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3 years ago

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CAN SOMEONE HELP ME PLEASE!? After chasing its prey, a cougar leaves skid marks that are 236 m in length. Assuming the cougar sk

idded to a complete stop with a constant acceleration of -2.87 m/s^2, identify the speed of the cougar before it began to skid.

1 answer:

malfutka [58]3 years ago

5 0

Answer:

u=36.8m/s

Explanation:

because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations

u^2=v^2-2ā*s. where:

u^2 stands for intial velocity

v^2 stands for final velocity

since the cougar skidded to a complete stop the final velocity is zero.

u^2=v^2-2ā*s

u^2=(0)^2 -2(-2.87 m/s^2)*236 m

u^2=0+5.74m/s^2* 236m

u^2=1354.64m^2/s^2

u=√1354.64m^2/s^2

u=36.8m/s (approximate value)

when ever the acceleration is constant you can use one of the following equation to find the required value.

1. v = u + at. (no s)

2. s= 1/2(u+v)t. (no ā)

3. s=ut + 1/2at^2. ( no v)

4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)

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Answer:

a. Vf = 39.24 [m/s]

b. Vf = 31.24 [m/s]

c. Vf = 47.24 [m/s].

Explanation:

To solve this problem we can use the following equation of kinematics. We have to keep in mind that the gravitational acceleration acts downwards, therefore when the rock falls towards the abyss it has the same direction of the acceleration and that is why the gravitational acceleration has a positive sign in the equation.

$v_{f}=v_{o}+g*t$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (when the rock is dropped)

g = gravitational acceleration = 9.81 [m/s²]

t = time = 4 [s]

a.

$v_{f}=0+9.81*4\\v_{f}= 39.24 [m/s]$

b.

In this particular situation, the acceleration will be taken as negative because the gravity is pointing in the opposite direction of the movement of the rock.

$v_{f}=8-(9.81*4)\\v_{f}=-31.24[m/s]$

The negative sign in the answer tells us that the rock no longer moves up instead it does downwards when 4 seconds have passed.

c.

$v_{f}=8+(9.81*4)\\v_{f}=47.24[m/s]$

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3 years ago

In the recent Olympics, many records were broken and history was re-written.

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Answer:

Yes

Explanation:

On the moon there is no gravity. So if there is no gravity pulling the weight down a weight can as light as a feather.

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3 years ago

A musician in a group plays a wrong note. would this note disrupt the harmony of the rhythm of the song being played? explain yo

Eddi Din [679]

If a musician plays a wrong note, he will disrupt the harmony of rhythm of the song. This is because, he will be able to produce a sound of different wavelength which will cause a possible destructive interference with the waves being produced by his groupmates.

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3 years ago

A sinusoidal wave travels along a stretched string. A particle on the string has a maximum speed of 2.0 m/s and a maximum accele

JulsSmile [24]

Answer:

The amplitude of the wave is 0.02 m.

Explanation:

Given that,

Maximum speed = 2.0 m/s

Maximum acceleration = 200 m/s²

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\dfrac{a_{max}}{v_{max}}$

Put the value into the formula

$\omega=\dfrac{200}{2.0}$

$\omega=100\ rad/s$

We need to calculate the amplitude of the wave

Using formula of velocity

$v_{max}=A\omega$

$A=\dfrac{v_{max}}{\omega}$

Put the value into the formula

$A=\dfrac{2.0}{100}$

$A=0.02\ m$

Hence, The amplitude of the wave is 0.02 m.

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4 years ago

At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is

Katyanochek1 [597]

Answer:

$4.47\ \text{km/h}$

Explanation:

$\dfrac{da}{dt}$ = Rate at which the distance between A and starting point of B is changing = -20 km/h

$\dfrac{db}{dt}$ = Rate at which the distance of B is changing = 15 km/h

$\dfrac{dc}{dt}$ = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = $20\times 4=80\ \text{km}$

a = Distance remaining to the start point of B = $110-80=30\ \text{km}$

b = Distance covered by B in 4 hours = $15\times 4=60\ \text{km}$

Distance between A and B after 4 hours

$c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}$

$c^2=a^2+b^2$

Differentiating with respect to time we get

$c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}$

The rate at which the distance between the ships is changing at 4 PM is $4.47\ \text{km/h}$ .

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3 years ago