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2 years ago

A ball thrown vertically upwards from ground

Physics

1 answer:

weqwewe [10]2 years ago

4 0

parabolic motion

$\tt t_{max}=\dfrac{2u~sin~\alpha }{g}$

$\tt h_{max}=\dfrac{u^2sin^2\alpha }{2g}$

vertically upwards ⇒ α = 90°⇒sin 90° = 1

$\tt 4=\dfrac{2u}{10}\rightarrow u=20~m/s$

$\tt h_{max}=\dfrac{20^2}{2.10}=\boxed{\bold{20~m}}$

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A 120-kg hollow spherical ball 1 m in diameter accelerates at a constant rate from rest to 5 rpm in 20 s and then continues to r

FromTheMoon [43]

Answer:

Explanation:

a )

moment of inertia of hollow ball

= 2 / 3 mR² , m is mass and R is radius of the ball

= 2 / 3 x 120 x .5²

= 20 kg m²

b )

5 rpm = 5 / 60 rps

n = .0833

angular velocity ω = 2πn= 2 x 3.14 x .0833= .523 rad /s

angular acceleration = increase in angular velocity / time

= .523 - 0 / 20

α = .02615 rad /s²

c )

Torque = moment of inertia x angular acceleration

= 20 x .02615

= .523 Nm

d )

θ = 1/2 α t²

= .5 x .02615 x 20²

= 5.23

2π n = 5.23 where n is required number

n = .83

3 0

3 years ago

4. A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the

Mrac [35]

As we know that speed of sounds is given as

$v = 332 + 0.6t$

here we know that

t = 23 degree C

now from above equation we will have

$v = 332 + (0.6)(23)$

$v = 345.8 m/s$

now we also know that distance between two consecutive resonance length is half of the wavelength

$L_2 - L_1 = \frac{\lambda}{2}$

$34 - 12 = \frac{\lambda}{2}$

$\lambda = 44 cm$

now we know that

$frequency = \frac{speed}{wavelength}$

$f = \frac{345.8}{0.44} = 786 Hz$

so frequency will be 786 Hz

7 0

3 years ago

If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?

Arisa [49]

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm$

Magnification,

$m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25$

Hence, the magnification of the image is 0.25.

7 0

3 years ago

The take-up reel of a cassette tape has a radius of 2.5 cm. Find the length of the tape that passes around the reel in 7.1 s whe

Jet001 [13]

Answer:

s = 0.337 m

Explanation:

First, we will find the angular displacement of the reel:

$\theta = \omega t$

where,

θ = angular displacement = ?

ω = angular speed = 1.9 rad/s

t = time taken = 7.1 s

Therefore,

θ = (1.9\ rad/s)(7.1 s)

θ = 13.5 rad

Now, we will find out the length of tape:

s = rθ

where,

s = length of tape = ?

r = radius of reel = 2.5 cm = 0.025 m

Therefore,

s = (0.025 m)(13.5 rad)

6 0

3 years ago

Seema knows the mass of basketball. What other information is needed to find the balls potential energy

Lelu [443]

Answer: The height (position) of the ball and the acceleration due gravity

Explanation:

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In the case of the Earth, in which the gravitational field is considered constant, the gravitational potential energy $U$ will be: