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attashe74 [19]
3 years ago
10

A block lies on a plane raised an angle θ from the horizontal. Three forces act upon the block: F⃗ w, the force of gravity; F⃗ n

, the normal force; and F⃗ f, the force of friction. The coefficient of friction is large enough to prevent the block from sliding
Physics
1 answer:
Verdich [7]3 years ago
5 0

Answer:

YFy = 0 = Ffsinθ + Fncosθ - Fw

Explanation:

From the base of the vector Fn, draw a vertical line. the small angle between this line and Fn is also theta. The component of Fn in the vertical direction is Fncos(theta).

Take a moment to picture extreme cases. Sine is 0 at 0 and 1 at 90. Cosine is 1 at 0 and 0 at 90.

Tilt the incline so that the box is on a flat surface. How much of the gravitational force is along the x direction of the floor.

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The main difference between a chest and a bounce pass is what?
snow_lady [41]

Answer: The main difference between the three is the mode of transmission. The chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate. Finally, the overhead pass is projected high in the air to avoid defenders.

Explanation:

4 0
3 years ago
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A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro
pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0
3 years ago
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A train moves from rest to a speed at 25m/s in 30.0 seconds. What is it’s acceleration?
sasho [114]

Answer:

25/30 = 5/6 m/s^2 5/6 meters per second squared

4 0
3 years ago
What is everyone's take on time travel?
Andrew [12]

Answer:

Its not really possible I don't think. UNLESS! You fall into a manhole then find a wirling vortex in the sewers! : )

Explanation:

7 0
3 years ago
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