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Kisachek [45]
2 years ago
14

A kayaker paddles at 4.0 m/s in a direction 30° south of west. He then turns and paddles at 3.7 m/s in a direction 20° west of s

outh.
What is the magnitude of the kayaker’s resultant velocity? Round your answer to the nearest tenth.

___m/s
What is the direction of the kayaker’s resultant velocity?

___ ° south of west
please help
Physics
2 answers:
user100 [1]2 years ago
7 0
7.2 m/s 

49 degrees south of west

Gre4nikov [31]2 years ago
5 0

Answer: The magnitude of the kayaker's resultant velocity = 7.23 m/s

and the direction is 49.26° south of west.

Explanation:

Initial velocity, u = 4.0 m/s 30° south of west.

Final velocity, v = 3.7 m/s 20° west of south °≈ 70° south of west.

Writing in terms of component:

u = -4.0 cos 30 i - 4.0 sin 30 j = -3.46 i -2.0 j

v = -3.7 cos 70 i - 3.7 sin 70 j = -1.26 i - 3.48 j

Resultant velocity = addition of the two velocity vectors

R = (-3.46 i -2.0 j)+(-1.26 i - 3.48 j) = -4.72 i - 5.48 j

Magnitude of the vector = √(-4.72)²+( - 5.48)² = 7.23 m/s

Direction of the resultant velocity vector, θ = tan⁻¹ (-5.48÷ (-4.72)) = 49.26° south of west.

Thus, the magnitude of the kayaker's resultant velocity = 7.23 m/s

and the direction is 49.26° south of west.

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Let \theta be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

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The current has velocity vector (relative to the Earth)

\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath

The swimmer's resultant velocity (her velocity relative to the Earth) is then

\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}

\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ

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6 0
2 years ago
A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it
vredina [299]

Answer:

The distance is 11 m.

Explanation:

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Using formula of friction force

F= \mu m g....(II)

Put the value of F in the equation (II) from equation (I)

ma=\mu mg....(III)

a = \mu g

Put the value in the equation (III)

a=0.24\times9.8

a=2.352\ m/s^2

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Using equation of motion

s = ut+\dfrac{1}{2}at^2

s=0+\dfrac{1}{2}2.352\times(3.0)^2

s=10.584\ m\ approx\ 11\ m

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3 0
2 years ago
A fly has a mass of 1 gram at rest. how fast would it have to be traveling to have the mass of a large suv, which is about 3000
Zigmanuir [339]

We solve this using special relativity. Special relativity actually places the relativistic mass to be the rest mass factored by a constant "gamma". The gamma is equal to 1/sqrt (1 - (v/c)^2). <span>

We want a ratio of 3000000 to 1, or 3 million to 1. 

</span>

<span>Therefore:
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8 0
2 years ago
The burning of fossil fuels contributes to the addition of greenhouse gases to the atmosphere. These gases trap thermal energy i
denpristay [2]

Answer:

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5 0
2 years ago
Read 2 more answers
A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. Wh
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

4 0
2 years ago
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