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2 years ago

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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

he horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Does the ball clear the net?

1 answer:

ipn [44]2 years ago

7 0

Answer:

ball clears the net

Explanation:

$v_{o}$ = initial speed of launch of the ball = 20 ms^{-1}

$\theta$ = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

$v_{ox}$ = initial velocity = $v_{o} Cos\theta = 20 Cos5 = 19.92 ms^{-1}$

$t$ = time of travel

$X$ = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

$X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}$ Eq-1

Consider the motion of the ball along the vertical direction

$v_{oy}$ = initial velocity = $v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}$

$t$ = time of travel

$Y_{o}$ = Initial position of the ball at the time of launch = 2 m

$Y$ = Final position of the ball at time "t"

$a_{y}$ = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

$Y = Y_{o} + v_{oy} t + (0.5) a_{y} t^{2}\\Y = 2 + (20 Sin5) (\frac{7}{20 Cos5}) + (0.5) (- 9.8) (\frac{7}{20 Cos5})^{2}\\Y = 2.00758 m\\$

Since Y > 1 m

hence the ball clears the net

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2 years ago

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A standard 1 kilogram weight is a cylinder 51.0 mm in height and 42.0 mm in diameter. Determine the density of the material

worty [1.4K]

Answer:

14160 kg/m^3

Explanation:

First of all, we need to find the volume of the cylinder.

The volume of the cylinder is given by:

$V=\pi r^2 h$

where:

$r=\frac{d}{2}=\frac{42.0 mm}{2}=21.0 mm=0.021 m$ is the radius

$h=51.0 mm=0.051 m$ is the height

Substituting, we find

$V=\pi (0.021 m)^2 (0.051 m)=7.1 \cdot 10^{-5} m^3$

And the density is given by

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2 years ago

In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a

fgiga [73]

(a) The diffraction decreases

The formula for the diffraction pattern from a single slit is given by:

$sin \theta = \frac{n \lambda}{a}$

where

$\theta$ is the angle corresponding to nth-minimum in the diffraction pattern, measured from the centre of the pattern

n is the order of the minimum

$\lambda$ is the wavelength

a is the width of the opening

As we see from the formula, the longer the wavelength, the larger the diffraction pattern (because $\theta$ increases). In this problem, since the wavelength of the signal has been decreased from 54 cm to 13 mm, the diffraction of the signal has decreased.

(b) $10.8^{\circ}$

The angular spread of the central diffraction maximum is equal to twice the distance between the centre of the pattern and the first minimum, with n=1. Therefore:

$sin \theta = \frac{(1) \lambda}{a}$

in this case we have

$\lambda=54 cm = 0.54 m$ is the wavelength

$a=5.7 m$ is the width of the opening

Solving the equation, we find

$\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.54 m}{5.7 m})=5.4^{\circ}$

So the angular spread of the central diffraction maximum is twice this angle:

$\theta = 2 \cdot 5.4^{\circ}=10.8^{\circ}$

(c) $0.26^{\circ}$

Here we can apply the same formula used before, but this time the wavelength of the signal is

$\lambda=13 mm=0.013 m$

so the angle corresponding to the first minimum is

$\theta = sin^{-1} (\frac{\lambda}{a})=sin^{-1} (\frac{0.013 m}{5.7 m})=0.13^{\circ}$

So the angular spread of the central diffraction maximum is twice this angle:

$\theta = 2 \cdot 0.13^{\circ}=0.26^{\circ}$

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2 years ago

Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol

klasskru [66]

The current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

<u>Explanation:</u>

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μn = 1200 cm^2 / V s

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J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

J = (1.6 $\times$ 10^-19) / 0.001 (104 $\times$ 1200 + 1016 $\times$ 500)

J = 1012480 $\times$ 10^-16 A / m^2.

or

J = 1.01 $\times$ 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 $\times$ 10^-9 $\times$ 0.002 = 2.02 $\times$ 10^-12

So, the current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

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2 years ago