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posledela
3 years ago
14

an air craft is travelling due north with velocity of 100m/s astrong wind blow from the west with in velocity of 25m/s the reshs

tant velocity​
Physics
1 answer:
OleMash [197]3 years ago
8 0

Explanation:

The aircraft is traveling north at 100 m/s.

The wind blows from the west (towards the east) at 25 m/s.

The two vectors form a right triangle.  The magnitude of the resultant velocity can be found with Pythagorean theorem:

v² = vx² + vy²

v² = (25 m/s)² + (100 m/s)²

v = 103 m/s

The direction can be found with trigonometry:

θ = atan(vy / vx)

θ = atan(100 / 25)

θ = 76.0°

The resultant velocity is 103 m/s at 76.0° north of east.

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GPE = 5 x 1
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GPE = 5 x 1.5
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4 years ago
Wich best describe the act of using senses or tools to gather information?
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The act of using senses or tools to gather information is called "observation." In scientific method, it is very essential to have this one. Through observation, we are able to get what we wanted to get using our senses and other tools. This process is very tedious if the field that you are studying is quite complicated.  Hope this Helps! PLEASE PUT BRAINLIEST!

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A ramp is needed to allow vehicles to climb a 2 foot wall. The angle of elevation in order for the vehicles to safely go up must
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Answer:

Yes the ramp can be safely used

Explanation:

Here, we have

Length of longest ramp = 5 ft

Height of wall = 2 ft

Therefore, the sine of the angle adjacent to the ramp which is equal to the angle of elevation is given by;

Sin\theta = \frac{Opposite \, side \, to\,  angle}{Hypothenus\, side \, of\,  triangle}

Where:

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Therefore,

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4 years ago
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
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wariber [46]

Answer:

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