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d1i1m1o1n [39]
3 years ago
8

The ground heats the air through what

Physics
1 answer:
Tema [17]3 years ago
4 0

Explanation:

The atmosphere is heated in several ways - heat from the core of the Earth, by radiation from the Sun, conduction from contact with warm land and water, convection to even out the temperature and by absorption of infrared radiation from the warm land and water. The core of the Earth is very hot.

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What is power the quotient of?
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Power is equal to work divided by time
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Standing at the same location, which organism would have the GREATEST gravitational attraction to the earth?
Nastasia [14]
To claculate the gravitational attraction between two bodies with mass 1(m1) and mass 2 (m2) you need to use the equation:

F= G ((m1*m2)/r^2)

Where;
G is the gravitational constant (6.67E-11 m^3 s-2 Kg-1) and
r is the distance between the two objects.


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3 years ago
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When a ball player throws a ball straight up, by how much does the speed of the ball decrease each second while ascending?
Alexxandr [17]
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6 0
3 years ago
Rhea kicks a soccer ball at 13km/h to Sean after kicking the ball the speed of the soccer ball from Rhea reference frame is 13km
Zinaida [17]

Answer:

Sean is standing still, and Rhea is running toward Sean while kicking the ball.

<em>Note: The question is in complete. The complete question is given below:</em>

<em>Rhea kicks a soccer ball at 13 km/h to Sean. After kicking the ball, the speed of the soccer ball from Rhea's reference frame is 13 km/h. The speed of the soccer ball from Sean's reference frame is 22 km/h.  Which conclusion is best supported by the information? </em>

<em>Sean is standing still, and Rhea is running toward Sean while kicking the ball.</em>

<em>Rhea is standing still after kicking the ball, and Sean is running away from Rhea.</em>

<em>Both soccer players are standing still after the ball is kicked.</em>

<em>Both soccer players are running while the ball is in motion.</em>

Explanation:

A reference frame is a position from which something is observed.

Since from the reference frame of Rhea, the ball is moving at a speed of 13 km/h after he kicks the ball at a speed of 13km/h whereas from the frame of reference of Sean, the ball is moving at a speed of 22km/h, there is a difference in the speed of the ball as seen from Rhea's and Sean's frame of reference of about 9 km/h. This difference can only be due to relative motion between the ball and Rhea.

Therefore, the best conclusion supported by the given information is that Sean is standing still, and Rhea is running toward Sean while kicking the ball.

4 0
3 years ago
Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As
Naddik [55]

Answer:

a

   \lambda = 1.667 nm

b

     \theta  =  0.8681^o

Explanation:

From the question we are told that

   The distance of separation is d  =  0.220 \ mm  =  0.00022 \ m

    The  is distance of the screen from the slit is  D   =  2.60 \ m

    The distance between the central bright fringe and either of the adjacent bright   y  =  1.97 cm  =  1.97 *10^{-2}\ m

Generally  the condition for constructive interference is  

      d sin \tha(\theta ) =  n \lambda

From the question we are told that small-angle approximation is valid here.

So    sin (\theta ) = \theta

=>        d \theta  =  n \lambda

=>        \theta =  \frac{n *  \lambda }{d }

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         y  =  D * sin (\theta )

From the question we are told that small-angle approximation is valid here.

So

       y  =  D * \theta

=>   \theta  =  \frac{ y}{D}

So

     \frac{n *  \lambda }{d } = \frac{y}{D}

     \lambda =\frac{d * y }{n * D}

substituting values

       \lambda =  \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }

        \lambda = 1.667 *10^{-6}

        \lambda = 1.667 nm

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     dsin (\theta ) =  n \lambda

    \theta  =  sin^{-1}[\frac{ n  *  \lambda }{d} ]

    \theta  =  sin^{-1}[\frac{ 2  *  1667 *10^{-9}}{ 0.00022} ]

    \theta  =  0.8681^o

7 0
4 years ago
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