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Mamont248 [21]
3 years ago
13

Why do some people suffer for nose bleed at higher altitudes?​

Physics
1 answer:
Novosadov [1.4K]3 years ago
8 0

Answer:because of small amount of oxygen

Explanation:

If you get higher by the atmosphere the air will start to decrease

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Evaluate (x +y)0 for x= -3 and y=5.<br> 0 1<br> 2<br> 01
frosja888 [35]

Answer:2.01201

Explanation:

8 0
3 years ago
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in
wel

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?  

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:  

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:  

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

3 0
3 years ago
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

5 0
3 years ago
A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion in
Rashid [163]

Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J  

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J  

But  ,

4 * 1875000 = 7500000

so the KE has increased by 4 times.

8 0
3 years ago
Read 2 more answers
Find the ratio of the diameter of iron to copper wire, if they have the same resistance per unit length (as they might in househ
Natasha_Volkova [10]

Answer:

The ratio of the diameter of iron to Cu is;

\frac{d{Fe}   }{ d{Cu}   } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}

Explanation:

R=(ρL)/A

  • R is resistance,
  • L is length,
  • A is area,
  • ρ is resistivity
  • d is diameter

from the question the two materials have the same resistance per unit length.

\frac{R}{L}= \frac{p}{A}

\frac{R}{L}   for iron = \frac{R}{L}  for copper

This means we can equate ρ/A for both materials.

\frac{p_{Fe} }{A_{Fe} } =\frac{p_{Cu} }{A_{Cu} }

re-arranging the equation we have,

\frac{A_{Fe}}{A_{Cu} } =\frac{p_{Fe} }{ p_{Cu} }

A=\pi \frac{d^{2} }{4}

\frac{A_{Fe}}{A_{Cu} } =\frac{d^{2}{Fe}   }{ d^{2}{Cu}   }

\frac{d^{2}{Fe}   }{ d^{2}{Cu}   } =\frac{p_{Fe} }{ p_{Cu} }

\frac{d{Fe}   }{ d{Cu}   } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}

5 0
3 years ago
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