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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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A balloon filled with helium occupies 20.0 l at 1.50 atm and 25.0◦
bija089 [108]
At stp (standard temperature and pressure), the temperature is T=0 C=273 K and the pressure is p=1.00 atm. So we can use the ideal gas law to find the number of moles of helium:
pV=nRT
where p is the pressure (1.00 atm), V the volume (20.0 L), n the number of moles, T the temperature (273 K) and R=0.082 atm L K^{-1} mol^{-1} the gas constant. Using the numbers and re-arranging the formula, we can calculate n:
n= \frac{pV}{RT}= \frac{(1.00atm)(20.0L)}{(0.082 LatmK^{-1}mol^{-1})(273 K)}=0.89 mol
5 0
3 years ago
Can someone help on this I'm really stuck
castortr0y [4]
Here, "Wavelength is same for both waves" it is the distance between two crests or two consecutive troughs, so, it is constant for both of them, you can easily figure it out.

In short, Your Answer would be "Wavelength"

Hope this helps!
5 0
3 years ago
What distance will a car cover while uniformly accelerating from 12m/s to 26m/in 14 seconds​
VikaD [51]

Answer:

Distance = 266m

Explanation:

6 0
3 years ago
A ball is thrown vertically upward with an initial velocity
ivolga24 [154]

Answer:

D

Explanation:

First we define our variables

V0=29.4

a=-9.8

V=0

We have to find the maximum displacement , which I will define as X

We use formula v^2=v0^2+2aX

All we do is substitute our values

0=29.4^2-19.6X

29.4^2=19.6X

X=29.4^2/19.6=44.1

5 0
2 years ago
A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl
Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}

R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}

Part C

The error in % is given by

\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%

4 0
3 years ago
Read 2 more answers
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