Ask question

Ask question

All categories

3 years ago

9

Help please an observer at the North Pole sees the moon phase shown below. What Moon phase will be observed approximately one we

ek later?

1 answer:

ziro4ka [17]3 years ago

4 0

Answer:

it should be a new moon

Explanation:

You might be interested in

Anyone going to be my friend

Artemon [7]

Explanation:

I'd love to but we cant talk right now cause its 12:22 am here and I'm gonna sleep now lol.

but let's follow each other.

who knows we might be able to help each other.

whaddya say?

have a good day ♡

5 0

2 years ago

A 86 kg human stands on the surface of Venus. The mass of Venus is 4.9 × 1024 kg and its radius is 6.1 × 106 m. Collapse questio

iris [78.8K]

Answer:

755.37 N

Explanation:

We are given that

Mass of venus= $m_1=4.9\times 10^{24}kg$

Radius= $r=6.1\times 10^6$ m

Mass of human= $m_2=86 kg$

We know that the gravitational force between two bodies

$F=G\frac{m_1m_2}{r^2}$

Where G=Gravitational constant= $6.67\times 10^{-11}Nm^2/kg^2$

Using the formula

The magnitude of the gravitational force exerted by Venus on the human= $F=\frac{6.67\times 10^{-11}\times 86\times 4.9\times 10^{24}}{(6.1\times 10^6)^2}$

The magnitude of the gravitational force exerted by Venus on the human=F= $755.37N$

7 0

2 years ago

EASY MULTIPLE CHOICE QUESTIONS! PLEASE HELP!

skad [1K]

1.A and 2.B there the answers

8 0

2 years ago

An astronaut is moving in space when a big explosion occurs about 50 metres BEHIND him. How will the astronaut come to know abou

kakasveta [241]

The matter from the explosion can reach him, hitting him. He should be able to feel that.

6 0

2 years ago

Read 2 more answers

A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.

fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

(c) 352.9 Hz

Explanation:

For an open pipe, the velocity is given by

$v=\frac {2Lf}{n}$

Making L the subject then

$L=\frac {nV}{2f}$

Where f is the frequency, L is the length, n is harmonic number, v is velocity

Substituting 1 for n, 320 Hz for f and 331 m/s for v then

$L=\frac {1*331}{2*320}=0.5171875\approx 0.52m$

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, $f2=2\times \frac {v}{2L}$ and $f3=3\times \frac {v}{2L}$

$f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz$

(c)

When v=367 m/s then

$f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz$

5 0

3 years ago