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2 years ago

12

Stars set about how many minutes earlier each day? A. 8 B. 6 C. 4 D. 2

1 answer:

Furkat [3]2 years ago

5 0

It is c which is four minutes per day

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A ferry is travelling at 10 m/s relative to the river towards the south. The river is flowing at 2 m/s relative to

mixer [17]

Answer:

11 m/s south

Explanation:

The velocity of the passenger relative to the river bank is equal to the velocity of the passenger relative to the ferry, plus the velocity of the ferry relative to the river, plus the velocity of the river relative to the river bank.

v_passenger,bank = v_passenger,ferry + v_ferry,river + v_river,bank

If we take north to be positive and south to be negative:

v = 1.0 m/s + (-10 m/s) + (-2 m/s)

v = -11 m/s

v = 11 m/s south

8 0

3 years ago

The first model of the atom was developed through

Ilya [14]

It was developed through Democritus who was a greek philosopher.

Hope this helps

7 0

2 years ago

Read 2 more answers

A jet airliner moving initially at 548 mph

sasho [114]

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

$v_1 =548 mph$ along the x-direction

$v_2 = 343 mph$ in a direction $67^{\circ}$ north of east.

To find the resultant, we must resolve both vectors on the x- and y- axis:

$v_{1x}= 548 mph$

$v_{1y}=0$

$v_{2x} = (343 mph)( cos 67^{\circ})=134.0 mph$

$v_{2y} = (343 mph)( sin 67^{\circ})=315.7 mph$

So, the components of the resultant velocity in the two directions are

$v_{x}=548 mph+134 mph=682 mph$

$v_{y}=0 mph+315.7 mph=315.7 mph$

So the new speed of the aircraft is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(682 mph)^2+(315.7 mph)^2}=751.5 mph$

3 0

2 years ago

An object at free fall near the Earth’s surface experiences constant velocity.

mariarad [96]

False it accelerates.

7 0

2 years ago

A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses the buildi

Vlada [557]

Explanation:

Given that,

Distance, d = -56.3 m

It strikes the ground 4.00 seconds after being thrown.

Using second equation of motion to find the speed was the rock thrown. So,

$d=ut+\dfrac{1}{2}at^2$

Here, a = -g

$d=ut-\dfrac{1}{2}gt^2\\\\-56.3=u(4)-4.9(4)^2\\\\-56.3=4u-78.4\\\\u=5.52\ m/s$

Let it will cover a distance of s meters. So,

$s=\dfrac{v^2-u^2}{-2g}\\\\s=\dfrac{0^2-(5.52)^2}{-2\times 9.8}\\\\s=1.55\ m$

4 0

2 years ago