The motion of an object parallel to the earth's surface is

If a substance has a high specific heat, is it easy or difficult to change the temperature of that

zmey [24]

It's difficult because the temperature is obviously maintained well if it's designated a specific heat.

7 0

4 years ago

It is difficult to observe 1-nanometer, 1 millimeter, and 100 meter radiation with ground-based telescopes. What are the reasons

Paul [167]

Explanation:

1 nano-meter radiations are very difficult to observe from the ground based telescope because most of this range of radiation is absorbed through ozone layer. A very small amount of this range escape out of the ozone layer. This remaining few radiations are very difficult to track from the ground base telescope.

1 millimeter range of radiation comprises of infrared. It has range from 710 nano-meter to 1 millimeter. Infrared radiation can be easily absorbed from water and carbon di oxide molecules present in the atmosphere. So, it is absorbed by water and carbon di oxide molecules in the atmosphere. Thus, it is difficult to observe from the ground based telescope.

100 meter radiations are are radio-waves. The charged particle present in the uppermost layer of atmosphere absorbs these radio waves. So, these waves are absorbed by charged particle in the upper atmosphere. Thus, it is difficult to observe from the ground based telescope.

7 0

4 years ago

What occurs during an exothermic change

Likurg_2 [28]

In an exothermic reaction, there is a transfer of energy to the surroundings in the form of heat energy. The surroundings of the reaction will experience an increase in temperature. Many types of chemical reactions are exothermic, including combustion reactions, respiration & neutralization reactions of bases & acids.

6 0

3 years ago

Read 2 more answers

A magnesium surface has a work function of 2.65 eV. Electromagnetic waves with a wavelength of 280 nm strike the surface and eje

Neko [114]

Answer:

Explanation:

Energy of radiation in eV = 1237.5/280 = 4.42 eV.

Work function = 2.65 eV.

Maximum kinetic energy = 4.42 - 2.65 = 1.77 eV.

8 0

4 years ago

Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of one’s birth. The on

astraxan [27]

Answer: (a) $F=7(10)^{-7}N$

(b) $F=1.344(10)^{-6}N$

(c) The force of Jupiter on the baby is slightly greater than the the force of the father on the baby.

Explanation:

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

Knowing this, let's begin with the answers:

<h2 /><h2>(a) Gravitational force Father exertes on baby</h2>

Using equation (1) and taking into account the mass of the father $m_{1}=100kg$ , the mass of the baby $m_{2}=4.20kg$ and the distance between them $r=0.2m$ , the force $F_{F}$ exerted by the father is:

$F_{F}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(100kg)(4.20kg)}{(0.2m)^2}$ (2)

$F_{F}=0.0000007N=7(10)^{-7}N$ (3)

<h2>(b) Gravitational force Jupiter exertes on baby</h2>

Using again equation (1) but this time taking into account the mass of Jupiter $m_{J}=1.898(10)^{27}kg$ , the mass of the baby $m_{2}=4.20kg$ and the distance between Jupiter and Earth (where the baby is) $r_{E}=6.29(10)^{11}m$ , the force $F_{J}$ exerted by the Jupiter is:

$F_{J}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.898(10)^{27}kg)(4.20kg)}{(6.29(10)^{11}m)^2}$ (4)

$F_{J}=0.000001344N=1.344(10)^{-6}N$ (5)

<h2>(c) Comparison</h2>

Now, comparing both forces:

$F_{J}=0.000001344N=1.344(10)^{-6}N$ and $F_{F}=0.0000007N=7(10)^{-7}N$ we can see $F_{J}$ is greater than $F_{F}$ . However, the difference is quite small as well as the force exerted on the baby.

7 0

3 years ago