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raketka [301]
3 years ago
12

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.​
Physics
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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Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen
bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

f_{a} =(\frac{v+v_{c} }{v} )f_{s} (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s} (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s

b) Substitute eq. 1 in eq. 2:

f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz

7 0
3 years ago
An airplane cruises at 900 km/h relative to the air. It is flying from Denver, Colorado, due west to Reno, Nevada, a distance of
Ulleksa [173]

Answer:

26.5 minutes

Explanation:

When the airplane is flying due West from Denver to Reno, the due-East wind with speed of 80km/h would reduce the ground speed by 80 km/h.

Its Denver to Reno ground speed is 900 - 80 = 720 km/h

The time it takes to cover 1200km at this speed is 1200 / 720 = 1.67 hours

On the other hand, when it returns from Reno to Denver in the due-East direction, the due-East wind with speed of 80km/h would add to the ground speed by 80 km/h

Its Reno to Denver ground speed is 900 + 80 = 980 km/h

The time it takes to cover 1200 km at this speed is 1200 / 980 = 1.22 hours

The difference it flight time would be 1.67 - 1.22 = 0.44 hours or 26.5 minutes

4 0
3 years ago
How do you calculate the total resistance of a series circuit
CaHeK987 [17]
Add all the resistances across the circuit together the calculate the total resistance
5 0
3 years ago
John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2
Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function cos(\omega t +\phi) is defined between -1 and 1, So it is not possible obtain a value 2\pi greater.  

In addition, if you  move the function cosine a T Value, and T is the Period,  the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have f=1+cos(\omega t +\phi), the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the f=1+cos(\omega t +\phi), and its mean that the function f=cos(\omega t +\phi), in general, is not greater than cos(\omega t +\phi).

3 0
2 years ago
According to newtos first law of motion. what is required to make a object slow down.
ladessa [460]

Answer:

Friction

Explanation:

Friction is a force that slows down moving objects. If you roll a ball across a shaggy rug, you can see that there are lumps and bumps in the rug that make the ball slow down. The rubbing, or friction, between the ball and the rug is what makes the ball stop rolling. External Force is required.

5 0
3 years ago
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