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3 years ago

A soccerball is kicked straight out from a hill at 15 m/s and lands 42m away. how tall is the hill

Physics

1 answer:

ozzi3 years ago

4 0

Answer:

The height of the hill is, h = 38.42 m

Explanation:

Given,

The horizontal velocity of the soccer ball, Vx = 15 m/s

The range of the soccer ball, s = 42 m

The projectile projected from a height is given by the formula

S = Vx [Vy + √(Vy² + 2gh)] / g

Therefore,

h = S²g/2Vx² (Since Vy = 0)

Substituting the values

h = 42² x 9.8/ (2 x 15²)

= 38.42 m

Hence, the height of the hill is, h = 38.42 m

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6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t

zubka84 [21]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

$d=v\times t$

$d=9.1\times t$ ....(I)

The distance of longitudinal wave

$d=5.7\times (t+71)$ ....(II)

From the first equation

$t=\dfrac{d}{9.1}$

Put the value of t in equation (II)

$d =5.7\times(\dfrac{d}{9.1}+71)$

$\dfrac{9.1d-5.7d}{9.1}=71\times5.7$

$d0.3736=404.7$

$d =1083.24\ km$

Hence, The distance away the center of the earthquake is 1083.24 km.

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A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

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3 years ago