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marissa [1.9K]
3 years ago
11

Which unit do astronomers use for angular measurement?

Physics
1 answer:
Alex787 [66]3 years ago
8 0
C and D are units of length or distance.
A is a measured angle.
B is a unit of angular measurement.
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KiRa [710]

Answer:

momentum in a body can be calculated using

<em><u>Mome</u></em><em><u>ntum</u></em><em><u>=</u></em><em><u>Mass×</u></em><em><u>V</u></em><em><u>e</u></em><em><u>l</u></em><em><u>o</u></em><em><u>s</u></em><em><u>i</u></em><em><u>t</u></em><em><u>y</u></em><em><u> </u></em>

<em><u>i</u></em><em><u>e(</u></em><em><u>p</u></em><em><u>=</u></em><em><u>m×</u></em><em><u>v</u></em><em><u>)</u></em>

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3 years ago
A towrope is used to pull a 1750-kg car, giving it an acceleration of +1.35 m/s ^2. What force does the rope exert?
Klio2033 [76]
Newton's 2nd law of motion:

               Force  =  (mass) x (acceleration)

                           =  (1,750 kg) x (1.35 m/s²)

                           =      2,362.5 Newtons

                              (about 531 pounds)
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Rosný bod závisí od ?
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Answer:

what did u say and what language are you speaking in

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a baby carriage is sitting at the top of a hill that is 21m high. The carriage with the baby weighs 12kg. The carriage has _____
Ket [755]
  • Height=h=21m
  • Mass=m=12kg

\\ \rm\longmapsto P.E=mgh

\\ \rm\longmapsto P.E=12(10)(21)

\\ \rm\longmapsto P.E=120(21)

\\ \rm\longmapsto P.E=2520J

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3 years ago
A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle
Sidana [21]

Answer:

B = 0.135T

Explanation:

To find the magnitude of the magnetic field you use the following formula, for the torque produced by a magnetic field B in a loop:

\tau=NIABsin\theta   (1)

τ: torque = 1.51*10^-5 Nm

I: current = 2.47mA = 2.47*10^-3 A

B: magnitude of the magnetic field

A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

N: turns = 181

θ: angle between B and the magnetic dipole (same as the direction  of the normal to the plane)

You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T

3 0
3 years ago
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