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34kurt
3 years ago
8

Which is not a medium through which a mechanical wave could travel?

Physics
2 answers:
Setler [38]3 years ago
8 0

Mechanical Wave cannot travel in space vaccumes, which would be considered "Empty Space" in your situation.

Tems11 [23]3 years ago
8 0

A mechanical cannot travel through empty space. So option (d) is correct.

A mechanical wave is a wave which needs a material medium for its propagation. For example sound, water waves etc . The medium required by the wave can be a solid, liquid or a gas. Empty space doesn't have any medium, so a mechanical wave cannot travel through empty space.

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For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with d
Sedaia [141]

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material (\sigma) = 200 M Pa

Diameter of the bar = 4.5 cm = 45 mm

We know that yield stress of the bar is given by the formula

                Yield Stress = \frac{Maximum load}{Area of the bar}

⇒                                \sigma = \frac{P_{max} }{A}  ---------------- (1)

⇒ Area of the bar (A) = \frac{\pi}{4} ×D^{2}

⇒                            A  = \frac{\pi}{4} × 45^{2}

⇒                            A = 1589.625 mm^{2}

Put all the values in equation (1) we get

⇒ P_{max} = 200 × 1589.625

⇒ P_{max} = 317925 N

In this bar the P_{max} is equal to the weight of the bar.

⇒ P_{max} = M_{max} × g

Where M_{max} is the maximum mass the bar can support.

⇒ M_{max} = \frac{P_{max} }{g}

Put all the values in the above formula we get

⇒ M_{max} = \frac{317925}{9.81}

⇒ M_{max} = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg

3 0
3 years ago
A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is dou
scoray [572]

Answer:

Explanation:

Charge on uranium ion = charge of a single electron

= 1.6 x 10⁻¹⁹ C

charge on doubly ionised iron atom = charge of 2 electron

= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²

force on electron at distance 61.10 x 10⁻⁹ - r from iron  ion

= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

1.414 r = 61.10 x 10⁻⁹ - r

2.414 r = 61.10 x 10⁻⁹

r = 25.31 nm .

4 0
3 years ago
By counting the number of waves that pass a certain point each second, we will know the _______ of the wave.
Travka [436]
The answer is A. Frequency
6 0
3 years ago
Read 2 more answers
A student estimated a mass to be 325 g, but the actual mass is 342 g. What is the percent error?
Dovator [93]

Answer:

Answer:

5.2307 %

Explanation:

(acutal mass- estimated mass) / ( estimated mass)

3 0
3 years ago
The speed of a boat in still water is 20 mph. it travels from one pier to another with the current in 4 hours and back against t
saw5 [17]
The answer is, "the speed of the current is 5 miles per hour."

To calculate the speed of the current,
let's assume speed of current =  xmph. Time taken to travel from one pier to another with the current = 100/(20+x)h


But the time taken to travel from one pier to another with the current, which is given is = 4 hours. So,  4=100/(20+x) 80+4x = 100

4x = 20

x = 5 Thus, the speed of the current is 5 miles per hour.
3 0
3 years ago
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