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lisabon 2012 [21]

3 years ago

11

What were the physical activities in your childhood that you still do today? Do you spend more time now in doing these activitie

s as compared before?

1 answer:

user100 [1]3 years ago

5 0

I used to play tag or run around a lot and sometimes i do these activities but its a lot less now that i am older

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The figure below shows a shooting competition, where air rifles fire soft metal at distant targets

bekas [8.4K]

Answer:

where is the figure?

Explanation:

6 0

3 years ago

Helppppppppppp im dieing

Otrada [13]

It would be 1. B 2. A 3. A

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3 years ago

A photon of wavelength 192 nm strikes an aluminum surface along a line perpendicular to the surface and releases a photoelectron

alex41 [277]

Answer:

$KE=3.529\times10^{−27}\ J$

Explanation:

Given that

Wavelength λ=192 nm

So energy of photon,E

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

Now by putting the values

$h=6.6\times 10^{-34}\ m^2.kg/s$

$C=3\times 10^{8}\ m/s$

$E=\dfrac{6.6\times 10^{-34}\times 3\times 10^{8}}{192\times 10^{-9} }$

$E=1.03\times 10^{-18} J$

We know that

Kinetic energy given as

$KE=\dfrac{P^2}{2m}$

$KE=\dfrac{E^2}{2mC^2}$

$KE=\dfrac{(1.03\times 10^{-18})^2}{2\times 1.67\times 10^{-27}(3\times 10^8)^2}$

$KE=3.529\times10^{−27}\ J$

5 0

3 years ago

In the diagram, q1= +8.0 C, q2= +3.5 C, and q3 = -2.5 C. q1 to q2 is 0.10 m, q2 to q3 is 0.15 m. What is the net force on q2? La

yulyashka [42]

Answer:

f(t) = 28,7 [N]

Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.

The net force on +q₂ is the sum of the force of +q₁ on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)

The two forces have the same direction to the right of charge q₂, we have to add them

Then

f(t) = f₁₂ + f₃₂

f₁₂ = K * ( q₁*q₂ ) / (0,1)²

q₁ = + 8 μC then q₁ = 8*10⁻⁶ C

q₂ = + 3,5 μC then q₂ = 3,5 *10⁻⁶ C

K = 9*10⁹ [ N*m² /C²]

f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻² [ N*m² /C²]* C*C/m²

f₁₂ = 252*10⁻¹ [N]

f₁₂ = 25,2 [N]

f₃₂ = 9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²

f₃₂ = 78,75*10⁻³/ 2,25*10⁻²

f₃₂ = 35 *10⁻¹

f₃₂ = 3,5 [N]

f(t) = 28,7 [N]

5 0

3 years ago

Read 2 more answers

an ice Puck travels 18 m in 3 S before it slides into the goal what is the speed of the traveling puck

DiKsa [7]

Answer:

$\boxed {\boxed {\sf 6 \ meters \ per \ second}}$

Explanation:

Speed can be found by dividing the distance by the time.

$s=\frac{d}{t}$

The distance is 18 meters and the time is 3 seconds.

$d= 18 \ m \\t= 3 \ s\\$

Substitute the values into the formula.

$s=\frac{18 \ m }{ 3 \ s }$

Divide.

$s= 6 \ m/s$

The speed of the puck is <u>6 meters per second.</u>

4 0

2 years ago

Read 2 more answers