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4 years ago

Two bodies of masses 1000kg and 2000kg are separated 1km which is the gravitational force between them

Physics

1 answer:

denpristay [2]4 years ago

3 0

Answer:

1.33×10⁻¹⁰ N

Explanation:

F = GMm / r²

where G is the gravitational constant,

M and m are the masses of the objects,

and r is the distance between them.

F = (6.67×10⁻¹¹ N/m²/kg²) (1000 kg) (2000 kg) / (1000 m)²

F = 1.33×10⁻¹⁰ N

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Which of the following are examples of a force? (choose as many as apply)

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3. What is the difference between an emission spectrum and an absorption spectrum? What is the Bohr model of the atom, and how d

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8 0

3 years ago

Young Jeffrey is bored, and decides to throw some things out of the window for fun. But since he is also very curious, he decide

Inga [223]

Answer:

a) Stone v_{y} = - 7.25 ft / s , vₓ = 0.362 ft / s

b) tennis ball v_{y} = -3.16 ft / s , vₓ = 0.634 ft / s

c) golf ball v_{y} = - 1,536 ft / s, vx = 0.634 ft / s

2) golf ball

Explanation:

1) The average speed is defined with the displacement interval in the given time interval

v =( $x_{f}$ -x₀) / Δt

let's use this expression for each object

a) Stone

It tells us that it is released from y₀ = 10 ft and reaches the floor at

t = 0.788 s, but in the problem they tell us that the calculation is for t = 1.38 s

$v_{y}$ = (0-10) / 1.38

v_{y} = - 7.25 ft / s

in this interval a distance of x_{f} = 0.500 ft was moved away from the building (x₀ = 0 ft)

vₓ = (0.500- 0) / 1.38

vₓ = 0.362 ft / s

In my opinion it makes no sense to keep measuring the time after the stone has stopped.

b) tennis ball

It leaves the building at a height of y₀= 10ft and at the end of the period it is at a height of y_{f} = 5.63 ft, all this in a time of t = 0.788 + 0.591 = 1.38 s

the average vertical speed is

$v_{y}$ = (5.63 - 10) / 1.38

v_{y} = -3.16 ft / s

for horizontal velocity the ball leaves the building xo = 0 reaches the floor

x₁ = 0.500 foot and when bouncing it travels x₂ = 0.375 foot, therefore the distance traveled

x_{f} = x₁ + x₂

x_{f} = 0.500 + 0.375

x_{f} = 0.875 ft

we calculate

vₓ = (0.875 - 0) / 1.38

vₓ = 0.634 ft / s

c) The golf ball

the vertical displacement y₀ = 10 ft, and y_{f} = 7.88 ft

v_{y} = (7.88 - 10) / 1.38

v_{y} = - 1,536 ft / s

the horizontal displacement x₀ = 0 ft to the point xf = 0.875 ft

vₓ = (0.875 -0) / 1.38

vₓ = 0.634 ft / s

2) in this part we are asked for the instantaneous speed at the end of the time interval

a) the stone is stopped so its speed is zero

v_{y} = vₓ = 0

b) the tennis ball

It is at its maximum height so its vertical speed is zero

v_{y} = 0

horizontal speed does not change

vₓ = 0.634 ft / s

c) The golf ball

they do not indicate that it is still rising. Therefore its vertical speed is greater than zero

v_{y} > 0

horizontal speed is constant

vₓx = 0.634 ft / s

the total velocity of the object can be found with the Pythagorean theorem

v = √ (vₓ² + v_{y}²)

When reviewing the results, the golf ball is the one with the highest instantaneous speed at the end of the period

4 0

3 years ago

Two airplanes leave an airport at the same time.The velocity of the first airplane is 700 m/h at a heading of 31.3 the velocity

MArishka [77]

Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity

$v_{12} = v_1 - v_2$

here

speed of first plane is 700 mi/h at 31.3 degree

$v_1 = 700 cos31.3\hat i + 700 sin31.3\hat j$

$v_1 = 598.12\hat i + 363.7\hat j$

speed of second plane is 570 mi/h at 134 degree

$v_2 = 570 cos134 \hat i + 570 sin134 \hat j$

$v_2 = -396\hat i + 410\hat j$

now the relative velocity is given as

$v_{12} = (598.12 + 396)\hat i + (363.7 - 410)\hat j$

$v_{12} =994.12\hat i -46.3 \hat j$

now the distance between them is given as

$d = v* t$

$d = (994.12 \hat i - 46.3 \hat j)* 3$

$d = 2982.36\hat i - 138.9\hat j$

so the magnitude of the distance is given as

$d = \sqrt{2982.36^2 + 138.9^2}$

$d = 2985.6$ miles

so the distance between them is 2985.6 miles

5 0

3 years ago

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