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3 years ago

What is the least possible initial kinetic energy kmin the oxygen atom could have and still excite the cesium atom?

1 answer:

5 0

If perfectly elastic there would be no energy left over for exciting the atom. if the collision were partially elastic, then some of the initial kinetic energy would be converted into internal energy,

hope this help

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A block of wood is 4 cm wide, 5 cm long, and 10 cm high. It weighs 100 grams. Calculate its volume. Calculate its density. Will

77julia77 [94]

Answer:

$V=200cm^3\\\\\rho =0.500g/cm^3$

It will float.

Explanation:

Hello.

In this case, given the width, length and height, we can compute the volume as follows:

$V=W*L*H\\\\V=4cm*5cm*10cm\\\\V=200cm^3$

Moreover, since the density is computed via the division of the mass by the volume:

$\rho =\frac{m}{V}$

We obtain:

$\rho =\frac{100g}{200cm^3} \\\\\rho =0.500g/cm^3$

In such a way, since the solid has a lower density than the water, we infer it will float.

Best regards.

3 0

4 years ago

A ball is projected vertically upward from the surface of the Earth with an initial speed of +49 meters per second. The ball rea

fenix001 [56]

Answer: 245 meters

Explanation: 49 ? Meters
over over over
1 5 seconds

Cross multiply- 49 times 5 divided by 1 which equals 245

8 0

3 years ago

Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

ruslelena [56]

Answer:

The orbital speed is $v= 2.6*10^{3} m/s$

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

7 0

3 years ago

The gas sample has now returned to its original state of 1.00 atm, 20.0 ∘c, and 1.00 l. what will the pressure become if the tem

steposvetlana [31]

At a constant volume and number of moles of the gas the ratio of T and P is equal to some constant. At another set of condition, the constant is still the same. Calculations are as follows:

T1/P1 = T2/P2

P2 = T2 x P1 / T1

P2 = 473.15 x 1.00 / 293.15

3 0

3 years ago

What force must the worker exert to get the box moving & what force must the worker exert to accelerate the box at 0.1 meter

photoshop1234 [79]

Since static friction is the minimum force required to just start the motion of a stationary object.

Here if we need to start an object from rest then we required F = 700 N

So for the first part of the above problem Force will be F = 700 N

Now if the box is already moving then we will have to use kinetic friction force between box and floor

now we can write the equation of net force as