Which of the following statements correctly compares the relationship between the earth, its atmosphere and radiation?1. The earth is cooled and its atmosphere is heated by solar radiation.

2. The earth is heated and its atmosphere is cooled by terrestrial radiation.

3. The earth is cooled and its atmosphere is heated by terrestrial radiation.

4. The earth is heated and its atmosphere is cooled by solar radiation.

Don't answer unless you know for sure. Thank you so much!

Answers

Answer 1

Answer:

Answer: The option 4 is correct answer.

Explanation:

Terrestrial radiation is a long wave electromagnetic radiation. It originates from the earth and its atmosphere.

The sun emits a huge amount of energy. It travels across the space. The atmosphere is not directly heated by the solar radiation. It is heated by the terrestrial radiation that the planet itself emits.

When the land is heated then it emits radiation which heats up the atmosphere.

The earth is cooled and its atmosphere is heated by terrestrial radiation.

Therefore, the relationship between the earth, its atmosphere and radiation is correctly compared by statement 4.

Answer 2

Answer: The earth is cooled and its atmosphere is heated by terrestrial radiation.

Answer

The Columb's law is the same as Gravitational law

Explanation

As we see the formula of both Coulomb and Gravitational Law,

$F_g = GM_1M_2/R_2$ (1)

$F_c = kq_1q_2/r_2$ (2)

The masses (M) in formula (1) experiencing the force of gravitational pull with each other which varies with changing the distance. In the formula (2), the charges also are felling the forces on each other which varies with distance. The charges and masses are just like the objects which are experiencing the forces which have a common factor as distance. The gravitational force is also called the mutual forces.

A square coil (length of side = 24 cm) of wire consisting of two turns is placed in a uniform magnetic field that makes an angle of 60° with the plane of the coil. If the magnitude of this field increases by 6.0 mT every 10 ms, what is the magnitude of the emf induced in the coil?

Answers

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

$\epsilon=N(d\phi)/(dt)$

$(d\phi)/(dt)$ is the rate of change if magnetic flux.

$\phi=BA\ cos\theta$

$\theta$ is the angle between the magnetic field and the normal to area vector.

$\theta=90-60=30$

$\epsilon=NA(dB)/(dt)* cos30$

$\epsilon=2* (0.24\ m)^2* (6\ mT)/(10\ mT)* cos(30)$

$\epsilon=0.0598\ T$

$\epsilon=59.8\ mT$

EMF = 60 mT

So, the magnitude of emf induced in the coil is 60 mT. Hence, this is the required solution.

A student goes skateboarding a few times a week what is the dependent variable

Answers

The dependent variable in this scenario is the outcome or result that you are trying to measure or analyze based on the student's skateboarding activity.

Since the student goes skateboarding a few times a week, the dependent variable could be any aspect related to their skateboarding experience or its effects.

Examples of possible dependent variables could include:

1. Improvement in skateboarding skills (e.g., measured by tricks learned, levels of proficiency).

2. Physical fitness (e.g., measured by changes in endurance, strength, or flexibility).

3. Time spent skateboarding per session.

4. The number of skateboarding injuries or accidents.

5. Overall enjoyment or satisfaction with skateboarding.

6. Changes in stress levels or mood before and after skateboarding sessions.

7. Social interactions and friendships formed through skateboarding.

The specific dependent variable would depend on the research question or hypothesis you are investigating in relation to the student's skateboarding activity.

11.
A current of 67 amps runs through a resistor of 37 ohms, how much voltage is lost?

Answers

You divide them. 67/37 is 1.5 and so you subtract it with the 67 and multiple it by the coherent integer from the multiplication and you would get 20 volts lost roughly.

Answer: 20 volts

A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignore and using conservation of mechanical energy, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.

Answers

Answer:

a) The initial speed of the rock is approximately 14.607 meters per second.

b) The greatest height of the rock from the base of the cliff is 42.878 meters.

Explanation:

a) The rock experiments a free-fall motion, that is a vertical uniform accelerated motion due to gravity, in which air friction and effects of Earth's rotation. By Principle of Energy Conservation we have the following model:

$U_(g,1)+K_(1) = U_(g,2)+K_(2)$ (Eq. 1)

Where:

$U_(g,1)$ , $U_(g,2)$ - Initial and final gravitational potential energies, measured in joules.

$K_(1)$ , $K_(2)$ - Initial and final translational kinetic energies, measured in joules.

By definitions of gravitational potential and translational kinetic energies we expand and simplify the equation above:

$m\cdot g\cdot (y_(1)-y_(2))= (1)/(2)\cdot m\cdot (v_(2)^(2)-v_(1)^(2))$

$g\cdot (y_(1)-y_(2)) = (1)/(2)\cdot (v_(2)^(2)-v_(1)^(2))$ (Eq. 2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$y_(1)$ , $y_(2)$ - Initial and final height, measured in meters.

$v_(1)$ , $v_(2)$ - Initial and final speed of the rock, measured in meters per second.

If we know that $g = 9.807\,(m)/(s^(2))$ , $y_(1) = 32\,m$ , $y_(2) = 0\,m$ and $v_(2) = 29\,(m)/(s)$ , then the equation is:

$\left(9.807\,(m)/(s^(2)) \right)\cdot (32\,m-0\,m) = (1)/(2)\cdot \left[\left(29\,(m)/(s) \right)^(2)-v_(1)^(2)\right]$

$313.824 = 420.5-0.5\cdot v_(1)^(2)$

$0.5\cdot v_(1)^(2) = 106.676$

$v_(1) \approx 14.607\,(m)/(s)$

The initial speed of the rock is approximately 14.607 meters per second.

b) We use (Eq. 1) once again and if we know that $g =9.807\,(m)/(s^(2))$ , $y_(1) = 32\,m$ , $v_(1) \approx 14.607\,(m)/(s)$ and $v_(2) = 0\,(m)/(s)$ , then the equation is:

$\left(9.807\,(m)/(s^(2)) \right)\cdot (32\,m-y_(2)) = (1)/(2)\cdot \left[\left(0\,(m)/(s) \right)^(2)-\left(14.607\,(m)/(s) \right)^(2)\right]$

$313.824-9.807\cdot y_(2) = -106.682$

$9.807\cdot y_(2) = 420.506$

$y_(2) = 42.878\,m$

The greatest height of the rock from the base of the cliff is 42.878 meters.

What is the angular width of a person's thumb viewed at arm's length? Assume that the width of the thumb is 17.3 mm and that the distance between the eyes and the thumb is 71.9 cm. Use the small-angle approximation and then convert the answer to degrees.

Answers

Answer:

$\theta_(degrees) =0.024\°=0\°1'26.62''$

Explanation:

To solve the problem it is necessary to take into account the concepts related to arc length and the radius that make up the measurements of an angle.

An angle is given by the length of arc displaced as a function of the radius, that is

$\theta = (Arc_(length))/(Radius)$