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Kaylis [27]
3 years ago
11

The speed of light in a vacuum is 3 x 108 miles/hour. true/false?

Physics
2 answers:
GarryVolchara [31]3 years ago
8 0
I'm guessing they rounded because it is technically 2.998 X 108 miles/hour

so I would go with true!
Leya [2.2K]3 years ago
3 0
That is false.

The speed of light in a vacuum is 3 x 10^8 m/sec.

hope this helps and have a great day :)



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In Example 2.7, we investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on sol
igor_vitrenko [27]

Answer:

a) 20 seconds

b) No.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s (given in the question)

v = Final velocity = 0 (because the jet will stop at the end)

s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)

a = Acceleration = -5.00 m/s² (slowing down, so it is negative)

a) Equation of motion

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.

5 0
2 years ago
A 12.0-g bullet is fired horizontally into a 109-g wooden block that is initially at rest on a frictionless horizontal surface a
kykrilka [37]

Answer:

v₀ = 280.6 m / s

Explanation:

we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,

We write the mechanical energy when the shock has passed the bodies

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We write the mechanical energy when the spring is in maximum compression

Em_{f} = K_{e} \\= \frac{1}{2} kx^2\\    Em_0 = Em_{f}

½ (m + M) v² = ½ k x²

Let's calculate the system speed

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   v = √[152 ×0.78² / (0.012 +0.109) ]

   v = 27.65 m / s

This is the speed of the bullet + Block system

Now let's use the moment to solve the shock

Before the crash

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After the crash

p_{f} = (m + M) v

The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved

 p_0 =  p_{f}

  m v₀ = (m + M) v

  v₀ = v (m + M) / m

let's calculate

v₀ = 27.83 (0.012 +0.109) /0.012

  v₀ = 280.6 m / s

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