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2 years ago

A footballer kicks a football of mass 430 g and it flies off at a speed of 8 m/s. What is the KE of the football?

Physics

1 answer:

klasskru [66]2 years ago

3 0

Answer:

13.76 Joules

Explanation:

The SI units for joules are defined to be kilogram*meters per second. Therefore, we must turn the grams to kilograms by dividing the answer by 1,000. This is due to 1,000 grams being in a single kilogram. Kinetic energy is defined to be:

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

Therefore:

$KE=\frac{1}{2} (0.43kg)(8m/s)^2=13.76J$

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In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

3 years ago

WILL MARK BRAINLIEST FOR CORRCET ANSWER!!!!!!!!!

jenyasd209 [6]

Answer:

It’s C

Explanation:

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3 years ago

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A projectile is launched vertically at 100 m/s. If air resistance can be ignored, at what speed will it return to its initial le

eduard

<h2>Speed with which it return to its initial level is 100 m/s</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 100 m/s

Acceleration, a = -9.81 m/s²

Final velocity, v = ?

Displacement, s = 0 m

Substituting

v² = u² + 2as

v² = 100² + 2 x -9.81 x 0

v² = 100²

v = ±100 m/s

+100 m/s is initial velocity and -100 m/s is final velocity.

Speed with which it return to its initial level is 100 m/s

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3 years ago

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A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the

alexandr402 [8]

Answer:

r=6.05km/hr

z=59.1 degree to the horizontal

Explanation:

A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place

can be solved using pythagoras theorem

r^2=o^2+a^2

r^2=5.2^2+3.1^2

r^2=36.65

r=6.1km/hr is te birds velocity relative to the ground

tanz=5.2/3.1

z=tan^-1(5,2/3.1)

z=59.1 degree to the horizontal

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3 years ago

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Two waves with amplitudes of 75 units and 74 units arrive at a point in a medium simultaneously. If the two waves are out of pha

avanturin [10]

If the two waves have the SAME FREQUENCY and are exactly
out of phase (180° apart), then the resultant wave will have the
same frequency and an amplitude of 1 unit.

If the two waves do not have the SAME FREQUENCY, then their
relative phase is meaningless.

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3 years ago

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