Answer:
A. 70 m OS the correct one
The speed of sound is greater in ice (4000 m/s), then in water (1500 m/s), then in air (340 m/s). The explanation for this is the differente state of the matter in the three cases.
In fact, sound waves travel faster in solids (like ice), then in liquids (like water), then in gases (like air). This is because the speed of the sound wave depends on the density of the medium: the greater the density, the faster the sound wave. This can be easily understood by thinking at how a sound wave propagates: a sound wave is a vibration of molecules, which is transmitted throughout the medium by collision of the molecules. Therefore, the smaller the spacing between the molecules (such as in solids), the more efficient is the propagation, and so the sound wave is faster. On the contrary, there is a large spacing between molecules in gases (such as in the air), so there are less collisions between the molecules and so the wave is not transmitted efficiently, and so it has less velocity.
<h2>
Option 2 is the correct answer.</h2>
Explanation:
Elastic collision means kinetic energy and momentum are conserved.
Let the mass of object be m and M.
Initial velocity object 1 be u₁, object 2 be u₂
Final velocity object 1 be v₁, object 2 be v₂
Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s
Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M
Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J
Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M
We have
Initial momentum = Final momentum
24 = 3v₁ + 6M
v₁ + 2M = 8
v₁ = 8 - 2M
Initial kinetic energy = Final kinetic energy
96 = 1.5 v₁² + 18 M
v₁² + 12 M = 64
Substituting v₁ = 8 - 2M
(8 - 2M)² + 12 M = 64
64 - 32M + 4M² + 12 M = 64
4M² = 20 M
M = 5 kg
Option 2 is the correct answer.
He will be a pilot and he will fly the plane over bridges fewwww
The equation for electrical power is<span>P=VI</span>where V is the voltage and I is the current. This can be rearranged to solve for I in 6(a).
6(b) can be solved with Ohm's Law<span>V=IR</span>or if you'd like, from power, after substituting Ohm's law in for I<span>P=<span><span>V2</span>R</span></span>
For 7, realize that because they are in parallel, their voltages are the same.
We can find the resistance of each lamp from<span>P=<span><span>V2</span>R</span></span>Then the equivalent resistance as<span><span>1<span>R∗</span></span>=<span>1<span>R1</span></span>+<span>1<span>R2</span></span></span>Then the total power as<span><span>Pt</span>=<span><span>V2</span><span>R∗</span></span></span>However, this will reveal that (with a bit of algebra)<span><span>Pt</span>=<span>P1</span>+<span>P2</span></span>
For 8, again the resistance can be found as<span>P=<span><span>V2</span>R</span></span>The energy usage is simply<span><span>E=P⋅t</span></span>
