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2 years ago

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a footballer kicks a ball vertically upwards. initially, the ball is stationary. his boot is in contact with the ball for 0.05s.

the average resultant force on the ball during this time in 180N. the ball leaves his foot at 20m/s. calculate the height to which the ball rises. ignore air resistance.

2 answers:

taurus [48]2 years ago

8 0

Kick Ball force average 200

lozanna [386]2 years ago

4 0

Answer:

The ball rises to a height of 20.4 meters.

Explanation:

Since we know the speed at which the ball starts rising, we can ignore the other data. The height reached by the ball can be obtained using the kinematics equation:

$v^{2}=v_0^{2}-2gy$

Since the velocity at the highest point is zero, we can solve for the height y at that point:

$y=\frac{v_0^{2}}{2g}$

Plugging in the given value for the initial velocity, and assuming that the acceleration due to gravity g is 9.8m/s², we obtain:

$y=\frac{(20m/s)^{2}}{2(9.8m/s^{2})}\\\\y=20.4m$

It means that the ball rises to a height of 20.4 meters.

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In a harbor, you can see sea waves traveling around the edges of small stationary boats. Why does this happen?

faust18 [17]

Below are the choices that can be found in the other sources:

A. diffraction
<span>B. refraction </span>
<span>C. reflection </span>
<span>D. transmission
</span>
The answer is diffraction. It means that <span>the process by which a beam of light or other system of waves is spread out as a result of passing through a narrow aperture or across an edge, typically accompanied by interference between the wave forms produced.</span>

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2 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an

a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, $\mu=6\times 10^{-2}$

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

$KE=PE+W$

$\dfrac{1}{2}mv^2=mgh+W$

W is the work done by the friction.

$W=f\times d$

$f=\mu mg\ cos\theta$

$W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}$

$W=\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}$

$\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}$

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0

2 years ago

An airplane travels horizontally at a constant velocity v. An object is dropped from the plane and one second later another obje

Delvig [45]

Answer:

the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

Explanation:

Since airplane is moving horizontally with constant speed v

so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane

so we can say that two object when dropped after some interval of time then they always lie in same vertical line

now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other

So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time

8 0

2 years ago

Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p

worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

$U1/U2=6$
$U1/U2=6$

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy is mathematically given as

$U=\frac{-k*qQ}{r}$

Therefore

The Equations of U1 and U2 is

For U1

$U1=\frac{-k*q_1Q}{r}$

For U2

$U2=\frac{-k*q_1Q}{3*2r}$

Since

U is a function of q and q2=q1/3

Therefore

$U1/U2=6$

For Question 2

For U1

$U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}$

Therefore

$U1/U2=6$

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2 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Darya [45]

Answer:

$y = -8.37 cm$

Explanation:

As we know that the equation of SHM is given as

$y = A cos(\omega t)$

here we know that

$\omega = \frac{2\pi}{T}$

here we have

$T = 0.66 s$

now we have

$\omega = \frac{2\pi}{0.66}$

$\omega = 3\pi$

now we have

$y = (8.8 cm) cos(3\pi t)$

now at t = 2.3 s we have

$y = (8.8 cm) cos(3\pi \times 2.3)$

$y = -8.37 cm$

6 0

2 years ago

Read 2 more answers