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alexdok [17]
3 years ago
5

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45m diameter

semicircles. A greyhound can run around these turns at a constant speed of 16m/s .part a: What is its acceleration in m/s2?partb: What is its acceleration in units of g ?
Physics
1 answer:
sashaice [31]3 years ago
8 0

To solve this problem we will apply the concept related to centripetal acceleration. Normal acceleration or centripetal acceleration is responsible for changing the direction of the velocity vector. It is the only type of acceleration present in the uniform circular motion. Its mathematical formula is given by

a = \frac{v^2}{r }

Here,

v = Tangential velocity

r = Radius

Our values are,

v = 16m/s

r = \frac{45m}{2} = 22.5

PART A) Using the previous expression the acceleration will be

a= \frac{16^2}{22.5 }

a= 11.37 m/s^2

PART B) In the case of g units, we know that

9.8m/s^2 = 1g

Then performing the conversion we have to

a = 11.37m/s^2(\frac{1g}{9.8m/s^2})

a = 1.16g

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4 0
2 years ago
what is the mechanical advantage of a crowbar when a worker uses 10N of force to pry open a window that has a resistance of 500N
Oksana_A [137]

Answer:

50

Explanation:

The mechanical advantage of a machine is given by

MA=\frac{F_{out}}{F_{in}}

where

F_{out} is the output force

F_{in} is the input force

For the crowbar in this problem,

F_{in}=10 N is the force in input applied by the worker

F_{out}=500 N is the force that the machine must apply in output to overcome the resistance of the window and to open it

Substituting into the equation, we find

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3 0
3 years ago
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A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.
JulsSmile [24]

Answer:

\mu =0.75

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

F_c=m.a_c

The centripetal acceleration a_c is computed as

\displaystyle a_c=\frac{v^2}{r}

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

F_c=m.\frac{v^2}{r}

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

F_r=\mu N

The normal force N is equal to the weight of the car, thus

F_r=\mu .m.g

Equating both forces

\displaystyle \mu .m.g=m.\frac{v^2}{r}

Simplifying

\displaystyle \mu =\frac{v^2}{rg}

Substituting the values

\displaystyle \mu =\frac{19^2}{(49)(9.8)}

\boxed{\mu =0.75}

7 0
3 years ago
You wad up a piece of paper and throw it into the wastebasket. How far will
vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, g=9.8 m/s^2)

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

d=\frac{u^2 sin 2\theta}{g}

where

u is the initial velocity

\theta is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

\theta=65^{\circ}

Substituting,

d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
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