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2 years ago

Please can someome answer this question with working out

Physics

2 answers:

deff fn [24]2 years ago

7 0

Answer: Yess, why not

Explanation:

I think I just answered this question

Murljashka [212]2 years ago

3 0

What is the question? I can’t see it?

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What is in the crysphere

butalik [34]

I think you mean the Cryosphere?

But the answer is D- Earths Ice

This word Cryosphere comes from the greek word "kryos" which means cold

Many people think of the cryosphere as being the north and south poles but snow and ice can be found in a lot of places on the Earth

3 0

3 years ago

a boy runs from his house to school at speed 5m/s on a straight road and returns with speed of 10m/s.The distance between the ho

CaHeK987 [17]

Explanation:

time spent to run from house to school=100/5=20s

time spent to return from school=100/10=10s

average velocity=200m/(10+20)

=6.66m/s

3 0

3 years ago

A musical note has a frequency of 512 Hz. If the wavelength of the note is 0.685 m, what is the speed of the sound of that note?

riadik2000 [5.3K]

Answer:

350.72 m/s

Explanation:

Formula for velocity of wave is;

v = fλ

Where;

v is speed

f is frequency

λ is wavelength

We are given;

f = 512 Hz

λ = 0.685 m

Thus;

v = 512 × 0.685

v = 350.72 m/s

5 0

3 years ago

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

3 years ago

a cart has 30J or potential energy at the top of a 20m high hill. Assuming no energy is transformed to another form of energy, h

topjm [15]

Answer:

i would say 20J

Explanation:

3 0

3 years ago

Please can someome answer this question with working out​

Please can someome answer this question with working out