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Anni [7]
2 years ago
6

A migrating salmon heads in the direction n 45° e, swimming at 2 mi/h relative to the water. the prevailing ocean currents flow

due east at 4 mi/h. find the true velocity of the fish as a vector. (assume that the i vector points east, and the j vector points north.)
Physics
1 answer:
Sophie [7]2 years ago
5 0
Calculate for the x and y-components of the velocities involved in this item.

 2 mi/h (45° east)
  x-component = (2 mi/h)(sin 45°)
                     = 1.41 mi/h
  y-component = (2 mil/h)(cos 45°)
                     = 1.41 mi/h

4 mi/h (east)
  x-component = 4 mi/h
  y-component = 0 mi/h

Adding up the corresponding components:
     x-component = 5.4142 mi/h

     y-component = 1.4142 mi/h

Calculating for the resultant,
        R = sqrt ((x²) + (y²))
   
        R = sqrt ((5.4142 mi/h)² + (1.4142 mi/h)²)
             R = 5.60 mi/h

Answer: 5.6 mi/h
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Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

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7 0
2 years ago
Block 1 of mass m1 slides along a frictionless floor and into a one-dimensional elastic collision with stationary block 2 of mas
Elan Coil [88]
<span>Answers: (a) 2.0 m/s (b) 4 m/s

Method:

(a) By conservation of momentum, the velocity of the center of mass is unchanged, i.e., 2.0 m/s.

(b) The velocity of the center of mass = (m1v1+m2v2) / (m1+m2)

Since the second mass is initially at rest, vcom = m1v1 / (m1+m2)

Therefore, the initial v1 = vcom (m1+m2) / m1 = 2.0 m/s x 6 = 12 m/s

Since the second mass is initially at rest, v2f = v1i (2m1 /m1+m2 ) = 12 m/s (2/6) = 4 m/s </span>
7 0
2 years ago
You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an
a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, \mu=6\times 10^{-2}

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

KE=PE+W

\dfrac{1}{2}mv^2=mgh+W

W is the work done by the friction.

W=f\times d

f=\mu mg\ cos\theta

W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}

W=\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}

\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0
2 years ago
A particle of mass M is moving in the positive x direction with speed v. It spontaneously decays into 2 photons, with the origin
anygoal [31]

Solution :

Mass of the particle = M

Speed of travel = v

Energy of one photon after the decay which moves in the positive x direction = 233 MeV

Energy of second photon after the decay which moves in the negative x direction = 21 MeV

Therefore, the total energy after the decay is = 233 + 21

                                                                           = 254 MeV

So by the law of conservation of energy, we have :

Total energy before the decay = total energy after decay

So, the total relativistic energy of the particle before its decay = 254 MeV  

7 0
2 years ago
An amusement park ride spins you around in a circle of radius 2.5 m with a speed of 8.5 m/s. If your mass is 75 kg, what is the
zhannawk [14.2K]
B is the answer to this truly did the math
3 0
2 years ago
Read 2 more answers
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