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RoseWind [281]
3 years ago
9

An object is tracked by a radar station and found to have a position vector given by r = (4570-160 t) i + 2930j + 130k , with r

in meters and t in seconds. The radar station's x axis points east, its y axis north, and its z axis vertically up. If the object is a 300 kg meteorological missile, what are (a) its linear momentum (in unit vector notation) and (b) the magnitude of the net force on it?
Physics
1 answer:
Kipish [7]3 years ago
6 0

Answer:

a)P= -4800 i

b)F= 0

Explanation:

Given that

r = (4570-160 t) i + 2930 j + 130 k

We know that velocity is rate of change of the space vector.

V= dr/dt

r = (4570-160 t) i + 2930 j + 130 k

dr/dt= -160 i + 0 + 0

dr/dt= -160 i

V= -160 i

It means that velocity is in only x-direction

We also know that acceleration is the rate of change of velocity .

a= dV/dt

V= -160 i

dV/dt=0

So we can say that acceleration is zero.

 a= 0

From Second law of Newton'

Force =  Mass x acceleration

F= 300 x 0

F= 0

We know that linear momentum P

P = m V

Given that m= 300 kg

P = 300 x (-160 i)

P= -4800 i

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Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The mass of the
IRISSAK [1]

Answer:

a₁ = 0.63 m/s²  (East)

a₂ = -1.18 m/s²  (West)

Explanation:

m₁ = 95 Kg

m₂ = 51 Kg

F = 60 N

a₁ = ?

a₂ = ?

To get the acceleration (magnitude and direction) of the man we apply

∑Fx = m*a   (⇒)

F = m₁*a₁         ⇒      60 N = 95 Kg*a₁    

⇒  a₁ = (60N / 95Kg) = 0.63 m/s²  (⇒)  East

To get the acceleration (magnitude and direction) of the woman we apply

∑Fx = m*a   (⇒)

F = -m₂*a₂         ⇒      60 N = -51 Kg*a₂    

⇒  a₂ = (60N / 51Kg) = -1.18 m/s²  (West)

For every case we apply Newton’s 3 d Law

8 0
3 years ago
Someone please help
saul85 [17]

Based on the attached image:

  • The name of the longitude line that passes through point A is the International Date Line
  • The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
  • The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
  • Longitude 120° is B
  • Solar time at Point B is 4 pm
  • The location will correspond to any point on the same latitude as A

<h3>What are lines of longitude?</h3>

Lines of longitude are imaginary lines which run along the earth from the North pole. to the South pole.

Longitude lines divide the earth into semi-circles.

Longitude lines are known as meridians and each meridian measures one arc degree of longitude.

Considering the attached image:

  • The name of the longitude line that passes through point A is the International Date Line
  • The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
  • The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
  • Longitude 120° is B
  • Solar time at Point B is 4 pm
  • the location will correspond to any point on the same latitude as A

In conclusion, longitude lines are imaginary lines and run from North to South on the earth.

Learn more about lines of longitude at: brainly.com/question/1939015

#SPJ1

8 0
2 years ago
Suppose that the angular separation of two stars is 0.1 arcseconds, and you photograph them with a telescope that has an angular
Crank

Answer: the photograph will likely show only one star.

Explanation:

Since their angular separation is smaller than the telescope's angular resolution, the picture will apparently show only one star rather than two.

4 0
3 years ago
Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.012 m,
GREYUIT [131]

Answer:

The current of the outer coil is  I_o   =   3.99 \ A

Explanation:

From the question we are told that

    The number of turns of the inner coil is  N_i  = 170 \ turn

    The radius of the  inner coil is R_i =  0.012 \ m

     The current of the inner coil is  I_i  =  6.2 \ A

      The number of turns of the outer coil is N_o  =  220 \ turns

      The radius of the  outer coil is R_o  =  0.02 0 \ m

       

Generally the net magnetic field is mathematically represented as

              B  =  \frac{N \ mu I }{ 2 * R  }

Now from told that the net magnetic field is common

So  

           \frac{N_i  \mu I_i} {2 * R _i} =  \frac{N_o  \mu I_o} {2 * R _o}

Here  \mu is the permeability of free space

making  I_o the subject

            I_o   =   \frac{ N_i  I_i *2 * R _o}{N_o  *2 * R _i}

substituting values

           I_o   =   \frac{ 170 *6.2 *2 * 0.020}{220   *2 * 0.012}

         I_o   =   3.99 \ A

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3 years ago
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