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2 years ago

Where might water be found on the moon

Physics

2 answers:

Cerrena [4.2K]2 years ago

8 0

Water Could be Found in the frozen Ice on the moon. it would most likely be underground however

NNADVOKAT [17]2 years ago

4 0

I think water would be found in the middle of the moon aka the core of the moon. If there was water on the moon it could either be found in rocks or in the core. The reason i said core instead of water is because the moon doesn't really have rocks like Mars does it's sandy or ashy on the moon.

Hope this helped :)

You might be interested in

Convert 1.5 radian into milliradians.

Alex

Answer:

1500 milliradians

Explanation:

Data provided in the question:

1.5 radians

Now,

1 radians consists of 1000 milliradians

1 milli = 1000

thus for the 1.5 radians, we have

1.5 radians = 1.5 multiplied by 1000 milliradians

1.5 radians = 1500 milliradians

Hence, after the conversion

1.5 radians equals to the value 1500 milliradians

4 0

2 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

3 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

3 0

1 year ago

What average force is required to stop a 1400 kg car in 6.0 s if the car is traveling at 90 km/h ? Express your answer to two si

devlian [24]

Answer:

F=5833.3 N N

Explanation:

Newton's second law applied to the car

F= m*a Formula (1)

F: Force in Newtons (N)

m : mass in kg

a: acceleration ( m/s²)

kinematics car

vf= v₀ + a*t Formula (2)

vf : final velocity (m/s)

v₀ : final velocity (m/s)

a : acceleration ( m/s²)

t : time t

Equivalences

1 km= 1000m

1 h = 3600 s

Data

m= 1000kg

v₀ = 90 km/h = 90*1000/3600 m/s = 25 m/s

vf= 0

t= 6 s

Problem Development

We calculate the acceleration replacing the data in the formula (2) :

0 = 25 + a*6

a= -25/6 = -4.16 m/s² ( The negative sign indicates that the car is braking)

We calculate the force is required to stop the car replacing the data in the formula (1)

-F = 1400 kg*(-4.16 m/s²)

F=5833.3 N

8 0

2 years ago

¿Cuánto tardará un automóvil, con movimiento uniforme, en recorrer una distancia de 195 Km si su velocidad es de 30Km/h.?

kykrilka [37]

Answer:

あなたのポイントを無駄にして申し訳ありませんが、あなたの質問がその言語を日本語にすることがわかりません

7 0

2 years ago