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krok68 [10]
1 year ago
10

Is the statement "An object always moves in the direction of the net force acting on it" true or false

Physics
1 answer:
Trava [24]1 year ago
7 0

Answer:

False

Explanation:

This is because according to newtons second law which says the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. So take for example a net a net force in opposite direction will cause an object to slow down.

velocity vector here is not the same as acceleration vector

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A solid spherical insulator has radius r = 2.5 cm, and carries a total positive charge q = 8 × 10-10 c distributed uniformly thr
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At r = 2R> R The expression for the electric field will be given by: (2R)^2*E=kQ. Where, k=(9*10^9)N.m/C^2, Q=(8*10^-10)C and R=0.025m.  So substituting and clearing, we have that the magnitude of the electric field will be:  E=(9*10^9)*(8*10^-10)/((2*0.025)^2)=2880 N / C.
5 0
2 years ago
A friend of yours who takes her astronomy class very seriously challenges you to a contest to find the thinnest crescent moon yo
RUDIKE [14]

Answer:

after the sun sets or just as it is setting

Explanation:

a crescent moon is thin and reflects less sunlight during the daylight sky so it becomes difficult to spot, but can be spotted when the sun is setting or just sets.

8 0
2 years ago
A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward
AleksandrR [38]

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

          = 52 N  

5 0
2 years ago
A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53° above the hor
Volgvan

Answer:

h=23.67 m  : Building height

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 12.2 m/s  , at an angle  α=53° above the horizontal

x= 25 m , y=0

Calculation of the time it takes for the ball to hit the ground

We replace data in the equation (1)

x = xi + vx*t

x= 25 m   ,xi=0 ,  vx= v₀*cosα = (12.2 m/s)*cos(53°) =7.34 m/s

25 = 0 + 7.34*t

t= 25 / 7.34

t= 3.406 s

Calculation of the Building height

v₀y =  v₀*sinα = (12.2 m/s)*sin(53°) = 9.74 m/s

in t= 3.406 s, y=0

We replace data in the equation (2)

y= y₀ + (v₀y)*t - (1/2)*gt²

0=  y₀ + (9.74)*(3.406 )- (1/2)*(9.8)(3.406 )²

0=  y₀ + 33.17- -56.84

0=  y₀ - 23.67

y₀ =  23.67 m =h: Building height

5 0
2 years ago
What is the depth of the crater if the time is 6.3 s? An astronaut stands by the rim of a crater on the moon, where the accelera
ollegr [7]

Complete Question

An astronaut stands by the rim of a crater on the moon, where the acceleration of gravity is 1.62 m/s2. To determine the depth of the crater, she drops a rock and measures the time it takes for it

to hit the bottom. If the time is 6.3 s, what is the depth of the crater?

Answer:

The depth is 32 m

Explanation:

From the question we are told that

  The  time is  t =  6.3 s

  The acceleration due to gravity is  g =  1.62 \  m/s^2

 

Generally from kinematic equation

    s = ut + \frac{1}{2}  - at^2

Here the u is the initial  velocity and the value is  0 m/s

      s = 0 + \frac{1}{2}  - (1.62) * (6.3)^2

        s = 32 \ m

   

8 0
2 years ago
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