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crimeas [40]
2 years ago
6

1. Los Angeles lies on Pacific plate, San Francisco lies on North American plate. When will the two cities meet if the distance

between them is 600 km and Pacific plate is moving (NW) at 5 cm/yr?
Physics
1 answer:
spin [16.1K]2 years ago
8 0

Los Angeles lies on the Pacific plate, San Francisco lies on the North American plate, and the meeting point of the two cities is mathematically given as

T = 120 x 105 years

<h3>What is the meeting point of the two plates?</h3>

Generally, the equation for Distance is mathematically given as

D = Rate x Time

Therefore

T = D/R

T = (600 x 105) / 5

T = 120 x 105 years

In conclusion, the meeting point of the two plates will be

T = 120 x 105 years

Read more about Arithmetic

brainly.com/question/22568180

#SPJ1

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How long would it take a plane to fly 1,054 km with an average speed of 886 km/h?
Leviafan [203]

Answer:

1.19 hours

Explanation:

divide distance by speed. hope this helps

8 0
2 years ago
Starting at (0,0) an object travels 36 meters north and then it covers 20 meters east. What is
Svetradugi [14.3K]

Answer:

Explanation:

Using the pythagoras theorem, the displacement is expressed as;

d² = x²+y²

y = 36m (north)

x = 20m east

Substitute;

d² = 36²+20²

d² = 1296+400

d² = 1696

d = √1696

d = 41.18m

For the direction;

theta = tan^-1(y/x)

theta = tan^-1(36/20)

theta = tan^-1(1.8)

theta = 60.95°

Hence the magnitude is 41.18m and the direction is 60.95°

8 0
3 years ago
Jupiter's moon Io has a radius of 1.82 ✕ 106 m and a measured gravitational field of 1.80 m/s2. What is its mass (in kg)?
AveGali [126]

Answer:

8.94*10^22 kg

Explanation:

Given that

Mass of Lo, M = ?

Radius of Lo, r = 1.82*10^6 m

Acceleration on Lo, g = 1.80 m/s²

Gravitational constant, G = 6.67*10^-11

Using the formula

g = GM/r²

Solution is attached below

Answer is 8.94*10^22 kg

7 0
3 years ago
How does the sun’s gravitational attraction impact Earth’s motion?
Tasya [4]
The Sun's gravitational pull keeps our planet orbiting the Sun <span>in a nice nearly-circular orbit.</span>
7 0
3 years ago
A 5kg bag falls a verticle height of 10m before hitting the ground.
g100num [7]

Answer:

u = 7m {s}^{ - 1}

Explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

me(1) = me(2)

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

m \times g \times h   =   \frac{1}{2}  \times m \times u {}^{2} \\

we divide both sides of the equation with mass to make the math easier.

9.8 \times 10 =  \frac{1}{2}  \times u {}^{2}  \\  \frac{98}{2}  = u {}^{2}  \\ u { }^{2} = 49 \\ u = 7

7 0
3 years ago
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