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mart [117]
3 years ago
5

A bartender slides a beer mug at 1.3 m/s towards a customer at the end of a frictionless bar that is 1.3 m tall. The customer ma

kes a grab for the mug and misses, and the mug sails off the end of the bar. (a) How far away from the end of the bar does the mug hit the floor
Physics
1 answer:
sineoko [7]3 years ago
6 0

Answer:

x = 0.67 m

Explanation:

For this problem, let's use the projectile launch equations, as the jug goes through the bar, it comes out with horizontal speed vx = 1.3 m / s, which does not decrease as there is no friction.

Let's find the time or it takes to get to the floor

          y = y₀ + v_{oy} - ½ g t²

in this case I go = 0 and when I get to the floor y = 0

         0 = y₀ + 0 - ½ g t²

          t² = 2y₀ / g

          t² = 2 1.3 / 9.8 = 0.2653

          t = 0.515 s

now let's find the distance traveled in this time

         x = vx t

         x = 1.3  0.515

         x = 0.6696 m

         x = 0.67 m

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just olya [345]

Answer: Im not entirly sure but I think It's D all the above. I think all but B because I never really heard of that but if you look in our history I think that happen im not sure I would wait untill you know that somone knows for sure.

Explanation:

3 0
2 years ago
Two light bulbs are 2.0 m apart. From what distance can these light bulbs be marginally resolved by a small telescope with a 4.5
andrezito [222]

Answer:

R = 1.2295 10⁵  m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

            θ = 1.22 λ / D

where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture

             

how angles are measured in radians

          θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

            \frac{y}{R} = 1.22 \frac{ \lambda}{D}

            R = \frac{ y \ D}{1.22 \lambda}

let's calculate

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3 0
3 years ago
Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration
eimsori [14]

The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

<em />

<u>Solution:</u>

1) Acceleration: -2336 m/s^2 on the hardwood floor, -382 m/s^2 on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

v^2 - u^2 = 2ad

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

d = 1.8 mm = 0.0018 m

So we can find the acceleration by using again the equation

v'^2 - v^2 = 2ad

Solving for a,

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2

For the carpeted floor instead,

d=1.1 cm = 0.011 m

therefore the acceleration is

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

a=\frac{v'-v}{t}

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

a=-2336 m/s^2

So the duration of the collision is

t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms

For the carpeted floor,

a=-382 m/s^2

So the duration of the collision is

t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

8 0
3 years ago
Why can't there be a number lower than absolute zero
nexus9112 [7]

Absolute zero is not about numbers.  It's about temperature, and the
motion of molecules in gases. 

You know that the temperature we feel with our skin is the result of the
average speed of all the tiny molecules zipping around or vibrating in
the solid, liquid, or gas.

The faster they're all moving, the warmer the substance feels to us. 
The slower they're all moving, the cooler the substance feels to us.

When molecules slow down to zero and lose all of their kinetic energy,
that temperature is what we call 'absolute zero' ... if they're not moving
at all, then they can't move any slower.

5 0
3 years ago
Read 2 more answers
PLEASE HELP
enyata [817]

Answer:

29 m/s.

Explanation:

Hope this helps.

8 0
3 years ago
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