Explanation:
When the body temperature tends to rise, such as during physical exercise, the body begins to sweat. The sweat with high water content is secreted in the skin and when it evaporates into the environment, it cools the body. This is due to the property of water having high heat capacity. It carries with it a lot of heat per molecule (because water requires much energy – than most materials - for its temperature to rise by a degree) hence ideal for cooling. This is why on a hot day, sweating makes the skin feel cooler than the surrounding.
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<h2>
Answer: Pulsars</h2>
A <u>pulsar</u> is a neutron star that emits very intense electromagnetic radiation at short and periodic intervals ( rotating really fast) due to its intense magnetic field that induces this emission.
Nevertheless, it is important to note that all pulsars are neutron stars, but not all neutron stars are pulsars.
Let's clarify:
A neutron star, is the name given to the remains of a supernova. In itself it is the result of the gravitational collapse of a massive supergiant star after exhausting the fuel in its core.
Neutron stars have a small size for their very high density and they rotate at a huge speed.
However, the way to know that a pulsar is a neutron star is because of its high rotating speed.
Answer:
The required pressure is 6.4866 atm.
Explanation:
The given data : -
In the afternoon.
Initial pressure of tire ( p₁ ) = 7 atm = 7 * 101.325 Kpa = 709.275 Kpa
Initial temperature ( T₁ ) = 27°C = (27 + 273) K = 300 K
In the morning .
Final temperature ( T₂ ) = 5°C = ( 5 + 273 ) K = 278 K
Given that volume remains constant.
To find final pressure ( p₂ ).
Applying the ideal gas equation.
p * v = m * R * T
= 657.2615 Kpa = 6.486 atm
The new period will be 2.486 days.
<h3>What is the period?</h3>
The period is found as the ratio of the angular displacement and the angular velocity. Its unit is the second and is denoted by t. The value of time needed to complete the rotation is the total period.
Given data;
Mass of a star,m= 1.210×10³¹ kg
The time period for one rotation of the star, T = 20.30 days
D' = 0.350 D
R' = 0.350 R
From the law of conservation of angular momentum;
Hence, the new period will be 2.486 days.
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