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All the planets orbit the Sun

aivan3 [116]

Answer:

Yes, they do

Explanation:

7 0

3 years ago

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A small sphere with a mass of 441 g is moving upward along the vertical +y-axis when it encounters an electric field of 5.00 N/C

sweet-ann [11.9K]

In your question where the ask is to calculate the charge that the small sphere carries which is the mass of it is 441g moving at an acceleration of 13m/s^2 nad having and electric field of 5N/C. So the formula in getting the charge is mutliply the mass and the quotients of Acceleration and the Electric Field so the answer is 1,146.6

3 0

3 years ago

Anyone here like conspiracies? I personally do, and I want to know what you guys think of the Last Thursday conspiracy. This is

alexdok [17]

Answer:

nope dont agree with that i think it would b a lot harder to do on a mass scale like that

Explanation:

Only way to do that is if aliens with far superior technology wise came to earth and did it

6 0

3 years ago

An airplane is heading due south at a speed of 690 km/h . A) If a wind begins blowing from the southwest at a speed of 90 km/h (

Afina-wow [57]

Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h east. c) 16,5 km NE of the desired position

Explanation:

Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where

see fig 1

Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then

90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,

this because we have an isosceles triangle, then the cathetus length is

hypotenuse/ $\sqrt{2}$

using Pythagoras, here the hypotenuse is 90, then the cathetus are of length

90/ $\sqrt{2}$ km/h= 63,6396 km/h.

Now the total speed of the plane is

690km/h south + 63,6396 km/h north +63,6396 km/h east,

this is 626,3604 km/h south + 63,6396 km/h east, here north is as if we had -south.

then using again Pythagoras we get the magnitude of the total speed it is

$\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h$ ,

the direction is calculated with respect to the south using trigonometry, we know the

sin x= cathetus opposed / hypotenuse,

then

x= $sin^{-1}'frac{63,6396}{629,5851}$ =5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form