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2 years ago

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A player kicks a football from ground level with an initial velocity of 27.0 30.0 above the horizontal Find the ball's maximum h

eight

1 answer:

stepladder [879]2 years ago

5 0

Answer:

Given

initial velocity (u) =27.030

Force of gravidity g) =9.8

Rtc maximum height Hmix =?

$sln \\ \ hmix = u {}^{2} \div 2g \\ 27.030 \div 2 \times 9 \\ \\ hmix = 38.025m$

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The resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is:a) 180 d

r-ruslan [8.4K]

Answer:

0 degrees

Explanation:

Let $F_1\ and\ F_2$ are two forces. The resultant of two forces acting on the same point is given by :

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos\theta}$

Where $\theta$ is the angle between two forces

When $\theta=0$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos(0)}$

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2}$

When $\theta=90^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(90)}$

$F_R=\sqrt{F_1^2+F_2^2}$

When $\theta=180^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(180)}$

$F_R=\sqrt{F_1^2+F_2^2-2F_1F_2}$

It is clear that the resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is 0 degrees. Hence, this is the required solution.

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3 years ago

6. A football is kicked with an initial speed of 10.2 m/s at an angle of 40.00 above the horizontal. It lands on the ground 2.12

igor_vitrenko [27]

Answer:

7.81 m/s

Explanation:

Given,

initial speed, u = 10.2 m/s

angle of inclination, θ = 40°

time, t = 2.12 s

Horizontal component of the velocity:

$u_x = u cos \theta$

$u_x = 10.2\times cos 40^0$

$u_x = 7.81 m/s$

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So, Velocity at the Pinnacle is equal to 7.81 m/s

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3 years ago

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The answer is A) sunlight

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2 years ago

The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of

natita [175]

Answer:

3.258 m/s

Explanation:

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x = Displacement of spring = 0.7 m (Assumed, as it is not given)

$\mu$ = Coefficient of friction = 0.4

Energy stored in spring is given by

$U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J$

As the energy in the system is conserved we have

$\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s$

The speed of the 8 kg block just before collision is 3.258 m/s

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3 years ago