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svetlana [45]
3 years ago
8

Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. one force is 9.0 n, acting due east, an

d the other is 9.2 n, acting 58° north of west. what is the magnitude of the body's acceleration?
Physics
1 answer:
Zolol [24]3 years ago
6 0
The first thing you should do for this case is to find the horizontal and vertical components of the forces acting on the body.
 We have then:
 Horizontal = 9-9.2cos (58) = 4.124742769 N.
 Vertical = 9.2sin (58) = 7.802042485 N
 Then, the resulting net force is:
 F = √ ((4.124742769) ^ 2 + (7.802042485) ^ 2) = 8.825268826 N
 Then by definition:
 F = m * a
 Clearing the acceleration:
 a = F / m
 a = (8.825268826) / (3.0) = 2.941756275 m / s ^ 2
 answer:
 The magnitude of the body's acceleration is
 2.941756275 m / s ^ 2
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A plane traveled west for 4.0 hours and covered a distance of 4,400 meters what’s the velocity
tatuchka [14]

0.31m/s

Explanation:

Given parameters:

Time of travel = 4hrs = 4 x 60 x 60 = 14400s

Displacement  = 4400m due west

Unknown:

Velocity = ?

Solution:

Velocity is defined as the displacement  per unit of time. It is expressed in m/s or km/hr:

     Velocity =  \frac{displacement}{time}

     Velocity =   \frac{4400}{14400} = 0.31m/s

Learn more:

Velocity brainly.com/question/10883914

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8 0
3 years ago
What distance does light travel in water, glass, and diamond during the time that it travels 1.0 m in vacuum? the refractive ind
alexdok [17]
We are given with
distance traveled through vacuum = 1.0 m
refractive index of water = 1.33
refractive index of glass = 1.50
refractive index of diamond = 2.42

distance traveled through water is = 1.0/1.33 = 0.75 m
distance traveled through water is = 1.0/1.50 = 0.67 m
distance traveled through water is = 1.0/2.42 = 0.41 m
8 0
3 years ago
Read 2 more answers
Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?
MrMuchimi
 T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r 

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance. 
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect, 

Point 1: 
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r

Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m 
The distance between the two points then is equal to 7.07 m.


</span>
8 0
3 years ago
In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m
Ksenya-84 [330]

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

                                q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

F = \dfrac{kq_1q_2}{r^2}

F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}

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b) using newton's second law

F = m a

m a =  8.20 x 10⁻⁸

a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}

    a = 9 x 10²² m/s²

c) speed of the electron

 a =\dfrac{v^2}{r}

 9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}

   v² = 4.77 x 10¹²

  v = 2.18 x 10⁶ m/s

d) the period of the circular motion.

    T=\dfrac{2\pi}{\omega}

    T=\dfrac{2\pi r}{v}

    T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}

          T = 1.53 x 10⁻¹⁶ s

8 0
3 years ago
Read 2 more answers
A satellite is orbiting Earth in an approximately circular path. It completes one revolution each day (86,400 seconds). Its orbi
stira [4]
@AL2006 had answered this before: Well, first of all, wherever you got this question from has done 
a really poor job of question-writing.  There are a few assorted 
blunders in the question, both major and minor ones:

-- 22,500 is the altitude of a geosynchronous orbit in miles, not km.

-- That figure of 22,500 miles is its altitude above the surface, 
   not its radius from the center of the Earth. 

-- The orbital period of a synchronous satellite has to match 
    the period of the Earth's rotation, and that's NOT 24 hours.  
    It's about 3 minutes 56 seconds less ... about 86,164 seconds. 

Here's my solution to the question, using some of the wreckage 
as it's given, and correcting some of it.  If you turn in these answers
as homework, they'll be marked wrong, and you'll need to explain 
where they came from.  If that happens, well, serves ya right for
turning in somebody else's answers for homework.


The satellite is traveling a circle. The circle's radius is 26,200 miles
(not kilometers) from the center of the Earth, so its circumference 
is (2 pi) x (26,200 miles) = about 164,619 miles.

    Average speed = (distance covered) / (time to cover the distance)

                             = (164,619 miles) / day
                                (264,929 km)

                             =      6,859 miles per hour
                                  (11,039 km) 

                           =          1.91 miles per second
                                      (3.07 km)
6 0
3 years ago
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