

mass of body $m=400 kg$

Pull on chain is $F_1=6000g N=60 kN$

Pull get smaller at the rate of $F_2=360g N/m$

Net Upward Force $F=6000 g-360 g\times 10=24 kN$

net acceleration $a=\frac{F}{m}$

$a=\frac{24\times 1000}{m}$

$a=\frac{24000}{400}$

$a=60 m/s^2$

but g is acting downward

$a_{net}=a-g=60-10=50 m/s^2$

using $v^2-u^2=2 as$

here initial velocity is zero

$v^2=2\times 50\times 10$

$v=31.62 m/s$

7 0

4 years ago

Determine the speed of sound in air at 300 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 3

amm1812

Answer:

$c_3_0_0_K=347.19m/s$

$M=0.864$

Explanation:

The speed of sound in the air increases 0.6 m / s for every 1 ° C increase in temperature. An approximate speed can be calculated using the following empirical formula:

$c=331.5+0.6\vartheta$

Where:

$\vartheta=T-273.15K\\\\$

A more exact equation, usually referred to as adiabatic velocity of sound, is given by the following formula:

$c=\sqrt{k*R*T}$

Where:

$R= Gas\hspace{3}constant\hspace{3}of\hspace{3}air=0.287kJ/kg*K=287J/kg*K\\k=Specific\hspace{3}heat\hspace{3}ratio=1.4\\T=Temperature=300K$

Hence:

$c=\sqrt{(287)*(1.4)*(300)} =347.1887095\approx347.19m/s$

Now, the Mach number at which an aircraft is flying can be calculated by:

$M=\frac{u}{c}$

Where:

$u= Velocity\hspace{3}of\hspace{3}the\hspace{3}moving\hspace{3}aircraft\\c= Speed\hspace{3}of\hspace{3}sound\hspace{3}at\hspace{3}the\hspace{3}given \hspace{3}altitude$

Therefore:

$M=\frac{300}{347.19} =0.8640833984\approx0.864$

5 0

4 years ago