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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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A 4.5-kg, three legged stool supports a 89-kg person. If each leg of the stool has a cross-sectional diameter of 2.8 cm and the
kkurt [141]

Answer:

4.96 × 10⁵ Pa

Explanation:

F = mg

F = (m_{person}+m_{stool})g\\\\F =  (4.5 + 89)*9.8\\\\F = 916.3 N

This force is evenly distributed on the three leg

radius, r = d/2

= 2.8 / 2

= 1.4 cm = 0.014 m

total cross sectional area of the three legs, A = 3*pi*r^2

= 3\times\pi \times0.014^2\\\\= 1.847\times10^-^3m^2

Pressure due to weight,

P = Weight/A

P = / 1.847 × 10⁻³\\P = \frac{ 916.3N}{1.847\times10^-^3} \\\\P= 496032.9Pa

P = 4.96 × 10⁵ Pa

8 0
3 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a
Vesna [10]

Answer:

Tangential speed=5.4 m/s

Radial acceleration=88.6m/s^2

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution=2\pi rad

2.59 rev=2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s

By using \pi=3.14

Angular velocity=\omega=16.27rad/s

Distance from axis=r=0.329 m

Tangential speed=r\omega=16.27\times 0.329=5.4m/s

Radial acceleration=\frac{v^2}{r}

Radial acceleration=\frac{(5.4)^2}{0.329}=88.6m/s^2

6 0
4 years ago
A passenger in a train accelerating smoothly away from a station observes that a child’s yo-yo hanging by its string from a lugg
olchik [2.2K]

Answer:

1.73 m/s²

3.0 cm

Explanation:

Draw a free body diagram of the yo-yo.  There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.

Sum of forces in the y direction:

∑F = ma

T cos 10° − mg = 0

T cos 10° = mg

T = mg / cos 10°

Sum of forces in the x direction:

∑F = ma

T sin 10° = ma

mg tan 10° = ma

g tan 10° = a

a = 1.73 m/s²

Draw a free body diagram of the sphere.  There are two forces: weight force mg pulling down, and air resistance D pushing up.  At terminal velocity, the acceleration is 0.

Sum of forces in the y direction:

∑F = ma

D − mg = 0

D = mg

½ ρₐ v² C A = ρᵢ V g

½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g

3 ρₐ v² C = 8 ρᵢ r g

r = 3 ρₐ v² C / (8 ρᵢ g)

r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))

r = 0.030 m

r = 3.0 cm

3 0
3 years ago
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