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Alekssandra [29.7K]
2 years ago
10

A boat heads directly across a river. Its speed relative to the water is 2.6 m/s. It takes it 355 seconds to cross, but it ends

up 690 m downstream. How fast is the boat going relative to the bank of the river?
Physics
1 answer:
Zigmanuir [339]2 years ago
7 0

Answer:

Vr = 3.24m/s

The boat is going 3.24m/s relative to the bank of the river.

Explanation:

The relative speed of the boat to the bank Vr is the resultant of speed of boat relative to the water Vb and the speed of boat as a result of the water current or wind Vw

Vr = √(Vb^2 + Vw^2) .....1

Given;

Vb = 2.6m/s

Vw = distance downstream/time = 690m/355s

Vw = 1.94m/s

From equation 1 above; substituting the values

Vr = √(2.6^2 + 1.94^2)

Vr = 3.24m/s

The boat is going 3.24m/s relative to the bank of the river.

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Answer:

x=\dfrac{r}{\sqrt2}

Explanation:

Given that

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Q=Charge on the ring

The electric filed at distance x given as

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For maximum condition

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E=K{Q}{(r^2+x^2)^{-3/2}}

\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}

For maximum condition

\dfrac{dE}{dx}=0

K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0

r^2+x^2-3x^2=0

x=\dfrac{r}{\sqrt2}

At x=\dfrac{r}{\sqrt2} the electric field will be maximum.

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2 years ago
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When you're talking about gravity, it's easy to identify the equal
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The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it
MrRissso [65]

Answer:

Explanation:

From the question we are told that

   The moment of inertia is  I = 2.50 \ kg \cdot m^2

    The final  angular speed is w_f =  400 rev/min  =  \frac{400 * 2\pi}{60}  = 41.89 \ rad/s

     The time taken is  t =  8.0 s

      The initial angular speed is  w_i  =  0\ rad/s

Generally the average angular acceleration is mathematically represented as

        \alpha  =  \frac{w_f - w_i }{t}

=>     \alpha  =  \frac{41.89}{8}

=>      \alpha  = 5.24 \ rad/s^2

Generally the torque is mathematically represented as

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3 years ago
Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q
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Answer:

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In this case q_{1}=q_{2}=q, so we have:

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Replacing the given values:

q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}

q=4.2*10^{-5}C

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