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3 years ago

14

A golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0°. If the hang time of the golf ball is 3.44

seconds, what is the range of the golf ball?

1 answer:

Ray Of Light [21]3 years ago

5 0

Solving this using the time, we know that range = horizontal velocity x time of flight

<span>since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. I</span><span>f we use the correct time of flight given the launch parameters, we have </span>

<span>range = 36 cos 28 x 3.44 s = 109.3 m </span>

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3 years ago

I got part c right but idk why the other parts are wrong HELP!

dedylja [7]

a) The impulse is 76.5 Ns

b) The average force is 546.4 N

c) The final speed is 31.5 m/s

Explanation:

a)

The impulse exerted on an object is defined as

$J=\int F\Delta t$

where

F is the magnitude of the force exerted on the object

$\Delta t$ is the time interval during which the force is applied

If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.

Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are

$B=0.14-0 = 0.14s\\b=8-5=3s$

and whose height is

$h=900 N$

Therefore, the area (and the impulse) is

$J=\frac{(B+b)h}{2}=\frac{(0.14+0.03)(900)}{2}=76.5 Ns$

b)

In this problem, the force applied is not constant. However, we can rewrite the impulse also as

$J=F_{avg} \Delta t$

where

$F_{avg}$ is the average force exerted during the whole time $\Delta t$

In this problem we have

J = 76.5 Ns is the impulse (calculated in part a)

$\Delta t = 0.14 s$ is the time interval

Solving for the average force, we find

$\Delta t = \frac{J}{F_{avg}}=\frac{76.5}{0.14}=546.4 N$

c)

According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:

$J=\Delta p = m(v-u)$

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have

J = 76.5 Ns

m = 3.0 kg is the mass

u = 6.0 m/s is the initial velocity

Solving for v, we find the final velocity (and speed):

$v=u+\frac{J}{m}=6.0+\frac{76.5}{3}=31.5 m/s$

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

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3 years ago

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marta [7]

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3 years ago

he capacitor can withstand a peak voltage of 550 VV . If the voltage source operates at the resonance frequency, what maximum vo

anygoal [31]

Answer:

The maximum voltage is 39.08 V.

Explanation:

Given that,

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Suppose, In an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is $1.20\times10^{-2}\mu F$

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Using formula of resonant frequency

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Put the value into the formula

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$f=2356.8\ Hz$

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Using formula of current

$I=\dfrac{V_{c}}{X_{c}}$

$I=2\pi f C\times V_{c}$

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Using formula of impedance

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At resonant frequency , $X_{L}=X_{C}$

So, Z = R

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Put the value into the formula

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