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3 years ago

3. A model rocket is launched straight upward at 58.8 m/s.

Physics

1 answer:

SIZIF [17.4K]3 years ago

8 0

Let us assume that rocket only runs in initial energy and not using its own to flying.

Also , let upward direction is +ve and downward direction is -ve .

Initial velocity , u = 58.8 m/s .

Acceleration due to gravity , $g=-9.8\ m/s^2$ .

Final velocity , v - = 0 m/s .

We know , by equation of motion .

$v^2-u^2=2gh\\\\2gh_{max}=0^2-58.8^2\\\\h_{max}=\dfrac{0^2-58.8^2}{-2\times 9.8}\\\\h_{max}= 176.4\ m$

Hence, this is the required solution .

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In a microwave oven, electrons describe circular motion in a magnetic field within a special tube called amagnetron; as you'll l

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Answer:

The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.

Explanation:

Given that,

Diameter = 2.62 mm

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(A). We need to calculate the magnetic field strength

Using formula of the magnetic field strength

$B=\dfrac{2\pi mf}{e}$

Where, f = frequency

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$B=\dfrac{2\times3.14\times9.1\times10^{-31}\times2.15\times10^{9}}{1.6\times10^{-19}}$

$B=0.077\ T$

(B). We need to calculate the energy of electron

Using formula of energy

$E=\dfrac{1}{2}m(r\omega)^2$

$E=\dfrac{1}{2}\times9.1\times10^{-31}\times(1.31\times10^{-3}\times2\pi\times2.15\times10^{9})^2$

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The energy in eV

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3 years ago

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3 years ago

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t

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Answer:

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

$y=y_{0}+uSin\theta t-1/2 gt^{2}$

$y=2+20Sin5 t-4.9t^{2}$

As it hits the ground in time t, so put y = 0

$0=2+1.74 t-4.9t^{2}$

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The ball travels a horizontal distance x in time t

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X = 16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

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t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

$y'=y_{0}+uSin\theta t'-1/2 gt'^{2}$

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3 years ago

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