What is a covalent bond considered

A helicopter blade spins at exactly 110 revolutions per minute. Its tip is 4.50 m from the center of rotation. (a) Calculate th

NeX [460]

Answer:

(a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

Explanation:

Given that,

Angular velocity = 110 rev/m

Radius = 4.50 m

(a). We need to calculate the average speed

Using formula of average speed

$v=r\omega$

$v = 4.50\times110\times\dfrac{2\pi}{60}$

$v=51.83\ m/s$

(b). The average velocity over one revolution is zero because the net displacement is zero in one revolution.

Hence, (a). The average speed is 51.83 m/s.

(b). The average velocity over one revolution is zero.

8 0

3 years ago

Why does the area around the equator stay about the same temperature year round?

Verdich [7]

Axial Tilt and Sun Energy

This axial tilt means that during the Earth's journey around the sun the poles receive varying amounts of sunlight. The equator, however, receives relatively consistent sunlight all year. The consistency of energy means the equator's temperature stays relatively constant all year.

3 0

3 years ago

A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul

zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

$\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}$

$\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}$

$\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}$

$\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }$

$\mathbf{\omega^2=\dfrac{39.24 }{2}}$

$\mathbf{\omega=\sqrt{19.62 } \ rad/sec}$

$\mathbf{\omega=4.429 \ rad/sec}$

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

4 0

2 years ago

Read 2 more answers

The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur

Nimfa-mama [501]

The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

P₂ is the pressure in the hose

ρ is the density of the fluid

h₁ is the initial height

h₂ is the final height

v₁ is the initial velocity of the fluid

v₂ is the final velocity of the fluid and

g is the acceleration due to gravity

Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

= (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

= (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

= (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

= 5.92 × 10⁵ Pa

= 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

<u>brainly.com/question/9506577</u>

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6 0

1 year ago

Do atoms have full outer energy level combine with other atoms

Pavlova-9 [17]

All of the Noble Gases, which are on the right side of the periodic table, have a full outer energy level. The elements that are Noble Gases are the following: <span>Neon Argon Krypton Xenon Radon Ununoctium.

Hope this helps.</span>

8 0

3 years ago