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3 years ago

6

represents the space-time, speed-time and acceleration-time graphs for a ball pulled upwards with a speed of 10 m / s from 1 met

er of ground

1 answer:

alexandr1967 [171]3 years ago

5 0

Gyhugugygubuhuhuhubu

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A body moves a speed of 20km/hr in 15secs, what is the distance covered.

Taya2010 [7]

Answer:

aa

Explanation:

aa

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3 years ago

What is the wavelength, in nm, of a photon with energy

lara31 [8.8K]

Answer:

(a) λ = 4136 nm → infrared

(b) λ = 413.6 nm → visible light

(c) λ = 41.36 nm → ultraviolet

Explanation:

The wavelength of infrared is on the range of 700 nm to 1000000 nm

The wavelength of visible light is between 400 nm and 700 nm

The wavelength of ultraviolet ray on the range of 10 nm to 400 nm

The wavelength of photon is given by;

E = hf

f is the frequency of the wave = c / λ

$E = h\frac{c}{\lambda}\\\\ \lambda = \frac{hc}{E}$

Where;

c is the speed of light = 3 x 10⁸ m/s

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

(a) 0.3 eV = 0.3 x 1.602 x 10⁻¹⁹ J

$\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(0.3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-6} \ m$

λ = 4136 x 10⁻⁹ m

λ = 4136 nm → infrared

(b) 3.0 eV

$\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(3)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-7} \ m$

λ = 413.6 x 10⁻⁹ m

λ = 413.6 nm →visible light

(c) 30 eV

$\lambda = \frac{(6.626 * 10^{-34})(3*10^8)}{(30)*(1.602*10^{-19})}\\\\\lambda = 4.136 *10^{-8} \ m$

λ = 41.36 x 10⁻⁹ m

λ = 41.36 nm →ultraviolet

5 0

3 years ago

What is the work Sarah,60kg does against gravity is she goes up a 4m high stairwell in 3 seconds?

ladessa [460]

Work can be obtained from the following expression

W = F*d

Where F is the force (In this case the force it takes to lift Sarah, which is mass*gravity)

And d is the distance traveled.

W = 60kg*9.8m/s^2 * 4m = 2352 Joules

3 0

3 years ago

A 6.00 kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 5.0

Andre45 [30]

Answer:

The height of the other ball go after the collision is 2.304 m.

Explanation:

Given that,

Mass of ball = 6.00 kg

Height = 12.0 m

Mass of bar =5.00 kg

Length = 4.00 m

Suppose we need to calculate how high will the other ball go after the collision

We need to calculate the velocity of ball

Using formula of velocity

$v=\sqrt{2gh}$

$v=\sqrt{2\times9.8\times12.0}$

$v=15.33\ m/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{before}=mvr$

Put the value into the formula

$l_{before}=6.00\times15.33\times2.0$

$l_{before}=183.96\ kgm^2/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{after}=I_{t}\omega$

$l_{after}=(\dfrac{ml^2}{12}+m_{1}r^2+m_{2}r^2)\omega$

Put the value into the formula

$l_{after}=(\dfrac{5\times4.00^2}{12}+6.00\times2.0^2+6.00\times2.0^2)\omega$

$183.96=54.66\omega$

$\omega=\dfrac{183.96}{54.66}$

$\omega=3.36\ rad/sec$

After collision the ball leaves with velocity

We need to calculate the velocity after collision

Using formula of the velocity

$v= r\omega$

$v=2.0\times3.36$

$v=6.72\ m/s$

We need to calculate the height

Using formula of height

$h=\dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{(6.72)^2}{2\times9.8}$

$h=2.304\ m$

Hence, The height of the other ball go after the collision is 2.304 m.

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3 years ago

Newton’s law of universal gravitation.. . a.. is equivalent to Kepler’s first law of planetary motion.. b.. can be used to deriv

mixer [17]

By the help of newtons law of gravitation we can derive keplers third law of planetary motion.

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3 years ago

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