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weqwewe [10]
1 year ago
3

A passenger plane is flying above the ground. Describe two components of its mechanical energy.

Physics
1 answer:
Nookie1986 [14]1 year ago
3 0
Tail structure and propulsion system if I'm wrong correct me
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The acceleration due to gravity on the Moon's surface is
Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

g'=\frac{g}{6}

We know that weight of an object on Earth is,

w = m\times g

m = \frac{w}{g}

Similarly, weight on Moon will be

w' = m\times g'

w' = \frac{w}{g}\times\frac{g}{6}

w' = \frac{300}{6}

w' = 50

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0
1 year ago
Explain the coin in water appears to be raised..​
Lilit [14]

Answer:

defraction

Explanation:

when light passes from a less dense medium(air) into a dense medium(water) the light rays bend towards the normal. This makes the coin look closer

5 0
1 year ago
Bob is pushing a box across the floor at a constant speed of 1.7 m/s, applying a horizontal force whose magnitude is 70 N. Alice
OlgaM077 [116]

Answer:

a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.

b) Alice used initially a greater force to accelerate the box up to needed speed and later reduced the external force to keep speed constant. The right choice is option b.

3 0
1 year ago
The addition of electron shells results in<br><br> shielding of electrons from the nucleus.
Pani-rosa [81]

Answer:

Addition of shells increases the distance of outer electrons from the nucleus.

Explanation:

Shielding effect is known as the attraction between the nucleus and an electron of any atom. In other words, it is the reduction in effective nuclear charge on an electron cloud.

Addition of electron shells results in the shielding of electron from nucleus.  As the number of electron shells increases then farther will be the electrons placed from the nucleus and hence it will become easier to expel the electrons from outer shells with only little amount of ionization energy.

So, the amount of ionization energy require will be indirectly proportional to the shielding effect because more the shielding of electrons from the nucleus less will be the ionization energy require to expel the electrons.

4 0
1 year ago
(based on 15-103 in the text) At an initial instant, a 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a sp
lisov135 [29]

Answer:

a

 v_r =8.65 \ ft/s

b

  W_{1-2} =  3.24 \  ft \cdot lb

Explanation:

From the question we are told that

  The mass of the ball is m  =  4 \  lb

  The radius is  r= 3 \  ft

   The speed is v_B_1  =  4.8 \ ft /s

    The speed of the attached cord is  v_c =2.2 \  ft

   The position that is been considered is  r_1 =  2 \  fth

Generally according to the law of angular momentum conservation

   L_a =  L_b

Here L_a is the initial  momentum of the ball which is mathematically represented as

      L_a  =  m*  v_B_1 *  r

while  

L_b is the  momentum of the ball  at  r =  2 ft which is mathematically represented as

       L_a  =  m*  v_B_2 *  r_1

So

      m*  v_B_1 *  r = m*  v_B_2 *  r_1

=>      4.8 *  3 =  v_B_2 *  2

=>     v_B_2 =  7.2 \  ft/s

Generally the resultant velocity of the ball is  

      v_r = \sqrt{v_B_2^2 + v_B_1^2   }

=>   v_r = \sqrt{7.2^2 + 4.8^2   }

=>   v_r =8.65 \ ft/s

Generally according to equation for principle of work and energy we have that

    K_1 + \sum W_{1-2} = K_2

Here K_1 is the initial kinetic energy of the ball which is mathematically represented as

K_1  =  \frac{1}{2}  *  m* v_B_1^2

While  \sum W_{1-2} is the sum of the total  workdone by the ball

and  K_2 is the final kinetic energy of the ball  which is mathematically represented as  K_2  =  \frac{1}{2}  *  m* v_r^2

So

     \sum W_{1-2} =  \frac{1}{2}  *  m  (v_r^2  -  v_B_1^2)

Here  m is the mass which is mathematically represented as

     m = \frac{W}{g} here W is the weight in  lb and  g is the acceleration due to gravity which is g =  32 \ ft/s^2

So

    \sum W_{1-2} =  \frac{1}{2}  *  \frac{4}{32} *   (8.65^2  -  4.8^2)

=>   W_{1-2} =  3.24 \  ft \cdot lb

   

   

3 0
1 year ago
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