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dalvyx [7]
2 years ago
15

You kick a soccer ball across a field the ball travels across the field and slowly comes to a stop. Since the ball slowly comes

to a stop what does this mean about the net force on the ball as it rolls across the field?
Physics
1 answer:
marysya [2.9K]2 years ago
6 0

Answer:

It slowly decreases and the friction acting on it slowing it down becomes the bigger net force, if that makes sense :)

Explanation:

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Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from
DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²  

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

   u = 0 m/s

   g = 9.8 m/s²

   s = 0 + 0.5 × 9.8 × 2.2²

  s = 23.72 m

b) impact velocity

      v = √(2gh)

      v = √(2× 9.8 × 23.72)    

      v =  √464.912

      v = 21.56 m/s²  

6 0
3 years ago
What is another word for cycles per second??
Hatshy [7]
Frequensey or hertz, I looked this up on the internet!
5 0
2 years ago
Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po
zvonat [6]

Answer:

0.293I_0

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

I_1=\frac{I_0}{2}

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

I_2=I_1 cos^2 \theta

where

\theta is the angle between the axes of the two polarizer

Here we have

\theta=40^{\circ}

So the intensity after the 2nd polarizer is

I_2=I_1 (cos 40^{\circ})^2=0.587I_1

And substituting the expression for I1, we find:

I_2=0.587 (\frac{I_0}{2})=0.293I_0

5 0
2 years ago
An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi
hoa [83]

Answer:

a)       a = 3.72 m / s², b)    a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

             v = v₀ + at

as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

Having the force let's use Newton's second law

            F = m a

            -18.75 m = m a

             a = -18.75 m / s²

4 0
2 years ago
Which of the following objects have kinetic energy?
Montano1993 [528]
I think the Answer is A (An elephant walking 1.5 m/s along the ground), and B (A jet flying across the sky 4,000 m above the ground.
4 0
1 year ago
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