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andrew-mc [135]
3 years ago
12

A helicopter’s speed increases from 30 m/s to 40 m/s in 5 seconds.

Physics
1 answer:
Mandarinka [93]3 years ago
5 0
7 m over s² or 7ms⁻¹
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Give short definitions for the following terms: 1. All Atomic number 2. Mass number 3. Atomic weight 4. Isotope 5. Natural abund
son4ous [18]

Explanation:

Hello ! Let's solve this!

1-atomic number: the atomic number is the number of protons that the atom has

2-mass number: is the amount of protons plus the amount of neutrons in the atom

3-atomic weight: average of the masses of an atom per 1/12 of the mass of carbon

4-isotope: it is the same element that has the same number of protons but a different number of neutrons

5-natural abundance: measures the amount of isotopes of an element in nature

6-unit of atomic mass: (uma) is a unit that is equivalent to 1/12 of the carbon atom

8 0
3 years ago
A sports car accelerates from 0 to 25 meters per second in 4 seconds. What is its acceleration?
WITCHER [35]

Answer:

6.25 ms²

Explanation:

..................

3 0
3 years ago
________ in the ear change sound waves to electrical signals that the brain can interpret as sounds
Assoli18 [71]

Answer:

ear

Explanation:

7 0
3 years ago
Ruff, the 50 cm tall Labrador Retriever stands 3m from a plane mirror and looks at his image. What is Ruffs image position and h
GaryK [48]
Ruff's image is 50m behind the mirror surface and the image is also 3m tall.

This is because it is a plane mirror.
5 0
3 years ago
A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st
Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r

where,

\tau= shear stress at a distance 'r' from the center

T = is the applied torque

I_{p} = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

\tau _{max}=\frac{4}{3}\times \frac{V}{A}

Applying values we get

\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa

3 0
3 years ago
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