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2 years ago

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

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A student conducts an experiment in which a cart is pulled by a variable applied force during a 2 s time interval. In trial 1, t

fredd [130]

Answer:

The answer is "Including all three studies of 0s to 2s, that shift in momentum is equal".

Explanation:

Its shift in momentum doesn't really depend on the magnitude of its cars since the forces or time are similar throughout all vehicles.

Let's look at the speed of the car

$F = m a\\\\a =\frac{F}{m}$

We use movies and find lips

$\to v = v_0 + a t\\\\\to v = v_0 + (\frac{F}{m}) t$

The moment is defined by

$\to p = m v$

The moment change

$\Delta p = m v - m v_0$

Let's replace the speeds in this equation

$\Delta p = m (v_0 + \frac{F}{m t}) - m v_0\\\\\Delta p = m v_0 + F t - m v_0\\\\\Delta p = F t$

They see that shift is not directly proportional to the mass of cars since the force and time were the same across all cars.

5 0

3 years ago

What is the kinetic energy of an object with a mass of 50kg and is and it is traveling at a rate of 60m/s.

liq [111]

Answer:

90,000 J

Explanation:

Kinetic energy can be found using the following formula.

$KE=\frac{1}{2}mv^2$

where <em>m </em>is the mass in kilograms and <em>v</em> is the velocity in m/s.

We know the object has a mass of 50 kilograms. We also know it is a traveling at a rate of 60 m/s. Velocity is the speed of something, so the velocity of the object is 60 m/s.

<em>m</em>=50

<em>v</em>=60

Substitute these values into the formula.

$KE=\frac{1}{2}*50*60^2$

First, evaluate the exponent: 60^2. 60^2 is the same as multiplying 60, 2 times.

60^2=60*60=3,600

$KE=\frac{1}{2}*50*3,600$

Multiply 50 and 3,600

$KE=\frac{1}{2}*180,000$

Multiply 1/2 and 3,600, or divide 3,600 by 2.

$KE=90,000$

Add appropriate units. Kinetic energy uses Joules, or J.

$KE=90,000 Joules$

The kinetic energy of the object is 90,000 Joules

6 0

3 years ago

An unstrained horizontal spring has a length of 0.26 m and a spring constant of 180 N/m. Two small charged objects are attached

MakcuM [25]

Answer:

two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)

both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

Explanation:

Given that;

L = 0.26 m

k = 180 N/m

x = 0.039 m

we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.

Spring force F = kx

F = 180 × 0.039

F = 7.02 N

Now, Electrostatic force F = Keq²/r²

where r = L + x = ( 0.26 + 0.039 )

we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2

so from the equation; F = Keq²/r²

Fr² = Keq²

q = √ ( Fr² / Ke )

we substitute

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q = √ (0.627595 / 9×10⁹)

q = √(6.97 × 10⁻¹¹)

q = 8.35 × 10⁻⁶ C

Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

5 0

2 years ago

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

Obi Wan hears the destruction of a planet and all of its people through 'the force'. These sounds are only in his head and are n

trapecia [35]

Answer:

medium

Explanation:

<em>A sound </em><em>medium</em><em> is defined as channel through which sound can travel or be transmitted. </em>

Sound medium could be in the form gases, liquids, solids or plasmas. Space is made up of vacuum and therefore, has no medium within it. Hence, space cannot transmit sound in any form or allows sound to travel through it.

6 0

3 years ago