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3 years ago

Please help me on this!

Physics

1 answer:

strojnjashka [21]3 years ago

8 0

Answer:

Explanation:

Am not sure

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2 years ago

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muminat

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Arocket launches at an angle of 33.6 degrees from the horizontal at a

babymother [125]

Answer:

Y component = 32.37

Explanation:

Given:

Angle of projection of the rocket is, $\theta=33.6$

Initial velocity of the rocket is, $u=58.5$

A vector at an angle $\theta$ with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle $\theta$ with the horizontal, then the horizontal and vertical components are given as:

$A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})$

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

$u_y=u\sin \theta$

Plug in the given values and solve for $u_y$ . This gives,

$u_y=(58.5)(\sin 33.6)\\u_y=58.5\times 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})$

Therefore, the Y component of initial velocity is 32.37.

4 0

2 years ago

The component of the external magnetic field along the central axis of a 78-turn circular coil of radius 34.0 cm decreases from

grigory [225]

Answer:

Induced current, I = 18.88 A

Explanation:

It is given that,

Number of turns, N = 78

Radius of the circular coil, r = 34 cm = 0.34 m

Magnetic field changes from 2.4 T to 0.4 T in 2 s.

Resistance of the coil, R = 1.5 ohms

We need to find the magnitude of the induced current in the coil. The induced emf is given by :

$\epsilon=-N\dfrac{d\phi}{dt}$

Where

$\dfrac{d\phi}{dt}$ is the rate of change of magnetic flux,

And $\phi=BA$

$\epsilon=-NA\dfrac{dB}{dt}$

$\epsilon=-78\times \pi (0.34)^2\dfrac{(0.4-2.4)}{2}$

$\epsilon=28.32\ V$

Using Ohm's law, $\epsilon=I\times R$

Induced current, $I=\dfrac{\epsilon}{R}$

$I=\dfrac{28.32}{1.5}$

I = 18.88 A

So, the magnitude of the induced current in the coil is 18.88 A. Hence, this is the required solution.

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2 years ago