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3 years ago

If we decrease the distance an object moves we will (2 points)

Physics

2 answers:

Fantom [35]3 years ago

8 0

Decrease the amount of force applied

Amanda [17]3 years ago

6 0

Work done = force × displacement

If we decrease the distance an object moves

displacement will decrease.

Thus work done will decrease.

Force will not be affected.
____________________

Thus your answer is : decrease the amount of work done.

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How long would it take an object to reach the ground from the top of a building that is 470 feet tall? Round to the nearest tent

Zinaida [17]

Answer:

It would take the object 5.4 s to reach the ground.

Explanation:

Hi there!

The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).

t = time

Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:

h = h0 + 1/2 · g · t²

We have to find the time at which h = 0:

0 = 470 ft - 1/2 · 32.2 ft/s² · t²

Solving for t:

-470 ft = -16.1 ft/s² · t²

-470 ft / -16.1 ft/s² = t²

t = 5.4 s

3 0

4 years ago

Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one en

Tom [10]

Explanation:

Wavelength in an emission spectrum, $\lambda=435\ nm=435\times 10^{-9}\ m$

The energy of an electron is given by :

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

Where

h is the Planck's constant

c is the speed of light

For 435 nm, the energy of the electron will be :

$E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8\ m/s}{435\times 10^{-9}}$

$E=4.57\times 10^{-19}\ J$

We know that $1\ eV=1.6\times 10^{-19}\ J$

So, $E=\dfrac{4.57\times 10^{-19}}{1.6\times 10^{-19}}$

So, E = 2.86 eV

The energy of the electron dropping from one energy level is 2.86 eV. We know that,

$\dfrac{hc}{\lambda}=E_{n_2}-E_{n_1}$

From the given energy levels :

$E_5-E_2=-0.544-(-3.403)$

$E_5-E_2=2.859\ eV$

So, the transition must be from E₅ to E₂. Hence, this is the required solution.

5 0

4 years ago

Read 2 more answers

Which statement best describes an atom?

ycow [4]

A group of protons and neutrons that are surrounded by electrons

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3 years ago

The enthalpy of fusion of mercury is 2.292 kJ mol-1 at its normal freezing point of 234.3 K; the change in molar volume on melti

aev [14]

Answer:

T_2= 234.37 K

Explanation:

According to Claperyon, we know that

$P_2-P_1= \frac{\Delta H_{fus}}{\Delta V}\times\frac{T_1}{T_2}$

P_1= Atmospheric pressure 760 mm Hg

P_2 = pressure at the bottom of the column

= 10×10^3 mm of Hg+ 760 mm of Hg

= 10760 mm of Hg

now,

P_2-P_1= 10760-760= 10^4 mm

P_2-P_1 ( in pascals) = 10^4× 133.322= 1333220 mm

the enthalpy of fusion (ΔH-fus) of mercury is 2.292 KJ/mol

use the above equation to calculate ΔT as follows

$1333220= \frac{2292}{0.517\times10^{-6}}\times ln\frac{T_2}{234.3K}$

therefore, T_2= 234.37 K

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Andrei [34K]

Answer:

Stars emit colors of many different wavelengths, but the wavelength of light where a star's emission is concentrated is related to the star's temperature - the hotter the star, the more blue it is; the cooler the star, the more red it is

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3 years ago