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2 years ago

7

A friend of yours who takes her astronomy class very seriously challenges you to a contest to find the thinnest crescent moon yo

u can find just after new moon? What time of day is best for looking for this very thin crescent?

1 answer:

RUDIKE [14]2 years ago

8 0

Answer:

after the sun sets or just as it is setting

Explanation:

a crescent moon is thin and reflects less sunlight during the daylight sky so it becomes difficult to spot, but can be spotted when the sun is setting or just sets.

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you are sitting on a beach and a wave strikes the shore every 10 seconds. a surfer tells you that these waves travel at a speed

aev [14]

Answer:

50 meters

Explanation:

5 0

2 years ago

Sara and Saba are identical twins who are the same in every way, including their weights. One day, Sara and Saba decided to go f

Sphinxa [80]

As Saba was wearing high heels they are long from the bottom so they sank however Sana was wearing snow boots which means they were flat and so she didn’t sink.

3 0

2 years ago

A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume

dalvyx [7]

Explanation:

It is given that,

Mass of the car 1, $m_1=900\ kg$

Initial speed of car 1, $u_1=15i\ m/s$ (east)

Mass of the car 2, $m_2=750\ kg$

Initial speed of car 2, $u_1=20j\ m/s$ (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$900\times 15i +750\times 20j=(900+750)V$

$13500i+15000j=1650V$

$V=(8.18i+9.09j)\ m/s$

The magnitude of speed,

$|V|=\sqrt{8.18^2+9.09^2}$

V = 12.22 m/s

(b) Let $\theta$ is the direction the wreckage move just after the collision. It is given by :

$tan\theta=\dfrac{v_y}{v_x}$

$tan\theta=\dfrac{9.09}{8.18}$

$\theta=48.01^{\circ}$

Hence, this is the required solution.

4 0

2 years ago

If you drive through water, your brakes may become slippery and ineffective. To dry the brakes off, __________.

GenaCL600 [577]

Answer:

I am not a driver, but I think it's C.

Explanation:

6 0

2 years ago

Read 2 more answers

The muzzle velocity of a 50.0g shell leaving a 3.00 kg rifle is 400m/s what is the recoil velocity of the rifle

serg [7]

Here if we consider bullet and gun as a system then we can say that momentum of the system will remain conserved

so we will have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

now we know that

$m_1 = 50 g = 0.050 kg$

$m_2 = 3 kg$

$v_{1i} = v_{2i} = 0$

$v_{1f} = 400 m/s$

now we will have

$0.050(0) + 3(0) = 0.050(400) + 3(v)$

$20 + 3v = 0$

$v = - \frac{20}{3} = - 6.67 m/s$

so gun will recoil with speed 6.67 m/s

6 0

2 years ago