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Stells [14]
3 years ago
5

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

es horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 34.1 m/s2 with a beam of length 5.91 m, what rotation frequency is required?
Physics
1 answer:
Natali [406]3 years ago
7 0

centripetal acceleration is given by formula

a_c = \omega^2*R

given that

a_c = 34.1 m/s^2

R  =  5.91 m

now we have

\omega^2 R = 34.1

\omega^2 * 5.91 = 34.1

\omega^2 = 5.77

\omega = 2.4 rad/s

so the ratationa frequency is given by

\omega = 2 \pi f

2.4 = 2 \pi f

f = \frac{2.4}{2\pi}

f = 0.38 Hz

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lora16 [44]

planetary accretion is the term that is used

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You and your little cousin sit on a see-saw. You sit 0.5 m from the fulcrum, and your cousin sits 1.5 m from the fulcrum. You we
svetoff [14.1K]

Answer:

200N

Explanation:

0.5/1.5=x/600N

1/3=x/600

x=200N

7 0
3 years ago
Which of the following has potential but not kinetic energy?
gavmur [86]

The answer would be option D "a ball sitting on a shelf." Potential energy is the amount of energy a object has while it's at rest.. (or not moving) Kinetic energy is how much energy a object is while it's moving. So in this case it's option D because a ball sitting on a shelf isn't moving therefore it has potential energy. It's not option A because thats a example of kinetic energy since how the roller coaster is moving. It's not option B because it's kinetic energy because the bike is moving. It's also not option C because it's kinetic energy because the bird is moving.


Hope this helps!

7 0
3 years ago
. Calculate the efficiency of a bicycle if the input work to turn the pedal is 45J and the output work is 20J. * 1 point 2.25 2.
cestrela7 [59]

20/45=0.4*100= 44.4 so the answer is..................................................

Answer: 44.4%

8 0
3 years ago
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}.

3 0
2 years ago
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