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Murrr4er [49]
3 years ago
15

Suppose a boat moves at 12.0m/s relative to the water. If the boat is in a river with the current directed east of 2.50m/s, what

is the boat's speed relative to the groiund when it is heading (a) east, with the current, and (b) west, against the current?
Physics
1 answer:
inessss [21]3 years ago
8 0

Answer:

(a) 14.5 m/s

(b) 9.5 m/s

Explanation:

Let the speed of the boat in the still water (i.e relative to ground) is v_0.

Given that the speed of the boat relative to water, v_r=12.0 m/s.

Speed of the water current of the river relative to the ground, u_0=2.50 m/s towards east.

(a) When the boat is heading toward the east (in the same direction of the current of the river)

The relative velocity, v_r, of the boat with respect to ground is

v_r=v_0-u_0

\Rightarrow 12=v_0-2.5

\Rightarrow v_0=12+2.5=14.5 m/s

Hence, the boat's speed relative to the ground when it is heading east, with the current, is 14.5 m/s.

(b) When the boat is heading toward the west (in the opposite direction of the current of the river)

The relative velocity, v_r, of the boat with respect to ground is

v_r=v_0-(-u_0)

\Rightarrow 12=v_0+2.5

\Rightarrow v_0=12-2.5=9.5 m/s

Hence, the boat's speed relative to the ground when it is heading west, against the current, is 9.5 m/s.

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A team exerts a force of 10 N towards the south in a tug of war. The other, opposite team, exerts a force of 17 N towards the no
ikadub [295]

Answer:

Option C

Explanation:

According to the question:

Force exerted by the team towards south, F = 10 N

Force exerted by the opposite team towards North, F' = 17 N

Net Force, \vec{F_{net}} = \vec{F'} - \vec{F}

\vec{F_{net}} = \vec{F'} - \vec{F} = 17 - 10 = 7 N

Thus the force will be along the direction of force whose magnitude is higher

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Two lumps of clay having equal masses and speeds, but traveling in opposite directions, collide and stick together. Part A Which
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Answer:

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8 0
3 years ago
A constant force of 3.2 N to the right acts on a 18.2 kg mass for 0.82 s. (a) Find the final velocity of the mass if it is initi
zheka24 [161]

Explanation:

The given data is as follows.

      F = 3.2 N,      m = 18.2 kg,

      t = 0.82 sec

(a)  Formula for impulse is as follows.

          I = Ft = \Delta P

        Ft = m(v_{f} - v_{i})

or,    v_{f} = \frac{Ft}{m} + v_{i}

Putting the given values into the above formula as follows.

      v_{f} = \frac{Ft}{m} + v_{i}

              = \frac{3.2 \times 0.82}{18.2} + 0

              = 0.144 m/s

Therefore, final velocity of the mass if it is initially at rest is 0.144 m/s.

(b)  When velocity is 1.85 m/s to the left then, final velocity of the mass will be calculated as follows.

           Ft = m(v_{f} - v_{i})

or,      v_{f} = \frac{Ft}{m} + v_{i}

                  = \frac{3.2 \times 0.82 sec}{18.2} - 1.85

                  = -1.705 m/s

Hence, we can conclude that the final velocity of the mass if it is initially moving along the x-axis with a velocity of 1.85 m/s to the left is 1.705 m/s towards the left.

4 0
3 years ago
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A 5.0 A electric current passes through an aluminum wire of 4.0~\times~10^{-6}~m^2 cross-sectional area. Aluminum has one free e
Serhud [2]

Answer: The electron number density (the number of electrons per unit volume) in the wire is 6.0 \times 10^{28} m^{-3}.

Explanation:

Given: Current = 5.0 A

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The electron density is calculated as follows.

n = \frac{density}{mass per atom}\\= \frac{\rho}{\frac{M}{N_{A}}}\\

where,

\rho = density

M = molar mass

N_{A} = Avogadro's number

Substitute the values into above formula as follows.

n = \frac{\rho \times N_{A}}{M}\\= \frac{2.7 g/cm^{3} \times 6.02 \times 10^{23}/mol}{27 g/mol}\\= \frac{16.254 \times 10^{23}}{27} cm^{3}\\= 0.602 \times 10^{23} \times \frac{10^{6} cm^{3}}{1 m^{3}}\\= 6.0 \times 10^{28} m^{-3}

Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is 6.0 \times 10^{28} m^{-3}.

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