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Nezavi [6.7K]
3 years ago
12

Salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2m from a waterfall that is 0.55m tall and jumps

at an angle of 32°. What must be the salmon's minimum speed to reach the waterfall?
Physics
1 answer:
mamaluj [8]3 years ago
7 0
The salmon can travel at a speed of 2894673763462m/s and slide into my microwave! PS- Don't judge the salmon, JUST BECAUSE IT ISN'T ATHLETIC!!!! Worst regards, Vivian Carolina Gartea Chiz Demetreo Perez Sanfrasio Smith
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Question: An increased number of rod cells increase the ability to see, A. during the day B. at night C. underwater or D. colors
Fynjy0 [20]
B- at night

In eyes of the humans, and many other animals there are two types of receptors, the rods, and the cones. The cones have an important role in color vision and they possess different types of pigment which enable color vision.
On the other hand, the rods are more sensitive to light, and they are responsible for black and white vision.
Therefore, an increase in the number of rods can enable the animal to see betther in the dark.
8 0
3 years ago
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For a given ideal gas if it temperature is constant the pressure of the gas is proportional ??
svetlana [45]

Answer:

C) Inversely to its volume

Explanation:

As the volume decreases (Container gets smaller) the pressure increases. Conversely, If the volume increases (Container gets larger) the pressure decreases.

5 0
2 years ago
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I what is the resistance of
Lynna [10]
The resistance of a conductor is given by:
R= \frac{\rho L}{A}
where
\rho is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area

We can use this formula to solve both parts of the problem.

a) The length of the copper wire is L=1.0 m. Its diameter is d=0.50 mm, so its radius is 
r= \frac{d}{2}=0.25 mm=0.25 \cdot 10^{-3} m
And its cross-sectional area is
A=\pi r^2 = \pi (0.25 \cdot 10^{-3}m)^2 = 1.96 \cdot 10^{-7} m^2
The copper resistivity is \rho=1.68 \cdot 10^{-8} \Omega m, therefore the resistance of this piece of wire is
R= \frac{\rho L}{A}= \frac{(1.68 \cdot 10^{-8} \Omega m)(1.0 m)}{1.96 \cdot 10^{-7} m^2}=  8.57 \cdot 10^{-2} \Omega

b) The length of this piece of iron is L=10 cm=0.10 m. Its cross-sectional size is L=1.0 mm=0.001 m, so its cross-sectional area is
A=L^2 = (0.001 m)^2 =1 \cdot 10^{-6}m^2
The iron resistivity is \rho = 9.71 \cdot 10^{-8} \Omega m, therefore the resistance of this piece of wire is
R= \frac{\rho L}{A}= \frac{(9.71 \cdot 10^{-8} \Omega m)(0.10 m)}{1.0 \cdot 10^{-6} m^2}=9.71 \cdot 10^{-3} \Omega

3 0
3 years ago
A long, hollow wire with inner radius a and outer radius b carries a uniform current density J. What is the magnetic field as a
sveta [45]

Answer:

The magnetic field in the region a < r < b is B=\frac{u_{0}I(r^{2}-a^{2})   }{2\pi r(b^{2}-a^{2})  }

Explanation:

If we have the a < r < b. The formula of current is:

J=\frac{I_{total} }{A}

Where:

A = area enclosed by the loop.

Itotal = total current in loop.

J=\frac{I}{\pi b^{2}-\pi  a^{2} }

I_{enclosed} =JA_{enclosed}

I_{enclosed} =\frac{I(\pi r^{2}- \pi a^{2})}{\pi b^{2}-\pi a^{2}  }

If we have the Ampere`s law:

\int\limits^a_b {B} \, ds  =u_{0} I_{enclosed} \\2B\pi r=u_{0} (\frac{I(\pi r^{2}-\pi  ^{2} }{\pi ^{2}-\pi  a^{2} } )\\B=\frac{u_{0}I(r^{2}-a^{2})   }{2\pi r(b^{2}-a^{2})  }

6 0
2 years ago
A machine which has an energy loss of 10% will have efficiency of​
larisa [96]

Answer:

90%

Explanation:

if you lose 10% of a 100 you get 90

6 0
2 years ago
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