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3 years ago

Which planet has a storm called the Great Dark Spot swirling in its atmosphere?

Physics

2 answers:

vesna_86 [32]3 years ago

5 0

The answer is Jupiter not neptune :)

Viktor [21]3 years ago

3 0

Im thinking D. Neptune..

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25 points! Please someone help me, this seems like a pretty easy question but I can’t seem to get it right, please list the give

elixir [45]

Answer:

1 my brother say that

Explanation:

i know my brother said it

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2 years ago

2 cannons fire projectiles upwards with the same velocity. The

dybincka [34]

Answer:

Explanation:

Given

Two projectile is fired vertically upward

One has 4 times the mass of other

When Projectile is fired their trajectory is independent of mass of object. Also if they launched with same speed then both achieved same maximum height in same time and will hit the ground at the same moment.

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2 years ago

What's going to happen in 3 weeks ?

Nana76 [90]

Answer:

god knows.

Explanation:

7 0

2 years ago

Read 2 more answers

A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this

baherus [9]

20 characters longer later and woah

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3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

2 years ago