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asambeis [7]
3 years ago
8

A cat is sleeping on the floor in the middle of a 2.8-m-wide room when a barking dog enters with a speed of 1.40 m/s. As the dog

enters, the cat (as only cats can do) immediately accelerates at 0.85 m/s2 toward an open window on the opposite side of the room. The dog (all bark and no bite) is a bit startled by the cat and begins to slow down at 0.10 m/s^2 as soon as it enters the room.
Required:
How far is the cat in front of the dog as it leaps through the window?
Physics
1 answer:
marysya [2.9K]3 years ago
7 0

Answer:

the cat is  0.4238 m in front of the dog as it leaps through the window

Explanation:

Given that;

acceleration a = 0.85 m/s²

speed v = 1.40 m/s

the cat is at rest, so initial velocity u = 0  

we know that, since the cat is sleeping on the floor in the middle of a 2.8-m-wide room, it needs to cover (2.8 m / 2 ) distance to get to the window;

using the second equation equation of motion;

s = ut + 1/2 at²  

we substitute

2.8/2 = 0×t + 1/2 × 0.85 × t²

1.4 = 0.425t²

t = √( 1.4 / 0.425 )

t = 1.81497 sec

now, at acceleration 0.10 m/s²

the dog has to cover the distance;

s = ut + 1/2 at²  

s = ( 1.4 × 1.81497) - 1/2 × 0.10 × 1.81497²

s =  2.540958 - 0.1647

s = 2.3762 m  

The cant in front of the dog as it leaps through the window;

distance = 2.8 m - 2.3762 m

distance = 0.4238 m

Therefore, the cat is  0.4238 m in front of the dog as it leaps through the window

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4 0
3 years ago
A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s.
Serggg [28]

Answer:

(a)  α = 35.20 rad/s^2

(b)  θ = 802°

(c)   v = 139.73 cm/s

(d)   a = 156.64 cm/s^2

Explanation:

(a) To find the angular acceleration of the disc you use the following formula:

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You replace the values of the parameters in the equation (1):

\alpha=\frac{31.4rad/s-0rad/s}{0.892s}=35.20\frac{rad}{s^2}

The angular acceleration of the disc, for the given time, is 35.20rad/s^2

(b) To calculate the angle describe by the disc in such a time you use:

\theta=\frac{1}{2}\alpha t^2         (2)

\theta=\frac{1}{2}(35.20rad/s^2)(0.892s)^2=14.00rad

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\theta=14.00rad*\frac{180\°}{\pi \ rad}=802\°

The angle described by the disc is 802°

(c) To calculate the tangential speed of the microbe for t=0.892s, you use the following formula:

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w: angular speed for t = 0.892s = 31.4rad/s

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v=(31.4rad/s)(4.45cm)=139.73\frac{cm}{s}

The tangential speed is 139.73 cm/s

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a=\alpha r

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The tangential acceleration is 156.64cm/s^2

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