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Westkost [7]
3 years ago
8

Why is water soaked cloth used on forehead during fever?please can you tell me answer​

Physics
1 answer:
olga_2 [115]3 years ago
5 0
<h3>Answer:</h3>

This is because the cloth soaked in water can absorb and evaporate the heat from your forehead.

<h3>Explanation:</h3>

This technique is used to cool down your fever, and to help you feel better.

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A string is stretched to a length of 376 cm and
sergey [27]

Answer: 53.09Hz

Explanation:

The fundamental frequency of an ideal taut string is:

Fn= n/2L(√T/μ)

Where:

F= frequency per second (Hz)

T= Tension of the string (cm/s sqr)

L= Length of the string (cm)

μ= Linear density or mass per unit length of the string in cm/gm

√T/μ= square root of T divided by μ

It is important to note:

Note: Typically, tension would be in newtons, length in meters and linear density in kg/m, but those units are inconvenient for calculations with strings. Here, the smaller units are used.

F1= 1/2(376cm)(0.01/1) × (√574/(0.036g/cm)(0.1kg/m÷1g/cm)

F1= 0.1329 × 399.30

= 53.09Hz

7 0
3 years ago
Determine an appropriate size for a square cross-section solid steel shaft to transmit 260 hp at a speed of 550 rev/min if the m
Phantasy [73]

Answer:46.05 mm

Explanation:

Given

Power=260\ hp\approx 260\times 746=193.96\ KW

speed N=550\ rev/min

allowable shear stress (\tau )_{max}=15\ kpsi\approx 103.421\ MPa

Power is given by

P=\frac{2\pi NT}{60}

193.96=\frac{2\pi 550\times T}{60}

T=3367.6\ N-m

From Torsion Formula

\frac{T}{J}=\frac{\tau }{r}-----1

where J=Polar section modulus

T=Torque

\tau=shear stress

For square cross section

r=\frac{a}{2}

where a=side of square

J=\frac{a^4}{6}

Substituting the values in equation 1

\frac{3376.6}{\frac{a^4}{6}}=\frac{103.421\times 10^6}{\frac{a}{2}}

a=0.04605\ m

a=46.05\ mm

7 0
3 years ago
A certain wave motion has a frequency of 10 cycles / s and a wavelength of 3 m. so its propagation speed is
kodGreya [7K]
3.5m is ur answer ask for more questions anytime
5 0
3 years ago
compare the distance time graph for a fast-moving object with the distance time graph of a slower moving object
soldi70 [24.7K]
The slope of the distance/time graph is the speed of the moving object. 
So the graph for a fast moving object will have a greater slope than the
graph for a slower moving object has.
8 0
4 years ago
A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec
stiv31 [10]

Answer:

Approximately 3.1 \times 10^4 \; \rm N (assuming that the acceleration due to gravity is g = 9.81\; \rm kg \cdot N^{-1}.)

Explanation:

Let A_1 denote the first piston's contact area with the fluid. Let A_2 denote the second piston's contact area with the fluid.

Similarly, let F_1 and F_2 denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N.  

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to \pi times its radius, squared:

  • \displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}.
  • \displaystyle A_2 = \pi \times \left(\frac{1}{2} \times 24\right)^2 = 144\, \pi\;\rm cm^{2}.

By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

\displaystyle P = \frac{F}{A}.

For the pressures on the two pistons to match:

\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}.

F_1, A_1, and A_2 have all been found. The question is asking for F_2. Rearrange this equation to obtain:

\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}.

Evaluate this expression to obtain the value of F_2, which represents the force on the piston with the larger diameter:

\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}.

6 0
3 years ago
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