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ikadub [295]
3 years ago
5

The iron-rich core of the Moon is thought to be about how many kilometers in diameter?

Physics
1 answer:
natka813 [3]3 years ago
6 0
The moon is thought to have an iron-rich core whose radius
is 330 km, plus or minus an uncertainty of 20 km.

That puts its diameter in the range of  620 km to 700 km.
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when light falls obliquely, there is a change in direction as it moves from one medium to another but when it falls perpendicula
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In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m
ikadub [295]

(a) -39.4^{\circ}

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

p_y =0

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2

where

m is the mass of each ball

v_1= 4.60 m/s is the velocity of the cue ball after the collision

v_2 = 3.40 m/s is the velocity of the second ball after the collision

\phi_1=28.0^{\circ} is the angle of the cue ball with the x-axis

\phi_2 is the angle of the second ball

Solving for \phi_2, we find the angle between the direction of motion of the second ball and the original direction of motion:

sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m

The initial total momentum along the x-direction as

p_x = m u

where

m is the mass of the cue ball

u is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

mu = 6.69 m\\u = 6.69 m/s

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Explanation:

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A speeding motorist traveling down a straight highway at 100 km/h passes a parked police car. It takes the police constable 1.0
Lubov Fominskaja [6]

Answer:

t = 7.5 s

Explanation:

The distance traveled by the car at the time of meeting of the two cars must be the same. First, we calculate the distance traveled by the police car. For that we use 2nd equation of motion. Here, we take the time when police car starts to be reference. So,

s₁ = Vi t + (0.5)gt²

where,

s₁ = distance traveled by police car

Vi = Initial Velocity = 0 m/s

t = time taken

Therefore,

s₁ = (0 m/s)(t) + (0.5)(9.8 m/s²)t²

s₁ = 4.9 t²

Now, we calculate the distance traveled by the car. For constant speed and time to be 1 second more than the police car time, due to car starting time, we get:

s₂ = Vt = V(t + 1)

where,

s₂ = distance traveled by car

V = Velocity of car = (100 km/h)(1000 m/1 km)(1 h/ 3600 s) = 27.78 m/s

Therefore,

s₂ = 27.78 t + 27.78

Now, we know that at the time of meeting:

s₁ = s₂

4.9 t² = 27.78 t + 27.78

4.9 t² - 270.78 t - 27.78 = 0

solving the equation and choosing the positive root:

t = 6.5 s

since, we want to know the time from the moment car crossed police car. Therefore, we add 1 second of starting time in this.

t = 6.5 s + 1 s

<u>t = 7.5 s</u>

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3 years ago
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