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avanturin [10]
3 years ago
9

If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi

ng at constant speed across the floor with a constant, horizontal force of 70 N. The puppy has a weight of 110 N. What is the coefficient of static friction between the puppy and the floor
Physics
1 answer:
netineya [11]3 years ago
5 0

Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

F_s = \mu_s*N

Where F_s is the static friction force, \mu_s is the coefficient of static friction and N is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, N = 110\text{ N}, to make the puppy moving we need to use a force of 80 N, therefore, F_s = 80 \text{ N}, so we can solve for the coefficient as shown below:

80 = \mu_s*110\\\mu_s = \frac{80}{110} = 0.7273\\

The coefficient of static friction between the puppy and the floor is 0.7273.

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A heat engine operating between energy reservoirs at 20?c and 600?c has 30% of the maximum possible efficiency.
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The actual efficiency of the engine is 30%, i.e. 
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On the other hand thermal efficiency is defined as the ratio of work done to the amount of heat absorbed from hot reservoir: 
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Which change is an example of transforming potential energy to kinetic energy?.
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there hope this helps

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2 years ago
A rifle that shoots bullets at 477 m/s is to be aimed at a target 45.5 m away. If the center of the target is level with the rif
Free_Kalibri [48]

Answer:

The rifle barrel must be pointed at a height of 4.45cm above the target so that the bullet hits dead center.

Explanation:

First, we need to sketch the situation so we can have a better idea of what the problem looks like (Refer to uploaded picture).

So as you may see in the drawing, when pointing the rifle to the target, we can see it as a triangle, but in reality, the bullet will have a parabolic trajectory. Both points of view will help us determine what the height must be. In order to find it, we need to first determine at what angle the bullet should be shot. In order to do so we can use the range formula, which looks like this:

R=\frac{v^{2}sin(2\theta)}{g}

Where R is the range of the bullet (this is how far it goes before it has the

same height it was shot from), v is the original speed of the bullet, θ is the angle at which the bullet is shot and g is the acceleration of gravity.

We can solve this equation for theta, so we get:

gR=v^{2}sin(2\theta)

\frac{gR}{v^{2}}=sin(2\theta)

sin^{-1}(\frac{gR}{v^{2}})=2\theta

\theta=\frac{sin^{-1}(\frac{gR}{v^{2}})}{2}

so now we can substitute the given data:

\theta=\frac{sin^{-1}(\frac{(9.8m/s^{2})(45.5m)}{(477m/s)^{2}})}{2}

so we get:

θ=0.05614°

once we get the angle, we can look at the triangle diagram. From the drawing we can see that we can use the tan function to find the height:

tan \theta = \frac{h}{45.5m}

so we can solve this for h, so we get:

h=45.5m*tan(0.05614^{o})

which yields:

h=0.0445m

or

h=4.45cm

5 0
3 years ago
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