What radiation do remote controls use?

on an unknown temperature scale, the freezing point of water is -15°U and the boiling point is +60°U. develop a linear conversio

shutvik [7]

Answer:

Since this is a linear equation

y = m x + b or

U = m F + b is a linear equation

when ΔF = (212 - 32) = 180

and ΔU = (60 - (-15)) = 75

m = 75 / 180 = 2.4 if converting F to U and a = .417

U = .417 F + b

If F = 32 then U = -15 and

-15 = .417 * 32 + b

b = -15 - 13.3 = -28.3 and our equation becomes

U = .417 F - 28.3

Check: let F = 212

U = .417 * 212 - 28.3 = 60 as it should

5 0

2 years ago

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

(c) $F_{se} = 3.521\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

(c) The force exerted by the Sun on the Earth is given by eqn (1):

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

Many physiological conditions are related to particular proteins in cell membranes. The number of insulin receptors in membranes

Afina-wow [57]

Answer: This is an example of downregulation and upregulation.

Explanation:

Downregulation is a process in which cells decrease the production of one of their components, responding to an external stimulus. Upregulation, on the other hand, is when cells increase the production of one of their components in response to an external stimulus.

In this case, the decrease in insulin receptors would make the cell less sensitive to the hormone.

If there's a lot of insulin around those cells, the cell would have to decrease its sensitivity, otherwise, it would metabolise more glucose than the body needs. The contrary would happen if there was too little insulin around those cells, they would have to become more sensitive to it by increasing the number of receptors.

8 0

3 years ago

An element's atomic number is 59. How many electrons would an atom of this element have?

katrin [286]

Hi,
An element’s atomic number tells how many protons an atom of this element has. Protons have a positive electric charge, while electrons have a negative one. If an atom is electrically neutral (as in this case, otherwise it should have been specified) the number of protons and electrons are equal. So, an atom of the element would have 59 electrons.

Hope this helps! If my answer was not clear enough or you’d like further explanation please let me know. Also, English is not my first language, so I’m sorry for any mistake.

6 0

3 years ago

A rock is thrown at an angle of 30 degrees above the horizontal with initial velocity 15m/s what is the displacement when the ro

Delvig [45]

The displacement of the rock will be the same as the total horizontal distance traveled. Here the rock's horizontal position is given by

$x=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\cos30^\circ\,t$

so to find the horizontal distance it traversed, we need to know the time it took for the rock to return to the ground. We use the rock's vertical position over time to figure that out:

$y=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)\sin30^\circ\,t-\dfrac g2t^2=0$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the acceleration due to gravity. Then we find that $t\approx1.5\,\mathrm s$ , at which point we find $x\approx20\,\mathrm m$ .

6 0

3 years ago