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777dan777 [17]
2 years ago
8

A player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

Physics
1 answer:
____ [38]2 years ago
7 0

At the highest point in its trajectory, the ball's acceleration is zero but its velocity is not zero.

<h3>What's the velocity of the ball at the highest point of the trajectory?</h3>
  • At the highest point, the ball doesn't go more high. So its vertical velocity is zero.
  • However, the ball moves horizontal, so its horizontal component of velocity is non - zero i.e. u×cosθ.
  • u= initial velocity, θ= angle of projection

<h3>What's the acceleration of the ball at the highest point of projectile?</h3>
  • During the whole projectile motion, the earth exerts the gravitational force with a acceleration of gravity along vertical direction.
  • But as there's no acceleration along vertical direction, so the acceleration along vertical direction is zero.

Thus, we can conclude that the acceleration is zero and velocity is non-zero at the highest point projectile motion.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

A- neither the ball's velocity nor its acceleration are zero.

B- the ball's acceleration points upward.

C- the ball's acceleration is zero but its velocity is not zero.

D- the ball's velocity points downward.

Learn more about the projectile motion here:

brainly.com/question/24216590

#SPJ1

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An object is five focal lengths from a concave mirror.how do the object and image heights compare?
enot [183]

An object distance is presented as s = 5f and we know that the mirror equation relates the image distance to the object distance and the focal length.

The mirror equation is 1/f = 1/s + 1/s’ where the variable f stands for the focal length of the mirror. Variable (s) represents the distance between the mirror surface and the object and the variable <span>(s’) represents the distance between the mirror surface and the image. </span>

In addition, a concave mirror will have a positive focal length (f) and a convex mirror will have a negative focal length (f).

Now, we then have 1/f = 1/5f + 1/s’ which is s’ = 5f/4

Then we get the magnification ratio that expresses the size or amount of magnification or reduction of the object or image and to get the magnification, we use this equation: M= s’/s

M= 5f/4x5f

s’ = 1/4s

Therefore, the image height is one fourth of the object height

7 0
3 years ago
in which type of wave are vibrations at right angles to the direction in which the wave is travelling​
Pani-rosa [81]

Answer:

longitudinal waves have those properties

7 0
3 years ago
Glycerin is poured into an open U-shaped tube until the height in both sides is 20cm. Ethyl alcohol is then poured into one arm
Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining \rho_A: Ethyl alcohol density and \rho_G: Glycerin density , we can write:

\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3

Simplifying:

\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)

On the other hand:

h_1 + h_2 = \Delta h + h_3

Rearranging:

h_1 - \Delta h = h_3 - h_2 (2)

Replacing  (2) in (1):

\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)

Rearranging:

\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h

Data:

h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}

\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}}  = \Delta h

7.5 cm =  \Delta h

7 0
3 years ago
State<br><br> What is the correct order of tasks for washing dishes in a three-compartment sink?
Nataliya [291]

Answer:

The summary of the given statement is explained below throughout the explanation segment.

Explanation:

  • Drain certain surfaces throughout warm water of such soap during the very first sink. This same sanitizing of bacteria would not destroy whether grime would be in the direction.
  • Exfoliate the plates throughout plain water during the secondary drain. As with grime, the residual soap could avoid the kill off bacteria and viruses by the sanitizer.
7 0
3 years ago
The current in a lamp is 0.5 ampere when it is plugged into a standard wall outlet. What is the resistance of the lamp when it i
lidiya [134]

The resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms)

Explanation:

In the United States Of America the standard voltage is 120 v and their frequency is 60 Hz

Standard wall outlet voltage is 120 V

The current in the lamp is 0.5 ampere

Resistance (R) = V/ I

      = 120/0.5

      = 240Ω (ohms)

Thus the resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms).

8 0
3 years ago
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