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Likurg_2 [28]
3 years ago
5

An element with _____ positioning is removed from the normal flow and keeps its position even when the user scrolls.

Physics
1 answer:
Colt1911 [192]3 years ago
7 0
An element with fixed positioning is removed from normal flow and keeps its place constant or even when user scrolls. This kind of positioning is fairly uncommon but truly has its uses of its own. A fixed role detail is placed relative to the viewport, or the browser window itself. The viewport would not exchange when the window is scrolled, so a set positioned element will stay proper in which it is when the page is scrolled, creating an impact a bit just like the antique college days.

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How much energy is stored in a spring with a spring constant of 390 N/m if the spring is compressed a distance of 0.45 m from it
masya89 [10]
F=-ks
F=-(390)(.45)
F=-175.5 N

Work=force x displacement
Work= 175.5(0.45)
Work= 78.98 J

Work = ∆E =78.98 J

Answer=79J (first option)
4 0
3 years ago
A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?
Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length f=\frac{R}{2}=\frac{12}{2}=6cm

Now for mirror we know that \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

So \frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}

16.66-0.333=\frac{1}{v}

v = 0.750 m

Now magnification of the mirror is m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025

5 0
4 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
Pertaining to simple machines and levers what changes when the fulcrum position is modified
ladessa [460]

Explanation :

Simple machines makes our work easier. Lever is one of the simple machine which consists of rigid rod that is pivoted at a fixed support called as Fulcrum.

There are three classes of lever.

Class 1 : In this type of class, fulcrum is placed in between effort and load. Hence the movement of load is in reverse direction of the movement of effort. (fig 1)

Class 2 : In this type of, the load is between the effort and the fulcrum. Hence, the movement of load is in same direction as that of the effort. (fig 2)

Class 3 : In this type of lever the effort between the load and the fulcrum. Hence, both the effort and load are in same direction. (fig 3)

Hence, when the position of fulcrum is modified the effort force changes.

6 0
4 years ago
Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 58.0º counter clockwise from the x-axis,
Inessa05 [86]

Answer:

option a is correct

net force = 28.7 N and 70º

Explanation:

given data

first force= 17.0 N

angle = 58.0º

2nd force = -19.0 N and 12.5 N

to find out

magnitude and direction net force

solution

we know here 1st force is at angle 58.0º is 17 N

so here x component of force is F1 cosθ and y component of force is F1 sinθ

so 1st force

x component of force =  F1 cosθ = 17 cos(58) = 9.0086 N    ...............1

y component of force = F1 sinθ = 17 sin(58) = 14.4168 N        ..............2

and 2nd force

x component of force =  -19 N                           .................3

y component of force = 12.5 N                          ..................4

so net force is

net force = √(Fx²+Fy²)

net force = √((9-19)²+(14.41+12.5)²)

net force = 28.7 N

and

magnitude

tanθ = Fy / Fx

tanθ = 26.92 / 10

θ = 70º

so option a is correct

3 0
3 years ago
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