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3 years ago

6

10. Cumulus and stratus clouds belong to which cloud group?

1 answer:

EastWind [94]3 years ago

5 0

Cumulus belongs to vertical clouds and status to low

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Two blocks are connected by a massless cable which goes through the center of a rotating turntable. The blocks have masses M1 =

Airida [17]

Answer:

If the final question is; at what velocity will the first block start to move outward in m/s?

$v = 3.5596 \frac{m}{s}$

Explanation:

The motion have the velocity that will make the block move using:

$F_{1}*F_{r}+ F_{2}= Ec \\Ec= \frac{M*v^{2} }{r}$

$m_{1} = 1.2 Kg$

$m_{2} = 2.8 Kg$

$r = 0,45 m$

μ $s= 0,54$

Resolving:

$\frac{m_{1}*v^{2} }{r} = us*m_{1}* g + m_{2} * g$

$v^{2} = \frac{((us *m_{1} + m_{2})*g )* r}{m_{1} }$

$v^{2} = \frac{((0.54 *1.2 + 2.8)*9.8 )* 0.45}{1.2}$

$v^{2} = 12.6714 \frac{m^{2} }{s^{2} }$

$v = 3.559 \frac{m}{s}$

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3 years ago

Please help It is Anatomy and Phys

kap26 [50]

Answer:

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3 years ago

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Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. If the incoming ball has a speed of 20

Bingel [31]

Answer:

Explanation:

Check the attachment for solution

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3 years ago

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Two pieces of cotton have ____ charges and they ___ each other

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Like . . . repel
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3 years ago

An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of 25.0° above

Greeley [361]

Answer:

$v_{i}= 19\times 10^5\ m/s$

Explanation:

given,

scattering angle of alpha particle = 25.0° above its initial direction of motion

oxygen nucleus recoils at = 50.0° below this initial direction.

final speed of the oxygen = 2.08×10⁵ m/s

mass of alpha particle = 4.0 u

mass o oxygen nucleus = 16 u

momentum conservation along x- axis

$m_{a}v_{i} = m_a v_a cos\theta + m_o v_o cos\theta$

$4v_{i} = 4\times v_a cos25^0 + 16\times 2.08 \times 10^5 cos50^0$

$v_{i}= \dfrac{3.625\times v_a+ 21.39\times 10^5}{4}$ ....(1)

Along y-direction

$0 = m_av_a sin \theta - m_ov_o sin\theta$

$0 = 4\times v_a sin 25 - 16\times 2.08 \times 10^5 sin50^0$

$v_a = \dfrac{25.49 \times 10^5}{1.69}$

$v_a = 15.082\times 10^5\ m/s$

putting value in equation (1)

$v_{i}= \dfrac{3.625\times 15.082\times 10^5+ 21.39\times 10^5}{4}$

$v_{i}= 19\times 10^5\ m/s$

5 0

3 years ago