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Ksenya-84 [330]
3 years ago
11

One plane departed from New York, and at the same time another plane departed from Mexico City. They flew toward each other at r

ate of 650 and 550 mph. If New York and Mexico City are 3000 miles apart, in how many hours did the planes pass each other?
Physics
2 answers:
gayaneshka [121]3 years ago
8 0

Answer:

Explanation:

We also know that, v = S/t which can be rewritten as v * t = S

Where,

v = velocity

S = distance

t = time

i. v1*t = S1

ii. v2*t = S2

iii. s1 + s2 = 3000

Substituting i and ii into equation iii,

St = S1 + S2

v1*t + v2*t = 3000

650*t + 550*t = 3000

1200*t = 3000

t = 3000/1200

t = 2.5 h.

sdas [7]3 years ago
3 0

Answer:

30 hours

Explanation:

Average speed = distance/time

Let the time taken for each plane to pass each other be t

Average speed of plane that left from New York = 650 mph

Distance covered = 650t miles

Average speed of plane that left from Mexico = 550 mph

Distance covered = 550t miles

Distance apart = 3000 miles

650t - 550t = 3000

100t = 3000

t = 3000/100 = 30 hours

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A child in a boat throws a 5.80-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 24.6kg and the ma
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Answer:

-0.912 m/s

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where m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg are the mass of the child, the boat and the package, respectively. , v_p = 10m/s, v_b are the velocity of the package and the boat after throwing.

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I am struggling on this physics question. Brainly is my last hope. Could somebody please provide an answer to this question, wit
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1) 29.4 N

The force of gravity between two objects is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M and m are the masses of the two objects

r is the separation between the centres of mass of the two objects

In this problem, we have

M=5.97\cdot 10^{24} kg (mass of the Earth)

m=3.0 kg (mass of the box)

r=R=6.37\cdot 10^6 m (Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)

Substituting into the equation, we find F:

F=\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})(3.0)}{(6.37\cdot 10^6)^2}=29.4 N

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Let's now calculate the ratio F/m. We have:

F = 29.4 N

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Subsituting, we find

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where a is the acceleration of the object. Re-arranging,

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which is exactly equal to the quantity we have calculated above.

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