A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4). What is the magnitude of her velocity ( v ) with respect to the shore?

Answers

Answer 1

Answer:

The magnitude of her velocity with respect to the shore will be 5.70 Km/h. The change of displacement with respect to time is defined as velocity.

What is velocity?

Velocity is defined as the change in displacement with respect to time. The quantity of velocity is a vector quantity. It is a component that is time-based. It is measured in meters per second.

The given data in the problem is;

Q is the magnitude of Velocity of the boat = 4.5 km/hr

P is the magnitude of Velocity of the river flowing = 3.5 km/hr

R is the resultant of velocity P and Q=?

θ is the angle between the two velocities = 90°

From the law of vector addition;

$\rm R= √(P^2+Q^2+2PQcos\theta) \n\n \rm R= √((3.5)^2+(4.5)^2+2* 3.5 * 4.5 cos90^0) \n\n \rm R=5.70 \ m/sec$

Hence the magnitude of her velocity with respect to the shore will be 5.70 Km/h.

To learn more about the velocity refer to the link;

brainly.com/question/862972

Answer 2

Answer:

Answer: The magnitude of her velocity ( v ) with respect to the shore is 5.70 km/h.

Explanation:

Magnitude of Velocity of the boat = Q

Magnitude of Velocity of the river flowing = P

R = Resultant velocity due to velocity of boat and velocity of river.

Applying Law of triangles of vector addition :

$R=√(P^2+Q^2+2PQCos\theta)$

From the figure attached:

P = 3.5 k/h, Q = 4.5 km/h

$\theta= 90^o$

$R=√(P^2+Q^2+2PQCos\theta)=√((3.5)^2+(4.5)^2+3.5* 4.5* cos90^o)=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^(-1)(Qsin\theta)/(P+Qcos\theta)=tan^(-1)(4.5 sin 90^o)/(3.5+4.5 cos90^o)=tan^(-1)(4.5)/(3.5)=52.12^o$

The magnitude of her velocity ( v ) with respect to the shore is 5.70 km/h.

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Answers

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If a ball is tossed straight up into the air, at what position is its potential energy the greatest?A.when it reaches the top of its flight
B.when it first leaves your hand
C.when it approaches the top of its flight
D.when it begins to fall back to your hand.
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Answers

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Explanation:

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where

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An object of mass 13kg is falling in air and experiences a force due to air resistance of 9N. Determine the magnitude of net force acting on the object.

Answers

Given data

mass (m) = 13 kg

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air resistance (F₂ ) = 9 N, acts in opposite direction

Net force is defined as sum of all forces

Fnet = F₁ - F₂

= 127.4 - 9

= 118.4 N

Net force acting on the object is 118.4 N

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