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3 years ago

The speed of a cruise ship is 50,000 m/hr. How fast is it moving in m/s?

Physics

1 answer:

solong [7]3 years ago

3 0

Answer:

13.888888 (on forever)

Explanation:

there are 3600 seconds in an hour if it is moving at an hourly rate then you just need to divide 50,000 buy 3600 in which case you get 13.88888

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Select all of the mixtures: oxygen gold trail mix soup air

Alexxandr [17]

Answer:

trail mix, soup, and gold

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Explanation:

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3 years ago

Read 2 more answers

Sometimes balance point may not be obtained on the potentiometer wire why

scoundrel [369]

Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage

.
The balance point is not on the potentiometer wire - this statement means that

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2 years ago

Ngan has a weight of 314.5 N on Mars and a weight 833.0 N on Earth. What is Ngan's mass on Earth?

mr_godi [17]

Ngan's mass on earth is 85kg.

Ngan has a weight on Mars = 14.5 N

Ngan’s weight on Earth = 833.0 N

Ngan’s mass on Earth = ?

Fg,earth = mg(earth)

M = Fg,earth / g(earth)

M = 833.0 N / 9.8 m/s2

M = 85 kg

4 0

3 years ago

Read 2 more answers

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 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

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.