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natulia [17]
3 years ago
12

A helicopter blade spins at exactly 180 revolutions per minute. Its tip is 10.00 m from the center of rotation. What is its aver

age velocity over one revolution
Physics
1 answer:
olya-2409 [2.1K]3 years ago
3 0

Answer:

5.8E-3m/s

Explanation:

Using

V= d/t

V= velocity

d= distance

t= time

But d= 2πr

But 180rev= 1min

So 1min/180= 60s/180

So

Vavr= 2π(10m)/180*60

=5.8E-3m/s

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The force that contributed to the formation of planets, determines the motion of bodies in the solar system, and pulls objects t
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The force of gravity

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Gravity was studied, by early scientists such as Copernicus and others, Galileo was the first to ensure that planets moved according to a physical equation that depended on a force that caused celestial bodies to move and interact with each other. But years later Newton based on studies conducted deciphering what Galileo assumed, he was able to find the equation of the force of gravity in any body in the universe. This equation depends on the masses of the two interacting bodies, the distance between them and a constant, which I call universal gravitation constant.

F_{g}=G*\frac{m_{1}*m_{2}}{r^2}

Fg = gravity force [N]

G =  universal gravitation constant = 6.67*10^(-11) [N*m^2/kg^2]

m1 = mass of the 1st body [kg]

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What are subtomic particles<br>​
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In the physical sciences, subatomic particles are smaller than atoms. They can be composite particles, such as the neutron and proton; or elementary particles, which according to the standard model are not made of other particles. Particle physics and nuclear physics study these particles and how they interact.

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Read 2 more answers
Find the quantity of heat needed
krok68 [10]

Answer:

Approximately 3.99\times 10^{4}\; \rm J (assuming that the melting point of ice is 0\; \rm ^\circ C.)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}

The energy required comes in three parts:

  • Energy required to raise the temperature of that 0.100\; \rm kg of ice from (-10\; \rm ^\circ C) to 0\; \rm ^\circ C (the melting point of ice.)
  • Energy required to turn 0.100\; \rm kg of ice into water while temperature stayed constant.
  • Energy required to raise the temperature of that newly-formed 0.100\; \rm kg of water from 0\; \rm ^\circ C to 10\;\ rm ^\circ C.

The following equation gives the amount of energy Q required to raise the temperature of a sample of mass m and specific heat capacity c by \Delta T:

Q = c \cdot m \cdot \Delta T,

where

  • c is the specific heat capacity of the material,
  • m is the mass of the sample, and
  • \Delta T is the change in the temperature of this sample.

For the first part of energy input, c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}.

Similarly, for the third part of energy input, c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}.

The second part of energy input requires a different equation. The energy Q required to melt a sample of mass m and latent heat of fusion L_\text{f} is:

Q = m \cdot L_\text{f}.

Apply this equation to find the size of the second part of energy input:

\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}.

Find the sum of these three parts of energy:

\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}.

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D

Explanation:

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