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maria [59]
3 years ago
5

If a circle was flattened by pushing down on it, it would most likely form into the shape of ----------- which has ----------- f

ocal points.

Physics
2 answers:
murzikaleks [220]3 years ago
8 0
The question is looking for "ellipse" and "two" to fill in the blanks.
labwork [276]3 years ago
5 0

Answer:If a circle was flattened by pushing down on it, it would most likely form into the shape of <em><u>ellipse  </u></em>which has <em><u>two </u></em>focal points.

Explanation:

Ellipse is curve in a plane with two focus points. The sum of the distance of  one point from both fixed points or focal points in the plane remain constant.

When circle was flattened by pushing it deform to plane shape likely to the shape of ellipse.

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Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction betw
jarptica [38.1K]

Answer:

a)  W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d)     the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

          W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

          W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

         θ = 90         cos 90 = 0

        W=0

In conclusion the work is zero

b) The friction force opposes the displacement whereby the angle is

       θ = 180      cos 190 = -1

        W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

            m g sin θ  L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀=  + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

               The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

               ΔU = -U₀

         the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

              W_Total = W_weight - W_roce

as the body moves down

              W_Total> 0

Therefore the total work is positive

3 0
3 years ago
if you push your chair across the floor at a constant velocity how does the force of friction compare with the force you exert?
Alik [6]
-- If velocity is constant, then there is no net force
on the chair.

-- If there is no net force on the chair, then friction
must exactly balance out your push.

-- The force of friction is exactly equal in magnitude
to your push, and in exactly the opposite direction.
5 0
3 years ago
Read 2 more answers
Which of the following is not an example of work, according to the scientific definition
bogdanovich [222]

Leaning against a brick wall.

All the others use scientific forces of work.

-Steel jelly.

6 0
3 years ago
Read 2 more answers
If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is
viktelen [127]
To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: V_{f}=V_{i}+gt
where
V_{f} is the final velocity 
V_{i} is the initial velocity 
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 
2. Kinematic equation for distance: d=V_{i}t+ \frac{1}{2} gt^2
where
d is the distance 
V_{i} is the initial velocity 
V_{f} is the final velocity
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 

First, we are going to convert 21.82 mi/h to ft/s:
21.82 \frac{mi}{h} =31.21 \frac{ft}{s}

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: V_{f}=0. We also know that it initial speed is 31.21 ft/s, so V_{i}=31.21. Lets replace those values in our formula to find t:
V_{f}=V_{i}+gt
0=31.21+(-32)t
-32t=-31.21
t= \frac{-31.21}{-32}
t=0.98seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
d=V_{i}t+ \frac{1}{2} gt^2
d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2
d=15.22ft 

Now we can add the height of the building and the maximum height of the object:
d=160+15.22=175.22ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
d=V_{i}t+ \frac{1}{2} gt^2
175.22=31.21t+ \frac{1}{2} (32)t^2
16t^2+31.21t-175.22=0
t=2.47 or t=-4.43
Since time cannot be negative, t=2.47 is the time it takes the object to fall to the ground. 

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
V_{f}=V_{i}+gt
V_{f}=31.21+(32)(2.47)
V_{f}=110.25 ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s


5 0
3 years ago
What do you need to know to be able to determine how far a projectile travels horizontally?
Effectus [21]
As for me, there are two suitable answers for the question represented above and here is a short explanation why I consider these two to be correct : 
D. The horizontal velocity of the projectile and <span>B. The length of time before it lands 
</span>v_x = v*cos(x) [&#10;&#10;d_x = v_x(t) + ½at² -- this led me to answers! Hope everything is clear! Regards!<span>
</span>
5 0
3 years ago
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