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3 years ago

With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 41 m

Physics

1 answer:

marishachu [46]3 years ago

6 0

Answer:

The speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

Explanation:

Given;

maximum vertical height of the throw, H = 41 m

Apply the following kinematic equation;

V² = U² + 2gH

where;

V is the final speed with which the ball will rise to a maximum height

U is the initial speed of the ball = 0

g is acceleration due to gravity = 0

V² = U² + 2gH

V² = 0² + 2gH

V² = 2gH

V = √2gH

V = √(2 x 9.8 x 41)

V = 28.35 m/s

Therefore, the speed must a ball be thrown vertically from ground level to rise to a maximum height is 28.35 m/s.

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A certain tuning fork vibrates at a frequency of 215 Hz while each tip of its two prongs has an amplitude of 0.832 mm. (a) What

Marysya12 [62]

Explanation:

It is given that,

Frequency of vibration, f = 215 Hz

Amplitude, A = 0.832 mm

(a) Let T is the period of this motion. It is given by the following relation as :

$T=\dfrac{1}{f}$

$T=\dfrac{1}{215}$

$T=4.65\times 10^{-3}\ s$

(b) Speed of sound in air, v = 343 m/s

It can be given by :

$v=f\times \lambda$

$\lambda=\dfrac{v}{f}$

$\lambda=\dfrac{343\ m/s}{215\ Hz}$

$\lambda=1.59\ m$

Hence, this is the required solution.

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3 years ago

What is the hang time when the person moves 6 m horizontally during a 1.25 m high jump?

AlekseyPX

Answer:

1 sec

Explanation:

Horizontal distance (x) = 6m

Vertical distance (y) = 1.25m

Hang time is the duration the object is in the air before it reaches maximum height.

The time of free fall is given by

t = √2y/g

g = acceleration due to gravity

t = √(2*1.25)/9.8

t = √2.5/9.8

t = 0.5secs

Hang time = 2*0.5

= 1 sec

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3 years ago

Monochromatic light of wavelength 385 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second

LiRa [457]

To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.

For this purpose we have that the constructive interference in waves can be expressed under the function

$a sin\theta = m\lambda$

Where

a = Width of the slit

d = Distance of slit to screen

m = Number of order which represent the number of repetition of the spectrum

$\theta =$ Angle between incident rays and scatter planes

At the same time the distance on the screen from the central point, would be

$sin\theta = \frac{y}{d}$

Where y = Represents the distance on the screen from the central point

PART A ) From the previous equation if we arrange to find the angle we have that

$\theta = sin^{-1}(\frac{y}{d})$

$\theta = sin^{-1}(\frac{1.4*10^{-2}}{3})$

$\theta = 0.2673\°$

PART B) Equation both equations we have

$a sin\theta = m\lambda$

$a \frac{y}{d} = m\lambda$

Re-arrange to find a,

$a = \frac{(2)(385*10^{-9})(3)}{(1.4*10^{-2})}$

$a = 1.65*10^{-4}m$

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3 years ago

If 41.5 mol of an ideal gas occupies 86.5 L at 27.00 °C, what is the pressure of the gas?

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Sharon throws a 0.20 Kg with an acceleration of 10 m/s/s.

OlgaM077 [116]

Force=A×M
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