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3 years ago

Identify whether the following statements describe a change in acceleration. Explain your response answer them all plz

Physics

2 answers:

weqwewe [10]3 years ago

5 0

Answer:c

Explanation:it picking up speed

stepladder [879]3 years ago

3 0

I got C
Hoped it was help byee:)

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Which of the following is not an assumption of stage theories?

olya-2409 [2.1K]

Answer:

B. Stages are continuous and gradual.

Explanation:

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3 years ago

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Global positioning satellites (GPS) can be used to determine your position with great accuracy. If one of the satellites is 20,0

o-na [289]

Answer:

0.00001 %

Explanation:

The distance from satellite = 20000 km

Position range = 2 m

The percentage uncertainty is given by dividing the distance from satellite by the position range of desired accuracy.

Percentage uncertainty is given by

$\dfrac{2}{20000\times 10^3}\times 100=0.00001\ \%$

The percentage uncertainty that is required is 0.00001 %

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4 years ago

Calculate the acceleration of a bus that speed up from 20ms-1 to 40ms-1 in 8 seconds?

Anastaziya [24]

Answer:

2.5 ms^-2

Explanation:

acceleration

= (final velocity - initial velocity)/time

= [(40m/s) - (20m/s)]/8s

= (20m/s)/8s

= 5/2 m/s²

= 2.5 m/s²

= 2.5 ms^-2

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3 years ago

Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with

IgorLugansk [536]

(a) $2.5\cdot 10^{-6}eV$

The energy of a photon is given by:

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

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

For the microwave photon,

$\lambda=50.00 cm = 0.50 m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV$

(b) $2.5 eV$

For the energy of the photon, we can use the same formula:

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

For the visible light photon,

$\lambda=500 nm = 5 \cdot 10^{-7}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV$

(c) $2500 eV$

For the energy of the photon, we can use the same formula:

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

For the x-ray photon,

$\lambda=0.5 nm = 5 \cdot 10^{-10}m$

So the energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J$

And converting into electronvolts,

$E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV$

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3 years ago

Which characteristic is common in mature rivers?

lbvjy [14]

Answer: its b

Explanation: A river with a gradient that is less steep than those of youthful rivers and flows more slowly. A mature river is fed by many tributaries and has more discharge than a youthful river. Its channels erode wider rather than deeper

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4 years ago

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