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Hoochie [10]
3 years ago
11

Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint

of the distance is marked to make it easier to see how the locations compare. Assume the spaceship has the same mass throughout the trip (that is, it is not burning any fuel). Rank the five positions of the spaceship from left to right based on the strength of the gravitational force that Earth exerts on the spaceship, from strongest to weakest.

Physics
1 answer:
djyliett [7]3 years ago
6 0

Answer:

F_5 >F_4>F_1 >F_2>F_3

Where F_i represent the force for each of the 5 cases i -1,2,3,4,5 presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

F = G \frac{m_{Earth} m_{Spaceship}}{r^2}

Where G is a constant G = 6.674 x10^{-11} m^2/ (ks s^2)

m_{Earth} represent the mass for the earth

m_{spaceship} represent the mass for the spaceship

r represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

F_5 >F_4>F_1 >F_2>F_3

Where F_i represent the force for each of the 5 cases i -1,2,3,4,5 presented on the figure attached.

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A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

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To Play Ice Hockey each player needs
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Answer:

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Determine which heat transfers below are due to the process of conduction. I) You walk barefoot on the hot street and it burns y
Rudik [331]

I) You walk barefoot on the hot street and it burns your toes.

The road is in direct contact with your skin. Thermal energy from the road will transfer to the bottom of your feet, then to the rest of your body. This is an example of conduction.


II) When you get into a car with hot black leather in the middle of the summer and your skin starts to get burned.

Just like in the previous example, the hot leather is in direct contact with your skin (I guess if you're going to drive naked). Thermal energy from the leather will transfe to your skin, then to the rest of your body. This is also conduction.


III) A flame heats the air inside a hot air balloon and the balloon rises.

The flame heats air directly at the bottom of the balloon. The warm air expands and becomes less dense. This will rise and let the unheated, denser air in the balloon fall down toward the flame. This is an example of the convection cycle.


IV) A boy sits to the side of a campfire. He is 10 feet away, but still feels warm.

The campfire heats air directly nearby. The warm air expands and moves away from the fire in all directions, leaving behind unheated, denser air to be heated up. Some of the warm air reaches the boy. This is another example of convection.


The answer is A) 1 and 2.

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3 years ago
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1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant
Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction (v_j) = 550 mph

Velocity of jet stream from west to east direction (v_s)=80\ mph

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph

Similarly, the velocity of the stream is, \vec{v_s}=80\vec{i}

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle \theta with the x axis in the third quadrant.

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

5 0
3 years ago
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