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vladimir1956 [14]
4 years ago
13

A sailboat is heading directly north at a speed of 20 knots (1 knot 50.514 m/s). the wind is blowing toward the east with a spee

d of 17 knots. (a) determine the magnitude and direction of the wind velocity as measured on the boat. (b) what is the component of the wind velocity in the direction parallel to the motion of the boat?
Physics
1 answer:
stellarik [79]4 years ago
5 0
(a) The magnitude of the wind as it is measured on the boat will be the result of the two vectors. Since they are at 90°, the resultant can be determined by the Pythagorean theorem.
 
                   R = sqrt ((20 knots)² + (17 knots)²)
                    R = sqrt (400 + 289)
                      R = 26.24 knots

The direction of the wind will have to be angle between the boat and the resultant.
                   cos θ = (20 knots)/(26.24 knots)
                        θ = 40.36°

Hence, the direction is 40.36° east of north.

(b) As stated, the wind is blowing in the direction that is to the east. This means that it only has one direction. Parallel to the motion of the boat, the magnitude of the wind velocity will have to be zero. 
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Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

F_{net} = \sqrt{F_1^2 + F_2^2}

here given that

F_1 = 130.0 N

F_2 = 4500.0 N

now we will plug in all data in the above equation

F_{net} = \sqrt{4500^2 + 130^}

F_{net} = 4501.9 N

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A body of mass 5 kg has a velocity<br><br><br>20 MS <br><br>-1 its Momentum is​
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Answer:

Momentum is 100 kg.m/s

Explanation:

given

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To find : momentum (P)

We know that momentum is given by equation:

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How do I remember the Mohs hardness scale?
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PRACTICE ANOTHER A cube of wood having an edge dimension of 19.7 cm and a density of 647 kg/m3 floats on water. (a) What is the
dedylja [7]

CHECK COMPLETE QUESTION

The cube of wood having an edge dimension of 19.7cm and a density of 647kg/m3 floats on water

(a) What is the distance from the horizontal top surface of the cube to the water level answer in cm

(b) What mass of lead should be placed on the cube so that the top of the cube will be just level with the water surface answer in kg

Answer:

a)6.29cm

b)2.78 kg

Explanation:

Given:

Let us calculate the volume first, we were given dimension as 19.7cm=0.197m

Volume is (0.197 meters)³ = 0.00764m³

Then we can calculate the mass as;

Given mass is 647 kg/m³ x 0.00774m³ = 4.947kg

The weight = mass × acceleration due to gravity

weight = 4.947 x 9.8 N/kg = 48.44N

By Floating we can say the the buoyancy force has to equal the weight (48.44 N) which has

which is equal to the weight of volume of the displaced water. Or the mass, the calculation is the same.

We know that density of fresh water at 20ºC is 998 kg/m³

Then we can calculate the volume of displaced water as

4.947 kg / 998 kg/m³ = 0.00496 m³

We know that the displaced water has a shape of a rectangular solid with 0.197 meters on the two horizontal dimensions, and h as the height to the surface then

V = 0.197²h = 0.00496

0.00496= 0.197²h

h = 0.1278 meters or 12.78 cm

Then the the distance exposed, would be 19.7–12.78 = 6.29 cm

b) ifthe cube is fully submerged, the volume of the displaced water is 0.00774m³

mass of displaced water is 0.00774m³ x 998 kg/m³ = 7.724 kg

Added mass is the mass of the displaced water – mass of block

= 7.724–4.947 = 2.78 kg

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3 years ago
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DaniilM [7]

Answer:

If the frequency of the source is increased the current in the circuit will decrease.

Explanation:

The current through the circuit is given as;

I = \frac{V}{Z}

Where;

V is the voltage in the AC circuit

Z is the impedance

Z = \sqrt{R^2 + X_L^2}

Where;

R is the resistance

X_L is the inductive reactance

X_L = ωL = 2πfL

where;

L is the inductance

f is the frequency of the source

Finally, the current in the circuit is given as;

I = \frac{V}{\sqrt{R^2 + (2\pi fL)^2} }

From the equation above, an increase in frequency (f) will cause a decrease in current (I).

Therefore, If the frequency of the source is increased the current in the circuit will decrease.

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