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2 years ago

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A naughty squirrel jumps from a tree limb onto a birdfeeder. The tree limb is 4.5 m above the ground and the top of the birdfeed

er is 1.5 m above the ground. If the squirrel jumps horizontally with a velocity of 2.5 m/s then how far away in the horizontal direction can the bird feeder be for the squirrel to just make it onto the top of the feeder.

1 answer:

andreev551 [17]2 years ago

7 0

Answer:

Distance of 1.956 m

Explanation:

Consider the vertical plane

$S=ut+\frac{1}{2} at^2$

where u=o, S=4.5-15=3 m

$3=0\times t+\frac{1}{2} 9.8 t^2\\t=\sqrt{0.612} \\t=0.782 s$

now consider horizontal plane

u=2.5 m/s, a=0

$S=2.5\times 0.782+0\\S=1.956 m$

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Un tractor halando un carretón cargado con caña de azucar, viaja por el camino recto de una finca a una velocidad de 20 km/h. Si

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Translation

A tractor pulling a cart loaded with sugar cane travels down the straight path of a farm at a speed of 20 km / h. If at 3:00 p.m.you pass the Finca Las Margaritas, what time will you arrive at the Las Ilusiones farm, located on the same road, if the distance between the two farms is 60 km

Answer:

6.00 pm

Explanation:

Speed is given by dividing distance by time and expressed as s=d/t. Making time the subject of the formula then t=d/s where s is the speed, d is distance covered and t is the time taken. Substituting 20 km/h for s and 60 km for d then t=60/20=3 hours

Adding 3 hours to 3 pm we get 6pm

Therefore, the time to reach the destination if the speed is constantly maintained is 6.00 pm

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A 0.250 kg block on a vertical spring with spring constant of 4.45 ✕ 103 N/m is pushed downward, compressing the spring 0.080 m.

Elza [17]

Answer:h=5.81 m

Explanation:

Given

Mass of block(m)=0.250 kg

Spring Constant $k=4.45\times 10^3 N/m$

Initial elongation =0.080 m=8 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block

$P_i=\frac{kx^2}{2}$

$P_i=\frac{4.45\times 10^3\times 64\times 10^{-4}}{2}=14.24 J$

$P_f=mgh=0.25\times 9.8\times h$

$P_i=P_f$

$14.24 =0.25\times 9.8\times h$

$h=\frac{14.24}{0.25\times 9.8}=5.81 m$

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2 years ago

You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be

Alex

Answer:

The amplitude is $A = 90.2 \ m$

Explanation:

From the question we are told that

The frequency of when sound is approaching observer is $f = 392 Hz$

The frequency as the move away from observer is $f_ a = 330 \ Hz$

The time between the pitch are $t = 10 \ s$

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

$T = 2 t$

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

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

Where $w$ is the angular velocity

=> $\frac{2 \pi}{w} = 2 * 10$

=> $w = 0.314 \ rad/sec$

Now using Doppler Effect,

The source of the sound is approaching the observer

The

$f = f_o (\frac{v}{v- wA} )$

$392 = f_o (\frac{v}{v- wA} )$

Where A is the amplitude

So when the source is moving away from the observer

$f_a = f_o (\frac{v}{v+ wA} )$

$330 = f_o (\frac{v}{v+ wA} )$

Here $f_o$ is the fundamental frequency

Dividing the both equation we have

$\frac{392}{330} = \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}$

$1.1878 = \frac{v+wA}{v-wA}$

$1.1878 v - 1.1878 wA = v+wA$

$1.1878 v = 2.1878 wA$

=> $A = \frac{(0.1878 * (330))}{(2.1878)* (0.314)}$

$A = 90.2 \ m$

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