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Sati [7]
3 years ago
11

The difference between speed and velocity is that

Physics
2 answers:
tekilochka [14]3 years ago
4 0

Answer:

velocity is a vector and requires a direction.

Explanation:

Speed is defined as the total distance covered per unit time and the velocity is defined as the total displacement per unit time.

Speed is a scalar quantity and velocity is a vector quantity. The quantities that have both magnitude and direction are vector quantity and the quantity that have only magnitude are scalar quantity.The SI unit of both velocity and speed is m/s.

So, the correct option is (c) "  velocity is a vector and requires a direction ".

kvv77 [185]3 years ago
3 0

Answer: C) velocity is a vector and requires a direction.

Explanation:

In physics, there are two types of quantities:

- scalars: these are quantities that have only a magnitude

- vectors: there are quantities that have both magnitude and direction

As an example, speed is a scalar while velocity is a vector. Therefore, speed has only a magnitude, while velocity has both magnitude and direction: therefore, the difference between the two quantities is that velocity is a vector and requires a direction, as stated in option C.

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2 years ago
A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate
kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid R=1.06 m

Emf induced E=8.50\times 10^{-6} V/m

no of turns per meter n=450

we know Induced EMF is given by

\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A

Magnetic Field is given by

B=\mu _0ni

thus \frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}

Area of cross-section

A=\pi R^2 where

solving integration we get

E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2

where r=distance from axis

R=radius of Solenoid

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s

4 0
3 years ago
A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity incr
____ [38]

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}

Where:

\omega_{0}: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

\omega_{f} = \omega_{0} + \alpha t

Where:

\omega_{f}: is the final angular velocity = 57 rad/s

\alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2}

Hence, the number of revolutions is:

\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev

Therefore, the number of revolutions is 44.6.

       

I hope it helps you!

4 0
2 years ago
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