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vodka [1.7K]
3 years ago
7

Which of the following corresponds to a supersonic speed?

Physics
2 answers:
IgorLugansk [536]3 years ago
8 0
Answer=C

Supersonic speed is a rate of travel that exceeds the speed of sound. The speed of sound is Mach 1

a. Mach 0.5

b. Mach 1

c. Mach 1.5

C is the correct answer. Mach1.5>Mach1

a is less than the speed of sound, therefore it doesn't exceed it.
Mach0.5<Mach1

b is exactly the speed of sound, so it doesn't exceed it.
Mach1=Mach1
oksian1 [2.3K]3 years ago
7 0

Answer:

The answer is MACH 1.5

Explanation:

Supersonic speed is anywhere from 1.5 Mach to 5 Mach.

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3. Which one of these has the most inertia?
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b. A Boeing 747 airplane

Explanation:

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Which best explains why the receiver of a signal must understand the code or language being used
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A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

v=u+at

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

v = 15.5 +(-9.8)(2.64)=-10.4 m/s

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

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