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KatRina [158]
4 years ago
9

A motorboat traveling from one shore to the other at a rate of 5m/s east encounters a current flowing at a rate of 3.5m/s north

a. What is the resultant velocity?
Physics
1 answer:
nikdorinn [45]4 years ago
6 0

Answer:

Resultant velocity will be equal to 6.10 m/sec

Explanation:

We have given a motorbike is traveling with 5 m/sec in east

And a current is flowing at a rate of 3.5 m /sec in north

We know that east and north is perpendicular to each other

So resultant velocity will be vector sum of both velocity

So resultant velocity v=\sqrt{5^2+3.5^2}=6.10m/sec

So resultant velocity will be equal to 6.10 m/sec

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Gauss's law combines the electric field over a surface with the area of the surface. From Coulomb's law we know that the electri
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The change in surface area of Gaussian surface with radius (r) is 8πr.

<h3>Electric field from Coulomb's law</h3>

The electric field experienced by a charge is calculated as follows;

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7 0
2 years ago
Does the speedometer of a car read average speed or instantaneous speed?
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8 0
3 years ago
ate around its central axis. A rope wrapped around the drum of radius 1.24 m exerts a force of 4.56 N to the right on the cylind
Mashcka [7]

Answer:

Magnitude the net torque about its axis of rotation is 1.3338 Nm

Explanation:

The radius of the wrapped rope around the drum, r = 1.24 m

Force applied to the right side of the drum, F = 4.56 N

The radius of the rope wrapped around the core, r' = 0.57 m

Force on the cylinder in the downward direction, F' = 7.58 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

\tau = F\times r\tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'\tau_{net} = - 4.56\times 1.24 + 7.58\times 0.57 \\\\= - 1.3338\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

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8 0
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