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3 years ago

Sami pops a helium balloon at a birthday party. What will happen to the particles of helium that were in the balloon?

1 answer:

4 0

They will mix with air. Air consists mostly of nitrogen and oxygen but there are also traces of CO2 and noble gases in air, and helium is one of those noble gases.

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Who can do my worksheet for me

timurjin [86]

Answer:

what is it on? like name one of the questions

Explanation:

5 0

3 years ago

Consider five atoms from the second period : lithium , beryllium , boron , carbon , and nitrogen , which of these elements has t

kap26 [50]

Answer:

Lithium has the lowest electronegativity.

Explanation:

Electronegativity measures the tendency of an atom to attract the bonding pair of electrons. As we move left to right in a period, the number of shell remains same but the number of electron increases(negative charge increases. Hence, electronegativity also increases.

As Lithium is the left most element in this period. It has the lowest electronegativity value which is equal to 0.98.

6 0

3 years ago

magine an astronaut on an extrasolar planet, standing on a sheer cliff 50.0 m high. She is so happy to be on a different planet,

Mama L [17]

Answer:

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

Explanation:

Given:

height above which the rock is thrown up, $\Delta h=50\ m$

initial velocity of projection, $u=20\ m.s^{-1}$

let the gravity on the other planet be g'

The time taken by the rock to reach the top height on the exoplanet:

$v=u+g'.t'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0=20-g'.t'$ (-ve sign to indicate that acceleration acts opposite to the velocity)

$t'=\frac{20}{g'}\ s$

The time taken by the rock to reach the top height on the earth:

$v=u+g.t$

$0=20-g.t$

$t=\frac{20}{g} \ s$

Height reached by the rock above the point of throwing on the exoplanet:

$v^2=u^2+2g'.h'$

where:

$v=$ final velocity at the top height = 0 $m.s^{-1}$

$0^2=20^2-2\times g'.h'$

$h'=\frac{200}{g'}\ m$

Height reached by the rock above the point of throwing on the earth:

$v^2=u^2+2g.h$

$0^2=20^2-2g.h$

$h=\frac{200}{g}\ m$

The time taken by the rock to fall from the highest point to the ground on the exoplanet:

$(50+h')=u.t_f'+\frac{1}{2} g'.t_f'^2$ (during falling it falls below the cliff)

here:

$u=$ initial velocity= 0 $m.s^{-1}$

$\frac{200}{g'}+50 =0+\frac{1}{2} g'.t_f'^2$

$t_f'^2=\frac{400}{g'^2}+\frac{100}{g'}$

$t_f'=\sqrt{\frac{400}{g'^2}+\frac{100}{g'} }$

Similarly on earth:

$t_f=\sqrt{\frac{400}{g^2}+\frac{100}{g} }$

Now the required time difference:

$\Delta t=(t'+t_f')-(t+t_f)$

$\Delta t=(\frac{20}{g'}+\sqrt{\frac{400}{g'^2}+\frac{100}{g'} } )-(\frac{20}{g}+\sqrt{\frac{400}{g^2}+\frac{100}{g} } )$

3 0

2 years ago

A train is approaching a town at a constant speed of 12 m/s. The town is 1.0 km distant. After 30 seconds, the conductor applies

Yuliya22 [10]

To solve this problem we will apply the linear motion kinematic equations. To determine the position in which the braking starts we will start from the definition of distance as a function of speed and time, that is

$x = x_0 - vt$

Here,

$x_0$ = Initial position

v = Velocity

t = time

Replacing we have that

$x = 1000-12*30$

$x = 640m$

Now the acceleration is given by the function,

$v_f^2=v_0^2 +ax$

Here,

$v_f$ = Final velocity

$v_0$ = Initial velocity

a = Acceleration

x = Displacement

Replacing we have that

$0 = 12^2+2a(640)$

$a = -0.1125m/s^2$

Therefore the acceleration necessary to bring the train to rest is $-0.1125m/s^2$

4 0

2 years ago

Why is it possible to remove a paper without moving the object on it.(apply newton's law)

klasskru [66]

This is possible due to inertia of motion. which is nothing but newton's first law.

according to this law , an object tries to retain its state of motion or rest unless acted upon by an external force.

consider an object placed on a paper, initially both the object and paper are at rest. to pull the paper , we apply force on the paper and paper gains velocity. but the object keeps its motion of rest and hence the paper can be removed without moving the object.

7 0

3 years ago